UUNIFORM CONVERGENCE AND KNOT EQUIVALENCE
FOREST KOBAYASHI
Dedicated to my grandparents, Albert and Elizabeth Kobayashi.
Abstract.
Given a uniformly convergent sequence of ambient isotopies ( (cid:72) n ) n ∈ N ,bijectivity of the limit function (cid:72) ∞ together with a minor compactness condi-tion guarantees that (cid:72) ∞ is also an ambient isotopy. By offloading the uniformconvergence hypothesis to a more diagrammatic condition, we obtain sufficientconditions for performing countably-many Reidemeister moves. We use this toconstruct examples of tame knots with countably-many crossings and discusswhat distinguishes these from similar-looking wild curves. Contents
1. Introduction 22. Background 42.1. Fundamental Definitions 42.2. Tameness and Wildness 62.3. Uniform Convergence 73. Uniform Convergence & Knots 83.1. Iteratively Constructing (Ambient) Homeomorphisms 93.2. Iteratively Constructing (Ambient) Isotopies 134. Various Applications of Theorem 3.8 155. Cases Where Theorem 3.8 Does Not Apply 276. Discussion 307. Acknowledgements 32References 32Appendix A. The Fox-Artin Tameness Invariant 33
E-mail address : [email protected] . Date : January 10 th , 2021. Key words and phrases.
Knots, wild knots, countable Reidemeister moves. a r X i v : . [ m a t h . G T ] J a n UNIFORM CONVERGENCE AND KNOT EQUIVALENCE Introduction
This work is based on results from the author’s undergraduate thesis [5].The goals of this document are to understand when one can apply countable se-quences of Reidemeister moves and preserve ambient isotopy in the limit. Themotivating examples are the following two curves. (a) A wild-looking unknot. L − L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L (b) An unknotted-looking wild knot. Figure 1.
Two scintillating curves.
Tameness of Fig. 1a.
As indicated by the caption, the curve in Fig. 1a is tame.In fact, it is the unknot. An explicit ambient isotopy taking it to the unknot canbe constructed as follows. From time t = 0 to t = , use a Reidemeister I moveto remove loop (cid:76) . Then, from time t = to t = , use a Reidemeister I move toremove loop (cid:76) . So on and so forth, removing loop (cid:76) n between time t = 1 − n − and t = 1 − n .If one is careful about exactly how the Reidemeister I moves are performed, thenthe result will be an ambient isotopy. We provide the details in Proposition 4.2. Wildness of Fig. 1b.
By contrast, the curve in Fig. 1b is wild, a result establishedby Ralph Fox in [3]. His argument uses techniques co-developed with Emil Artin in[4], namely, a sort of “invariant” for tameness of arcs. We summarize the relevantresult in Appendix A.
The Problem.
At first, the wildness of the curve in Fig. 1b can appear verycounterintuitive. As with the loops in Fig. 1a, any finite number of the “stitches”in Fig. 1b can be safely removed using Reidemeister II moves. The procedure is asfollows. First, use a Reidemeister II move to move (cid:76) − into (cid:76) . Next, use anotherReidemeister II move to remove both (cid:76) − and (cid:76) . NIFORM CONVERGENCE AND KNOT EQUIVALENCE 3 L − L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L (a) Moving (cid:76) − into (cid:76) . L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L (b) Removing (cid:76) − and (cid:76) . Figure 2.
Removing one stitch.Note, this leaves us with a shrunk copy of the same diagram we started with. Hencewe can repeat the process any finite number of times. In general, to perform step n we use a Reidemeister II move to slide loop (cid:76) n − into loop (cid:76) n − , then we remove (cid:76) n − and (cid:76) n by using another Reidemeister II move.What’s to stop us from using the same approach as in Fig. 1a, where we justperformed step n between t = 1 − n − and t = 1 − n ? Indeed, if we choose ourReidemeister II moves carefully, it’s possible to guarantee continuity of the resultinglimit function.The issue is with bijectivity on the ambient space. As we show in Section 5, thereis no way to choose our sequence of Reidemeister II moves without accidentallydragging points from the ambient space down to the wild point in the limit. Theproof is a simple geometric argument but it can be easy to overlook. The Upshot.
To clarify the situation, in Theorems 3.3 and 3.8 we generalizethe strategy we described for Fig. 1a and make it rigorous. Theorem 3.3 usesthe language of ambient homeomorphisms while Theorem 3.8 uses the language of ambient isotopies ; other than that, they are equivalent.The layout of the rest of the paper is as follows:( §
2) In Section 2, we go over the definitions of knots, ambient homeomorphisms,ambient isotopies, PL-ness, and tameness/wildness. It might be helpful toreview these concepts since we will be working directly with equivalencein this paper instead of by proxy through Reidemeister’s Theorem. Wealso give the definition of uniform convergence and recall two elementaryresults from first courses in Analysis and Topology, respectively; the maintheorems will follow from these.( §
3) In Section 3, we give the definition of uniform convergence and use it tobuild up our main results (Theorems 3.3 and 3.8). In most cases Theo-rem 3.3 tends to be more ergonomic than Theorem 3.8, but we will gener-ally employ Theorem 3.8 because it seems the language of ambient isotopyis more ubiquitous than that of ambient homeomorphism.( §
4) In Section 4, we apply Theorem 3.8 to various curves with countably-manycrossings.
UNIFORM CONVERGENCE AND KNOT EQUIVALENCE ( §
5) In Section 5, we show what goes wrong if we try to apply Theorem 3.8 totwo noteworthy examples; the second is the wild curve from Fig. 1b.Finally, some notation.
Symbol Interpretation f Embedding and/or knot (usually). h Homeomorphism. h PL Homeomorphism. (cid:104)
Homeomorphism constructed by composing other homeomorphisms. H Ambient isotopy. H PL Ambient isotopy. (cid:72)
Ambient isotopy constructed by iteratively gluing other ambient isotopies.(
X, d X ) Metric space X with metric d X .( X, (cid:84) X ) Topological space X with topology (cid:84) X . nk =1 f k The composite function f n ◦ f n − ◦ · · · ◦ f ◦ f . k Generally reserved for an “free” index (e.g., k ∈ { , . . . , n } above). n Generally reserved for a “particular” index (e.g., n is the “stop” index above).( f k ) ∞ k =1 A sequence f , f , . . . , f k , . . . . f k → f The f k converge to f pointwise. f k u −→ f The f k converge to f uniformly. Note, some authors use ⇒ instead of u −→ . A ◦ Interior of A . A Closure of A . A c X \ A (whenever X is understood) A \ A Set difference of A and A . A (cid:116) A Disjoint union of A and A . 〔 . . . 〕 Used to indicate parsing order in grammatically-ambiguous sentences. (cid:51)
Indicates the completion of a case in a proof by casework. Background
Today we will be working with knot equivalence directly instead of making ap-peals to
Reidemeister’s theorem . This is because we’re interested in knots thatmight be wild (Definition 2.4), but Reidemeister’s theorem assumes tameness (alsoDefinition 2.4). We begin with a reminder of some fundamental definitions.2.1.
Fundamental Definitions.
Let ( X, (cid:84) X ), ( Y, (cid:84) Y ) be topological spaces. An embedding of X into Y is a function f : X → Y such that restricting the codomainof f to f ( X ) gives us a homeomorphism (cid:101) f : X → f ( X ). f Figure 3.
Example of embedding an X into R . NIFORM CONVERGENCE AND KNOT EQUIVALENCE 5
Since embeddings must be injective, some authors choose to denote them by f : X (cid:44) → Y . Here we call Y the ambient space and refer to f ( X ) as X embedded by f in Y .Two embeddings f , f : X → Y are said to be equivalent (denoted f ∼ = f ) ifthere exists a homeomorphism h : Y → Y such that for all x ∈ X ,( h ◦ f )( x ) = f ( x ) . Since h is a homeomorphism on the ambient space, we refer to it as an ambienthomeomorphism . Remark . This definition requires pointwise equality for ( h ◦ f ), f . In general,this is stronger than requiring h ( f ( X )) = f ( X ) as sets. The interested readershould see [1, 6] for more.Geometrically, we think of h as “deforming” the ambient space to take f ( X ) to f ( X ). h Figure 4.
An example h taking f ( X ) to a distorted version rep-resenting f ( X ).A knot is an embedding f : S (cid:44) → R . One should note that some authors takethe codomain to be S instead of R because S is compact. Our proofs todayonly require that Y be a metric space, hence we are free to choose either option.We could even choose a thickened orientable surface in order to work with virtualknots. However, we will do neither of these things and instead choose Y = R because it is easier to represent graphically.For embeddings in R , defining equivalence through ambient homeomorphisms isequivalent to defining equivalence with ambient isotopy (defined below). We referto this fact as the equivalence of equivalences . For further discussion (as well as a listof references about the correspondence in each of the PL , C ∞ , and Topological categories), see [5], particularly § Definition 2.2 (Ambient Isotopy) . Let ( X, (cid:84) X ), ( Y, (cid:84) Y ) be topological spaces. Let f , f : X (cid:44) → Y be embeddings. Then a function H : [0 , × Y → Y is called an ambient isotopy iff The correspondence holds for tame knots and certain everywhere-wild knots. An example ofa knot for which it fails is Fig. 1b. The idea is that the ambient homeomorphism can only sendwild points to other wild points, and this makes it impossible to pull the strand “through” thewild point.
UNIFORM CONVERGENCE AND KNOT EQUIVALENCE (1) H is continuous,(2) H (0 , · ) is the identity on Y ,(3) For all t ∈ [0 , H ( t, · ) : Y → Y is a homeomorphism, and(4) For all x ∈ X , we have( H (1 , · ) ◦ f )( x ) = f ( x ) . We often refer to H as an ambient isotopy from f to f .Note that H (1 , · ) is an ambient homeomorphism from f to f . We can think ofthe t variable as describing a “time” parameter in a movie connecting H (0 , · ) to H (1 , · ). t = 0 t = t = t = t = 1 Figure 5. H (1 , · )is the h in Fig. 4We will prove our results today for both ambient homeomorphisms and ambientisotopies. Although this is not strictly necessary in light of the equivalence ofequivalences, we have chosen to include both arguments to illustrate the additionalstep required for working with ambient isotopy.2.2. Tameness and Wildness.
Oftentimes, knot theory is restricted to the studyof tame knots, which we define in a moment. We think of tame knots as being well-behaved because they belong to equivalence classes of knots that have representativeelements that can described with finite information, namely polygonal or PL knots . This is the loose intuition underpinning Reidemeister’s theorem.
Definition 2.3 (Polygonal Knot) . A polygonal knot is a knot that is comprised ofa finite union of straight line segments. Figure 6.
Examples of some polygonal knots PL stands for Piecewise Linear. For more on PL topology, the reader might look at J.L.Bryant’s
Piecewise Linear Topology . An online version can be found at
NIFORM CONVERGENCE AND KNOT EQUIVALENCE 7
Definition 2.4 (Tame & Wild Knots) . A tame knot is a knot that is ambientisotopic to a polygonal knot. A wild knot is a knot that is not tame. ∼ = Figure 7.
Example of a tame knot
Remark . There are many other common definitions for tameness and wildness.A discussion of these definitions (and some of the equivalences) can be found in [5].2.3.
Uniform Convergence.
Finally, we recall the definition of uniform conver-gence . Definition 2.6 (Uniform Convergence) . Let (
X, d X ), ( Y, d Y ) be metric spaces,and consider a sequence of functions f n : X → Y . Suppose that the f n convergepointwise to some f : X → Y . Then we say the f n converge to f uniformly iff forall ε >
0, there exists n ∈ N such that for all x ∈ X , n > n implies d Y ( f n ( x ) , f ( x )) < ε. We typically denote uniform convergence by f n u −→ f . Geometrically, we think ofthis in terms of pictures like the following: − − Figure 8.
Example of some f n : R → R satisfying (cid:107) f n − f (cid:107) ∞ < ε .Dashed lines indicate f ( x ) ± ε . Remark . It’s worth noting that, as with the definition of ambient home-omorphism, this definition is generally stronger than requiring convergence ofthe images as sets. In fact, we can construct simple examples of f n : [0 , (cid:44) → R such that for all n , m ∈ N , f n ([0 , f m ([0 , f n do not even con-verge pointwise. One way to achieve this is to alternate between two differentparameterizations of the same curve. UNIFORM CONVERGENCE AND KNOT EQUIVALENCE − . . . . − . − . . . . − − (a) One parameterization. . . − . . . . − . − . . . . − − (b) . . . And another. Figure 9.
Two parameterizations of the same curve.In the figures above, the colors correspond to value of t ∈ [0 ,
1] that yields thegiven point under f n . In order for f n u −→ f , not only do the curves have to takeon the same “shape,” but also every point of a given color in f n ([0 , ε away from the corresponding point in f ([0 , Proposition 2.8.
Let ( X, d X ) , ( Y, d Y ) be metric spaces. For each k ∈ N , let f k : X → Y be continuous. Suppose that there exists f : X → Y such that f k u −→ f .Then f is continuous. Proposition 2.9.
Let ( X, (cid:84) X ) , ( Y, (cid:84) Y ) be topological spaces. Suppose that X iscompact and Y is Hausdorff. Now suppose that f : X → Y is bijective and contin-uous. Then f is a homeomorphism. We are now ready to begin our discussion of these results in the context of knots.3.
Uniform Convergence & Knots
These two propositions give us the following simple result.
Corollary 3.1.
Let ( X, d X ) , ( Y, d Y ) be metric spaces, and suppose that X is com-pact. For each k ∈ N , let f k : X (cid:44) → Y be an embedding. Suppose that the f k converge uniformly to some f : X → Y . Then if f is injective, it follows that f isan embedding.Proof. By hypothesis, for all k ∈ N we have f k is an embedding and hence contin-uous. Since we also have f k u −→ f , Proposition 2.8 implies f is continuous.( Y, d Y ) is a metric space and thus Hausdorff. So f ( X ) with the subspace topologyis Hausdorff. Now, f is injective and thus f is a bijection between X and f ( X ). ByProposition 2.9 it follows that f is a homeomorphism between X and f ( X ). Thus f is an embedding. (cid:4) Among other things, Corollary 3.1 can be used to construct fractal-like knot dia-grams.
NIFORM CONVERGENCE AND KNOT EQUIVALENCE 9
Example 3.2.
Consider the knot constructed by the following iterative procedure,loosely inspired by the Koch Snowflake: (a) f (b) f (c) f (d) f (e) f (f) f Figure 10.
A “snowflake” knot.One can show that with the proper choice of parameterizations for the f k , Corol-lary 3.1 guarantees the limit function f ∞ = lim k →∞ f k is an embedding. The mainchallenge is explicitly proving injectivity; this can be done as long as the “shrink”factor for the twists is sufficiently small.3.1. Iteratively Constructing (Ambient) Homeomorphisms.
For the rest ofthis document we will be primarily interested in a special case of Corollary 3.1.Namely, when X = Y , the f k ’s become homeomorphisms from X to itself. When X is a compact subset of R this will give us a way to construct ambient homeomor-phisms (and later, ambient isotopies) by composing countably-many Reidemeistermoves.To that end we repackage Corollary 3.1 into a form that is more ergonomic whenworking with this special case. In particular, we’ll write the limit function in termsof a composition of homeomorphisms, which is more in line with our intuitionof applying multiple Reidemeister moves in succession. This will also allow usto offload the uniform convergence requirement to one of acting on a shrinkingcollection of neighborhoods, which is easier to interpret in terms of knot diagrams.Note, the following theorem is true for homeomorphisms in general, but we are onlyinterested in applying it to ambient homeomorphisms. Theorem 3.3. ( Y, d Y ) be a metric space. For all k ∈ N , let h k : Y → Y be ahomeomorphism, and for all n ∈ N , define (cid:104) n = nk =1 h k = ( h n ◦ h n − ◦ · · · ◦ h ◦ h ) . For each k let V k ⊆ Y such that h k is identity on V ck . Then provided(1) The V k satisfy lim n →∞ diam (cid:18) ∞ (cid:91) k = n V k (cid:19) = 0 , (2) There exists a compact A ⊆ Y such that (cid:18) ∞ (cid:91) k =1 V k (cid:19) ⊆ A ◦ , and(3) (cid:104) ∞ defined by (cid:104) ∞ = lim n →∞ (cid:104) n exists and is bijective,then (cid:104) ∞ is a homeomorphism. Before continuing to the proof, we make some remarks about the statement.
Remark . The hypothesis that the limit (cid:104) ∞ exists is superfluous as it is impliedby conditions (1) and (2). Remark . One can replace the somewhat-technical conditions on the V k withsimpler ones. E.g., conditions (1) and (2) could be substituted with(1) Requiring · · · ⊆ V k +1 ⊆ V k ⊆ · · · ⊆ V ⊆ A ◦ and(2) lim k →∞ diam( V k ) = 0.Another option would be to replace condition (1) with something like “diam(lim sup V n ) =0.” In any case, we avoided simplifications like these in an effort to make corre-spondence with the hypotheses of Corollary 3.1 more direct. Remark . We require lim n →∞ diam ( (cid:83) ∞ k = n V k ) = 0 instead of lim n →∞ diam ( V k ) =0 to avoid situations where the V k have diameters like k . If we were to allow caseslike these the divergence of the harmonic series would cause problems. Remark . Though perhaps tempting, it is not sufficient to do away with theconditions on the V k by requiring something like “ h k u −→ Y .” As a counterexample,consider Y = [0 ,
1] with the standard metric on R . For all k ∈ N , define h k = x ( k +1) /k . NIFORM CONVERGENCE AND KNOT EQUIVALENCE 11
Then h k u −→ [0 , . But note, (cid:104) k = x k , and thus (cid:104) ∞ ( x ) = (cid:40) x ∈ [0 , x = 1which is not a homeomorphism. Proof.
We will employ the gluing lemma. To that end, we need to partition Y intotwo closed sets and show that (cid:104) ∞ is a homeomorphism on both. A natural choiceis to consider A c and A . Note, the compactness of the latter will allow us to appealto Corollary 3.1. V V V V V V A V V V V V V A c A Figure 11.
Example A shown shaded on the left; example A c shown shaded on the right.We examine these two sets separately.(1) (On A c ): By construction, each h k is identity on V ck . Since each V k ⊆ A ◦ ,it follows (cid:104) ∞ is identity (and hence a homeomorphism) on ( A ◦ ) c = A c .(2) (On A ): Now, we show that (cid:104) ∞ is a homeomorphism on A . By Corol-lary 3.1, because (cid:104) ∞ was assumed to be bijective it suffices to show thatthe restrictions (cid:104) k | A converge uniformly to (cid:104) ∞ | A . We will suppress writingthe | A for now because it clutters the notation too much.Let ε > n →∞ diam (cid:18) ∞ (cid:91) k = n V k (cid:19) = 0 , hence there exists n ∈ N such thatdiam (cid:18) ∞ (cid:91) k>n V k (cid:19) < ε. We have the following claim.
Claim:
For all n > n , for all y ∈ A , we have d ( (cid:104) n ( y ) , (cid:104) ∞ ( y )) < ε. Proof of Claim:
Fix an n > n and let y ∈ A be arbitrarily chosen. Wehave two subcases.(a) First, suppose (cid:104) n ( y ) (cid:54)∈ (cid:83) ∞ k>n V k .Recall that we defined the h k such that each h k is identity outside V k .It follows that for all k > n we have h k ( y ) = y . Hence (cid:104) n ( y ) = (cid:104) ∞ ( y ) , so d ( (cid:104) n ( y ) , (cid:104) ∞ ( y )) = 0 < ε, as desired. (cid:51) h n ( y ) h h h y V V V V V V A Figure 12.
An example of this case with n = 4. The shadedportions represent (cid:83) ∞ k>n V k . Here, y starts in V , is mapped into V by h , into V by h , skipped by h , then finally mapped toanother point of V by h , before remaining fixed for all k > (cid:104) n ( y ) ∈ (cid:83) ∞ k>n V k . Note, since 〔 for all k > n , h k is bijective and is identity outside (cid:83) ∞ k>n V k 〕 , it follows that for all n > n , (cid:104) n ( y ) ∈ ∞ (cid:91) k>n V k and hence (cid:104) ∞ ( y ) ∈ ∞ (cid:91) k>n V k . NIFORM CONVERGENCE AND KNOT EQUIVALENCE 13
Note, the set in the second is just the closure of the set in the first;thus they have the same diameter. By definition of n , we havediam (cid:0)(cid:83) ∞ k>n V k (cid:1) < ε , hence d ( (cid:104) n ( y ) , (cid:104) ∞ ( y )) < ε as desired. (cid:51) In either case, we get that d ( (cid:104) n ( y ) , (cid:104) ∞ ( y )) < ε . Now (writing the restric-tions explicitly again) it follows that (cid:104) n | A is a sequence of homeomorphismswith ( (cid:104) n | A ) u −→ ( (cid:104) ∞ | A ).Finally, recall that by hypothesis, (cid:104) ∞ is a bijection. This implies (cid:104) ∞ | A is too, and since A is compact, Corollary 3.1 now guarantees (cid:104) ∞ | A is ahomeomorphism on A .Now, applying the gluing lemma to ( (cid:104) ∞ ) | A and ( (cid:104) ∞ ) | A c we conclude that (cid:104) ∞ iscontinuous. An identical argument shows (cid:104) − ∞ is continuous. It follows that (cid:104) ∞ isa homeomorphism, as desired. (cid:4) Iteratively Constructing (Ambient) Isotopies.
We now state the anal-ogous result for isotopies. As with Theorem 3.3, the theorem below is valid forisotopies in general but we are only interested in applying it to ambient isotopies.It might be easy to get bogged down by the additional details so we summarize themain ideas.Given a sequence of isotopies H k , if the associated homeomorphisms h k ( · ) := H k (1 , · ) satisfy the hypotheses of Theorem 3.3, then we can stitch the H k ’s to-gether into an isotopy (cid:72) ∞ as follows. First, define t = 0 and let ( t k ) ∞ k =1 be astrictly increasing sequence in (0 , (cid:72) ∞ to apply the effects of H overthe compressed time interval [ t , t ]. Then, do the same to apply H over [ t , t ].Continue this process, in general applying H k over the interval [ t k − , t k ].Stopping the construction above after n steps will give us an isotopy (cid:72) n . Taking n → ∞ we will get a function (cid:72) ∞ with (cid:72) ∞ (1 , · ) = (cid:104) ∞ . By Theorem 3.3, (cid:104) ∞ will bea homeomorphism. And since the (cid:72) n are all isotopies, we’ll see that (cid:72) ∞ ( t, · ) willbe a homeomorphism for all t ∈ [0 , (cid:72) n will then show (cid:72) ∞ is continuous overall and thus an isotopy! Theorem 3.8.
Let ( Y, d Y ) be a metric space. For all k ∈ N , let H k : [0 , × Y → Y be an isotopy, and let V k ⊆ Y such that H k is identity on [0 , × ( V ck ) . For each k define h k : Y → Y by h k ( y ) = H k (1 , y ) ; note that by definition of an isotopy, h k isa homeomorphism.Suppose that the h k ’s and V k ’s satisfy the hypotheses of Theorem 3.3, and for all n ∈ N define (cid:104) n = nk =1 h k . Let t = 0 and let ( t k ) ∞ k =1 be a strictly increasing sequence in (0 , converging to . Then (cid:72) ∞ : [0 , × Y → Y defined by (cid:72) ∞ ( t, y ) = H (cid:16) t − t t − t , y (cid:17) if t ∈ [ t , t ] H (cid:16) t − t t − t , (cid:104) ( y ) (cid:17) if t ∈ ( t , t ] H (cid:16) t − t t − t , (cid:104) ( y ) (cid:17) if t ∈ ( t , t ] ... H k (cid:16) t − t k − t k − t k − , (cid:104) k − ( y ) (cid:17) if t ∈ ( t k − , t k ] ... (cid:104) ∞ ( y ) if t = 1 , is an isotopy.Proof. To show (cid:72) ∞ is an isotopy we must show(1) (cid:72) ∞ (0 , · ) is identity,(2) For each t ∈ [0 , (cid:72) ∞ ( t, · ) is a homeomorphism, and(3) (cid:72) ∞ is continuous.We prove these in the order above.(1) For all y ∈ Y , (cid:72) ∞ (0 , · ) = H (0 , · ). Since H is an isotopy, H (0 , · ) isidentity (by definition) and this proves (1).(2) To prove (2) we break things up into three subcases.(a) Suppose t = 0. Then (cid:72) ∞ ( t, · ) = (cid:72) ∞ (0 , · ) which is the identity andthus a homeomorphism. (cid:51) (b) Suppose t ∈ (0 , k ∈ N such that t ∈ ( t k − , t k ].Recall that by construction, (cid:72) ∞ ( t, · ) = H k (cid:18) t − t k − t k − t k − , (cid:104) k − ( · ) (cid:19) . (3.1)Define (cid:103) ( · ) := H k (cid:16) t − t k − t k − t k − , · (cid:17) . By definition of an isotopy, (cid:103) is a home-omorphism. Also, (cid:104) k − is a finite composition of homeomorphisms andthus a homeomorphism. Rewriting Eq. (3.1) gives us that (cid:72) ∞ ( t, · ) = ( g ◦ (cid:104) k − )( · )which is a finite composition of homeomorphisms and hence itself ahomeomorphism, as desired. (cid:51) (c) Now suppose t = 1. Then (cid:72) ∞ ( t, · ) = (cid:104) ∞ ( · ). Since the h k ’s, V k ’s wereassumed to satisfy the hypotheses of Theorem 3.3 we get that (cid:104) ∞ is ahomeomorphism as desired. (cid:51) In either case, we see (cid:72) ∞ ( t, · ) is a homeomorphism.(3) Finally, it remains to show (cid:72) ∞ is continuous. We employ uniform conver-gence. NIFORM CONVERGENCE AND KNOT EQUIVALENCE 15
Define a sequence of isotopies (cid:72) n as follows: For each n ∈ N , let (cid:72) n : [0 , × Y → Y be given by (cid:72) n ( t, y ) = H (cid:16) t − t t − t , y (cid:17) if t ∈ [ t , t ] H (cid:16) t − t t − t , (cid:104) ( y ) (cid:17) if t ∈ ( t , t ]... H n (cid:16) t − t n − t n − t n − , (cid:104) n − ( y ) (cid:17) if t ∈ ( t n − , t n ] (cid:104) n ( y ) if t ∈ ( t n , . That is, we follow (cid:72) ∞ ( t, y ) until we reach t = t n and then we freeze. Onecan verify that the (cid:72) n ( t, y ) are indeed isotopies; of particular note, theyare continuous. We now show (cid:72) n u −→ (cid:72) ∞ .Let ε > V k there exists n ∈ N such that diam (cid:18) ∞ (cid:91) k>n V k (cid:19) < ε. Let n > n be arbitrarily chosen, and similarly let ( t, y ) ∈ [0 , × Y . Weshow d ( (cid:72) n ( t, y ) , (cid:72) ∞ ( t, y )) < ε . We have two subcases.(a) First, suppose t ∈ [0 , t n ]. Then (cid:72) n ( t, y ) = (cid:72) ∞ ( t, y ) and so we have d ( (cid:72) n ( t, y ) , (cid:72) ∞ ( t, y )) = 0 and the bound holds.(b) Now, suppose t ∈ ( t n , y ∈ (cid:0)(cid:83) ∞ k>n V k (cid:1) c , then (cid:72) n ( t, y ) = (cid:72) ∞ ( t, y ),so we have d ( (cid:72) n ( t, y ) , (cid:72) ∞ ( t, y )) = 0 and the bound holds. Else, notethat both of (cid:72) n ( t, y ), (cid:72) ∞ ( t, y ) ∈ (cid:83) ∞ k>n V k , hence d ( (cid:72) n ( t, y ) , (cid:72) ∞ ( t, y )) < ε as desired.In either case, we have d ( (cid:72) n ( t, y ) , (cid:72) ∞ ( t, y )) < ε . As ( t, y ) were arbitrarilychosen, this implies (cid:72) n u −→ (cid:72) ∞ . By Proposition 2.8, (cid:72) ∞ is continuous.It follows that (cid:72) ∞ is an isotopy as desired. (cid:4) In the next section, we apply this result to a variety of curves, beginning with theexample from Fig. 1.4.
Various Applications of Theorem 3.8
The first few examples will all make use of the following lemma, which allows us toremove the bijectivity hypothesis from Theorem 3.8.
Lemma 4.1.
Let all variables be quantified as in Theorem 3.8. Then if the V k ’sare all disjoint, (cid:72) ∞ (1 , · ) is guaranteed to be a bijection.Proof. If the V k are all disjoint, then defining U = (cid:83) ∞ k =1 V k we can write Y as thedisjoint union Y = ( U c ) (cid:116) (cid:18) ∞ (cid:71) k =1 V k (cid:19) . Note, (cid:72) ∞ (1 , · ) is identity on U c and hence a bijection. Since the V k are all dis-joint, (cid:72) ∞ (1 , · ) | V k = (cid:72) k (1 , · ) | V k , the latter of which is a homeomorphism and thusbijective, so (cid:72) ∞ (1 , · ) is a bijection on each of the V k .Thus (cid:72) ∞ (1 , · ) is a bijection overall. (cid:4) Proposition 4.2.
The curve shown in Fig. 13 below is tame. In particular, it isan unknot.
Figure 13.
A wild-looking unknot, redux.
Proof.
Let f : S (cid:44) → R be the standard unknot, and let f : S (cid:44) → R be anembedding yielding a diagram like Fig. 13. We apply Theorem 3.8 to construct anambient isotopy (cid:72) ∞ : [0 , × R → R taking f to f .Consider the sequence of V k ’s chosen as follows. V V V V V Figure 14. V k ’s. Strictly speaking we have not defined tameness for curves, only for knots. A tame curve is acurve that’s ambient homeomorphic (equivalently, ambient isotopic) to a polygonal curve. A parametrization can be found in [5], although it is given in the context of a “theorem”about tameness and parametrizations that turns out to be incorrect.
NIFORM CONVERGENCE AND KNOT EQUIVALENCE 17
One can verify that lim n →∞ diam ( (cid:83) ∞ k = n V k ) = 0 and that there exists a compactset A ⊆ R such that (cid:83) ∞ k =1 V k ⊆ A .For all k ∈ N , let H k : [0 , × R → R be an ambient isotopy inserting a Reide-meister I into the arc bound in V k . Use these H k to define (cid:72) ∞ as in Theorem 3.8.By Lemma 4.1, (cid:72) ∞ (1 , · ) is a bijection. Thus by Theorem 3.8, (cid:72) ∞ is an ambientisotopy from f to f . (cid:4) Proposition 4.3.
The following curve is tame.Proof (Sketch).
We apply Theorem 3.8 twice. This two-step method is not strictlynecessary, but the diagram is a bit less cluttered this way. Consider the followingstarting curve:Apply Reidemeister II moves within the dotted regions below:Since these V k are disjoint, we can again apply Lemma 4.1 to obtain an ambientisotopy. The result looks something like the following: Now, perform Reidemeister II moves in the following regions:Again, the V k here are all disjoint, hence one can apply Lemma 4.1 to show thatthis is an ambient isotopy. The end result iswhich is the desired diagram. (cid:4) We now consider a similar curve, this time constructed using Reidemeister I moves.This will be the most technical argument of the paper. We advise the reader toread through Example 5.2 in the next section before continuing. This is becauseExample 5.2 shows how we can lose bijectivity in the limit, and the bulk of thechallenge in Proposition 4.4 is addressing similar concerns. We have to addressbijectivity in a manner like this whenever we have points that are moved by infinitelymany of the H k ’s (whereas in Proposition 4.3, each point is moved by only finitelymany H k ). Proposition 4.4.
Let f : [0 , (cid:44) → R be NIFORM CONVERGENCE AND KNOT EQUIVALENCE 19
Figure 15.
Our starting arc. and let f : [0 , (cid:44) → R be Figure 16.
A different countable sequence of Reidemeister Imoves.
Then f ∼ = f .Proof (Sketch). We will construct the ambient isotopy from f to f by a recursiveprocess. We will repeatedly insert Reidemeister I moves like the following: Figure 17.
The general procedure.We must do so in such a way that we can still argue bijectivity of (cid:72) ∞ (1 , · ). Thekey idea is to choose our H k ’s so that only one point (denoted y ∞ ) gets movedinfinitely-many times. We explicitly guarantee this by constructing our H k ’s sothat for all y ∈ R \ { y ∞ } , there exists n such that for n > n , (cid:104) n ( y ) is unmovedby H n ( t, · ). y ∞ will be the point that gets sent to the limit of the twists in Fig. 16. In our construction, y ∞ will be the vertex in Fig. 15, but one can create other constructions where it is a differentpoint. To that end, define (cid:96) as shown in Fig. 18, and let ε > ε (cid:28) Define V tobe a closed rectangular prism of dimensions (6 + ε ) (cid:96) × (2 + ε ) (cid:96) × (2 + ε ) (cid:96) , and H tobe a PL ambient isotopy inserting the first loop such that H is identity off V . Note, even though we define V to be closed, we’ll draw it with dotted lines in thebelow. (6 + 2 ε ) ‘‘ (6 + 2 ε ) ‘‘ H Figure 18.
The same figure, now showing V .Now, we describe the general strategy for inserting the k + 1 st loop given the first k loops. We want V k +1 , H k +1 to be half-scale versions of V k , H k . The figure belowshows this for k = 1. (6 + 2 ε ) ‘‘ ‘ (6 + 2 ε ) ‘ (6 + 2 ε ) ‘‘ ‘ (6 + 2 ε ) ‘ H Figure 19. V and V , with application of H shown.But, to make things work, it will be important to first apply some intermediate ambient isotopy H k + . in between H k and H k +1 such that H k + . does not changethe diagram, but does help ensure that points in the ambient space aren’t lost inthe limit. Desired Properties of H k + . : We want H k + . to preemptively “unsquish” pointsthat might be compressed together by H k +1 . To determine exactly how muchunsquishing we have to do, we look at a sort of inverse Lipschitz condition.Let h k +1 = H k +1 (1 , · ). Since H k +1 is a PL ambient isotopy, h k +1 is a PL ambienthomeomorphism, and thus there exists c ∈ (0 ,
1) such that for all x , x ∈ R , c · d ( x , x ) ≤ d ( h k + ( x ) , h k + ( x )) . (4.1) Actually, ε > k ∈ N we have V k +1 ⊆ V ◦ k . Wejust choose ε (cid:28) We can assume PL-ness because the modifications can be realized by elementary moves . The 6 in our prism dimensions comes from the fact (cid:96) is defined to be 1 / rd of the length ofthe twist inserted in Fig. 18, and the moves halve in size at each iteration. This essentially pops out of the finiteness condition on our simplicial complexes for PL maps.
NIFORM CONVERGENCE AND KNOT EQUIVALENCE 21
Note that because the H k are all identical up to scaling, c is independent of k . We interpret Eq. (4.1) as giving us a bound on how much h k +1 can “squish” pointsin the space together. Let q k be the tip of the twist before applying H k +1 : q k Figure 20. q k labeled.For all p ∈ V k +1 , we want H k + . to be constructed to guarantee that either(1) h k +1 ( h k + . ( p )) ∈ V k \ V k +1 (i.e. p gets moved to the outer box), or(2) p gets “moved farther from q k than it can be squished in later”:1 c · d ( p, q k ) ≤ d ( h k + . ( p ) , h k + . ( q k )) . (4.2) Constructing H k + . : Let V k + . be a slightly-scaled-up version of V k +1 such that V k +1 (cid:40) V k + . (cid:40) V k . To make things easier, we will require that V k + . also onlyintersects with ( h k ◦ h k − ◦ · · · ◦ h ◦ f )([0 , V k + . and V k +1 share the same center of mass and have all sides parallel (see Fig. 21). q Figure 21. V k + . and V k +1 We can parameterize every point p ∈ V k + . in terms of a piecewise linear functionas detailed below. The construction is a bit unergonomic to formalize explicitly,but it is meant to capture the ideas of Fig. 22. One might ask why we can’t have c ≥
1. Note that V k +1 being bounded precludes c > c = 1, note that h k +1 is not a vector space isomorphism of R , and hence not an isometry on R ; since h k +1 is identity outside V k +1 , isometry must fail on V k +1 . V k + . V k +1 Figure 22.
The piecewise-linear parameterization, with some ofthe v k + . ’s (outer points; defined below) and v k +1 ’s (inner points;also defined below) shown in white.We construct it in two parts and then scale them by half and glue them together.(1) If p ∈ V k + . \ V k +1 , there exist unique points v k + . ∈ ∂V k + . and v k +1 ∈ ∂V k +1 such that v k + . is the point in V k + . “corresponding” to v k +1 in V k +1 , and p is on the line segment v k + . v k +1 . Thus there exists a unique s ∈ [0 ,
1] such that we can write p as a convex combination p = s · v k + . + (1 − s ) · v k +1 . (2) If p ∈ V k +1 , there exists a unique point v k +1 ∈ ∂V k +1 such that p is onthe line segment q k v k +1 . Analogously to the above, there exists a unique s ∈ [0 ,
1] such that we can write p as p = s · v k +1 + (1 − s ) · q k . We re-scale s to glue these two parameterizations into a single function which wecall H (cid:48) k + . : H (cid:48) k + . ( s, v k + . , v k +1 ) = (cid:40) s · v k +1 + (1 − s ) · q k s ∈ (cid:2) , (cid:3) (2 s − · v k + . + (2 − s ) · v k +1 s ∈ (cid:0) , (cid:3) . We’ll now do something a bit bizarre and rewrite the parameters in H (cid:48) k + . as func-tions of p . Note that with the re-scaling of s , we now have s uniquely determinedby p . Also recall that by construction, v k + . and v k +1 are each uniquely determinedby p . Hence, we can think of s , v k + . , v k +1 as being functions of p . One can showthat these are all continuous. As such, we can indeed think of H (cid:48) k + . as just beinga complicated way of writing the identity function on V k + . .To turn H (cid:48) k + . into our ambient isotopy H k + . , we now introduce time dependence in s . Define s c = c and s c = , and observe s c < s c . Define s c : [0 , → [ s c , s c ] By “corresponds,” we mean that given a linear function that scales up V k +1 to yield V k + . , v k +1 gets mapped to v k + . . The reason that c appears in this expression is because we’re trying to get Eq. (4.2) out inthe end. NIFORM CONVERGENCE AND KNOT EQUIVALENCE 23 by s c ( t ) = t · s c + (1 − t ) · s c , and use this to define s (cid:48) ( t, p ) = (cid:16) s ( p ) s c (cid:17) · s c ( t ) if s ( p ) ∈ [0 , s c ] (cid:16) s ( p ) − s c − s c (cid:17) · (cid:16) − (cid:16) s ( p ) − s c − s c (cid:17)(cid:17) · s c ( t ) if s ( p ) ∈ ( s c , s ( p ) represents aparameter in [0 ,
1] that tells us how to write p as a convex combination of otherpoints. One can verify that when t = 0, s (cid:48) ( t, p ) reduces to s ( p ). Then, as t increasesto 1, s (cid:48) ( t, p ) distorts the interval represented by s ( p ) until we end up with somethinglike the following, in which s c gets mapped to where s c was initially: s c s c Figure 23.
The interval [0 ,
1] represented by s (cid:48) ( t, p ) as t goes from0 to 1. The light portion represents the values where s ( p ) ∈ [0 , s c ]and the dark portion represents s ( p ) ∈ [ s c , H k + . is to take a diagram like the following V k + . V k +1 and turn it into V k + . V k +1 Figure 24.
The effects of H k + . .Finally, we have the following: Claim: H k + . : [0 , × R → R given by H k + . ( t, p ) = (cid:40) H (cid:48) k + . ( s (cid:48) ( t, p ) , p ) p ∈ V k + . p p (cid:54)∈ V k + . satisfies the desired properties of H k + . . Proof of Claim:
One can verify that H k + . satisfies all the properties of anambient isotopy. a It remains to show that h k + . = H k + . (1 , · ) satisfies theconditions stipulated near Eq. (4.2).Let p ∈ V k + . be arbitrary. We have two cases.(1) Suppose s ( p ) ∈ ( s c , h k + . ( p ) ∈ V k + . \ V k +1 , and hence h k +1 ( h k + . ( p )) ∈ V k \ V k +1 . (cid:51) (2) Suppose s ( p ) ∈ [0 , s c ]. One can verify that in this case, H k + . ( t, p ) onlyslides p along a ray segment originating from q k , with the sliding dictatedby s (cid:48) ( t, p ). Hence d ( p, q k ) d ( h k + . ( p ) , h k + . ( q k )) = d ( H k + . (0 , p ) , H k + . (0 , q k )) d ( H k + . (1 , p ) , H k + . (1 , q k ))= s (cid:48) (0 , p ) s (cid:48) (1 , p )= (cid:0)(cid:0) s ( p ) s c s c (cid:0)(cid:0) s ( p ) s c s c = c. Simplifying gives us1 c · d ( p, q k ) = d ( H k + . (1 , p ) , H k + . (1 , q k )) , as desired. a Intuitively, all it is doing is sliding all the points in V k +1 along the lines in Fig. 22 until theyare either in V k + . \ V k +1 or c times as far from q k as they were at the start. NIFORM CONVERGENCE AND KNOT EQUIVALENCE 25
Guaranteeing Bijectivity:
Observe that for all k ∈ N , for all p ∈ V k +1 , we have d ( h k +1 ( h k + . ( p )) , h k +1 ( h k + . ( q k ))) ≥ c · d ( h k + . ( p ) , h k + . ( q k )) ≥ c · c · d ( p, q k ) ≥ d ( p, q k ) . (4.3)For each n ∈ N , let (cid:104) n denote the composition of all these homeomorphisms: (cid:104) n = (cid:32) n − k =1 ( h k +1 ◦ h k + . ) (cid:33) ◦ h = ( h n ◦ h n − . ◦ · · · ◦ h ◦ h . ◦ h ) . Note that the sequence of points (cid:104) − n ( q n ) is constant, hence the limit lim n →∞ (cid:104) − n ( q n )exists; in particular, it is y ∞ . For all y ∈ R \ { y ∞ } , Eq. (4.3) shows that at eachstep, y is sent no closer to q k +1 than it was to q k . Since the boxes are shrinkingit follows that each such y will eventually leave the boxes and thus remain fixed atsubsequent steps. Explicitly: If n satisfies(6 + 2 ε ) (cid:96) n < d ( y, y ∞ ) , Then for all n > n we have (cid:104) n ( y ) (cid:54)∈ V n , and hence (cid:104) n ( y ) = (cid:104) n ( y ) . This implies (cid:104) ∞ is a bijection between R \ { y ∞ } and R \ { (cid:104) ∞ ( y ∞ ) } . So Theo-rem 3.3 implies (cid:104) ∞ is a homeomorphism between R \ { y ∞ } and R \ { (cid:104) ∞ ( y ∞ ) } .Thus (cid:104) ∞ is bijective on R , and Theorem 3.8 implies that (cid:72) ∞ (1 , · ) is an ambientisotopy. (cid:4) Finally, we have the following famous example.
Proposition 4.5.
The following curve is a tame arc.Sketch.
We apply Lemma 4.1. Consider a sequence of properly nested boxes V , V , . . . , as follows: Figure 25.
A countable connected sum of trefoils.
For all k ∈ N , define V (cid:48) k by V (cid:48) k = V k \ V k +1 . Note the V (cid:48) k are disjoint. We can definethe ambient isotopies H k such that H k performs the following modification on V (cid:48) k : NIFORM CONVERGENCE AND KNOT EQUIVALENCE 27
Taking the limit, we obtain an ambient isotopy unknotting the arc. (cid:4)
Having established the versatility of Theorem 3.8, we now discuss situations inwhich it cannot be applied. In a sense, we will see that each of the hypotheses ofthe theorem are sharp.5.
Cases Where Theorem 3.8 Does Not Apply
Example 5.1.
If we extend the right hand side of Fig. 25 with a straight linesegment, then we cannot apply Theorem 3.8. In fact, the curve is wild — see [2],Exercise 2.8.4.What breaks here is that if we try to apply the same argument as we did with thenon-extended version in Proposition 4.5, we can’t force lim n →∞ diam ( (cid:83) ∞ k = n V k ) =0. In particular, diam ( (cid:83) ∞ k = n V k ) is bounded below by the length of the straightline segment.And now, as promised, we discuss the curve from Fig. 1b. Example 5.2.
One cannot apply Theorem 3.8 to the following curve:
Figure 26.
Fox’s “Remarkable Knotted Curve.”Here, the V k ’s are not the problem; rather it is bijectivity on the ambient space.Consider a “line” of points in the ambient space passing through the first loop: Figure 27.
The curve, now with an imaginary “line” of pointsfrom the ambient space.As we remove the first loop, some points on the “line” get pulled through:
NIFORM CONVERGENCE AND KNOT EQUIVALENCE 29
Figure 28.
Removing the first loop. Note, by continuity, the“line” must remain unbroken.As we remove the second loop, a similar process occurs:
Figure 29.
Removing the second loop.As n → ∞ , the stitching process continues, with the number of lines doubling ateach iteration. No matter what we try, in the limit, a countable subset of theoriginal line gets mapped to the wild point. In fact, if we thicken the line in Fig. 27to a cylinder, we see that uncountably -many points are lost in the limit!This is reflected in the fact that if we were to try something like the approach takenin Proposition 4.4, we would not be able to define ambient isotopies that do thejobs of the H k + . ’s. Remark . Another perspective one might consider here is that there is no obviousway to “tie” the curve in Fig. 26 if starting with just an unknotted line. Indeed,there’s no “first loop” to insert — to tie one, we require infinitely-many of theothers to be present already! 6.
Discussion
We conclude with a discussion of directions for future work.Theorems 3.3 and 3.8 give us one direction to a loose “countable analogue” ofReidemeister’s theorem. The restrictions on the V k ’s have a nice diagrammaticinterpretation — “the total region we’re going modify has to shrink in the limit” —but so far, a similarly-concise description of the bijectivity requirement has eludedthe author.Qualitatively, it seems that problems tend to occur when the set of points thatget moved infinitely-many times is not topologically discrete; however, it’s beendifficult to find the right language to distinguish between cases when this gives riseto legitimate problems versus ones where the issue is superficial. As an example NIFORM CONVERGENCE AND KNOT EQUIVALENCE 31 of the former, consider Example 5.2, and as an example of the latter, consider asequence of homeomorphisms h k : R → R where each h k is defined by h k ( y ) = y + k . Then h ∞ ( y ) = y + , and so all points in R are moved infinitely-many times by h ∞ , yet we have noproblems.Thus, we have the following question: Question 1.
Is there a simpler way to guarantee bijectivity of the limit functionin Theorem 3.8? In particular, is there a purely diagrammatic condition?One of the things that makes the problem in Example 5.2 tricky to spot at firstis that the limiting process yields an isotopy , just not an ambient isotopy (this isreminiscent of
Bachelor’s unknotting ). We wonder whether a similar effect can beobtained using only Reidemeister I or Reidemeister III moves, as detailed in thefollowing questions:
Question 2.
Does there exist a knot f : S (cid:44) → R such that applying a count-able sequence of Reidemeister I moves to f yields an isotopy, but not an ambientisotopy? Question 3.
Same as Question 2, but with Reidemeister III moves instead ofReidemeister I moves.Of course, we want to avoid trivial examples like taking H to be Bachelor’s un-knotting and then taking the remaining H k ’s to be identity.Now, returning to the question of a countable analogue for Reidemeister’s theorem: Question 4.
If we restrict ourselves to Reidemeister moves, when do the conversesof Theorems 3.3 and 3.8 hold? That is, given an ambient isotopy between twoembeddings f , f : S (cid:44) → R , when can we guarantee the existence of the desired V k ’s and h k ’s, where each of the h k ’s represent single Reidemeister moves?We have a conjecture in this direction. In [5] the author employed Theorem 3.8 toargue the following result: Theorem 6.1.
Call a knot diagram a discrete diagram if it satisfies all theaxioms of a regular diagram except perhaps having finitely-many crossings.
The “discrete” in “discrete diagram” comes from the fact that the set of crossing points onlyneeds to be topologically discrete rather than finite.
Now, let f : S (cid:44) → R be an arbitrary knot. Then if f admits a discrete diagram, f is ambient isotopic to a representative comprised of countably-many line segments. This gave rise to the following conjecture (we thank Kye Shi for the insight ofadding the fourth move):
Conjecture 1.
Define the extended
Reidemeister moves to be the standard moveset together with a fourth move
A A
Figure 30.
Fourth move where in the above, A is a compact set whose interior remains fixed relative to itsboundary. Note, A can contain wild points.Let f , f : S (cid:44) → R be knots that admit discrete diagrams. Suppose f ∼ = f .Then there exists a countable sequence of extended Reidemeister moves satisfyingthe hypotheses of Theorem 3.8 and taking f to f . For more details on the proposed approach, see [5], § Acknowledgements
We thank Kye Shi for proofreading and for offering many helpful suggestions onhow to improve the exposition. We also thank Francis Su and Sam Nelson for themany insights they provided when advising the author’s undergraduate thesis [5],on which this work is based.
References
1. H. Bothe,
Homogeneously Wild Curves and Infinite Knot Products , Fundamenta Mathematicae (1981), no. 2, 91–111 (eng).2. Robert J. Daverman and Gerard A. Venema,
Embeddings in Manifolds , Graduate Studies inMathematics, vol. 106, American Mathematical Society, 2009.3. Ralph H. Fox,
A Remarkable Simple Closed Curve , Annals of Mathematics (1949), no. 2,264–265.4. Ralph H. Fox and Emil Artin, Some Wild Cells and Spheres in Three-Dimensional Space ,Annals of Mathematics (1948), no. 4, 979–990, Publisher: Annals of Mathematics.5. Forest D. Kobayashi, Where the Wild Knots Are , Bachelor’s thesis, Harvey Mudd College,Claremont, CA, May 2020.6. Arnold C. Shilepsky,
Homogeneity and Extension Properties of Embeddings of S in E , Trans-actions of the American Mathematical Society (1974), 265–276. NIFORM CONVERGENCE AND KNOT EQUIVALENCE 33
Appendix A. The Fox-Artin Tameness Invariant
We will now describe the invariant for tameness established by Fox and Artin.
Theorem A.1 (Fox-Artin) . Let f : [0 , (cid:44) → R be a tame arc. Let p = f (0) , andfor all k ∈ N , let V k ⊆ R be a closed set such that p ∈ V ◦ k . Suppose that · · · (cid:40) V k (cid:40) · · · (cid:40) V (cid:40) V , and { p } = ∞ (cid:92) n =1 V k . Then there exists n ∈ N such that the inclusion map ι ∗ : π ( V n \ f ([0 , (cid:44) → π ( V \ f ([0 , is the trivial homomorphism.Proof. f is tame implies that there exists an ambient homeomorphism h : R → R such that h ◦ f is a straight line segment. Since homeomorphism preserves thefundamental group (as well as all the conditions on the V n ), it suffices to prove theclaim for a straight line.Hence, without loss of generality, suppose f is a line segment. p Figure 31.
An example of f ([0 , p labeled. V V V V p Figure 32.
An example of some possible first few V n ’s.Since p ∈ V ◦ , there exists ε > B ε ( p ) (cid:40) V . Now, because (cid:84) ∞ n =1 V n = { p } , we have diam( V k ) →
0, and hence there exists n ∈ N such that V n (cid:40) B ε ( p ).This gives us the inclusions (of sets) ι : V n (cid:44) → B ε ( p ) and ι : B ε ( p ) (cid:44) → V . Then the inclusion (of sets) ι : V n (cid:44) → V is given by ι = ι ◦ ι . The result is identical if we replace V n , B ε ( p ), and V with V n \ f ([0 , B ε ( p ) \ f ([0 , V \ f ([0 , ι ) ∗ : π ( V n \ f ([0 , (cid:44) → π ( B ε ( p ) \ f ([0 , ι ) ∗ : π ( B ε ( p ) \ f ([0 , (cid:44) → π ( V \ f ([0 , , we see that ι ∗ : π ( V n \ f ([0 , (cid:44) → π ( V \ f ([0 , ι ∗ = ( ι ) ∗ ◦ ( ι ) ∗ .B ε ( p ) \ f ([0 , π ( B ε ( p ) \ f ([0 , ι ) ∗ is the trivial homomorphism, and thus ι ∗ isthe trivial homomorphism. (cid:4) In the case of the curve in Fig. 1b, one can find a sequence of V k such that no such n0