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This question came up again in an algebraic topology course, so I'm placing a bounty on it. The answers below use algebraic topology - I am asking for a proof that does not use algebraic topology, e.g. homology, higher homotopy groups, or the Jordan Curve/Separation Theorem. No smooth structures, just point-set stuff.

Original question:

Suppose that $X_n$ is the union of $n$ half closed, straight line segments, all joined at a common point (their boundary point). So $X_2$ is just an open interval, and otherwise is an $n$-pointed asterisk with open ends. How can one show that for $n \geq 2$ all the spaces $X_n \times \mathbb{R}$ are topologically distinct? These spaces are like $n$ sheets attached along a copy of $\mathbb{R}$.

More generally if we take $n$ copies of the $k$-dimensional closed half-space $\mathbb{H}^k$ and identify them along a $(k-1)$-plane, how do we show that this space is not homeomorphic to $\mathbb{R}^k$ for $n > 2$?

Discussion:

In the more general form, the $n=1$ case is known to be equivalent to the Invariance of Dimension Theorem, but for $n > 2$ it seems more geometrical that the spaces aren't homeomorphic.

Be careful trying to use a fact like "a homeomorphic image of a hyperplane disconnects Euclidean space into two pieces." This may be the key ingredient, but would need a point-set proof; I only know of algebraic ones.

There is a theorem of Moore, simplified by Bing, that there is no uncountable collection of copies of a simple triod (three arcs attached at a common vertix, i.e. a three-legged asterisk) contained in the plane. There is a Cantor Set of them along the spine of our three-finned 'plane.' However, the proofs of this theorem depend crucially on the Jordan Curve Theorem and its corollaries. It has higher-dimensional analogues, but they also depend on the higher-dimensional form of the Jordan Separation Theorem.

Similarly, evaluating the simplicial complex structures could differentiate them, but that is a lot of theory. If we look at neighborhoods on, say, $X_3 \times \mathbb{R}$ of the points along the 'spine' we will get connected neighborhoods whose boundaries are the theta curve. It can be shown that no connected set has boundary of a theta curve in $\mathbb{R}^2 = X_2 \times \mathbb{R}$ and this method also generalizes inductively, but again the details are not trivial and we are depending on at least the polygonal version of the Jordan Curve Theorem. Putting a circle into one of the legit-planes doesn't separate an $n$-finned plane; similarly (equivalently, as an aside), the complete bipartite graph $K_{3,3}$ embeds in it.

Yet another method would be counting the number of components that removing the spine creates, which could distinguish $X_2$ from the others but seems harder to generalize. This is one of the classic separation theorems, that a hyperplane cuts Euclidean space into two components, but we can't assume it.

And generalizing any of these methods (except for analyzing as complexes) to higher dimensions seem to have even more difficulties, for example by attaching three copies of threespace to a plane. There is also the possibility of using symmetry groups and relating those to the bundle structure, but again it would use a lot of theory.

What is the point-set way to prove this? Or is it equivalent to Jordan Curve Theorem, which would be really interesting?

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    $\begingroup$ The title doesn't seem to correspond to the question. $\endgroup$
    – freakish
    Commented Jun 1, 2018 at 7:13
  • $\begingroup$ Since $X_3$ is homeomorphic to a subset of $X_n$ for all $n\geq3$ you don't really need to modify the Moore-Bing argument as you suggested, but rather can apply it directly, showing that $X_n\times I$ cannot be homeomorphic to the plane. If you want more general non-homeomorphism results not mentioned in your title, you should ask such a question explicitly. $\endgroup$ Commented Jun 1, 2018 at 7:43
  • $\begingroup$ Also is $I$ an open interval? That's what $\mathbb{R}^2=X_2\times I$ suggests. Why aren't you using simply $\mathbb{R}$ instead of $I$? Typically $I$ is reserved for $[0,1]$. $\endgroup$
    – freakish
    Commented Jun 1, 2018 at 7:48
  • $\begingroup$ If $a_n\in X_n$ is the special glueing point and you can show that every homeomorphism $f:X_n\times I\to X_m\times I$ maps $f(\{a_n\}\times I)\subseteq \{a_m\}\times I$ then the thesis easily follows. I'm pretty sure this has to hold but I have trouble showing that formally. $\endgroup$
    – freakish
    Commented Jun 1, 2018 at 8:01
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    $\begingroup$ Another possible approach: isn't the one point compactification of $X_n\times\mathbb{R}$ a couple of spheres glued along a line segment (actually along a big circle minus a point)? This does not seem to have the homotopy type of the one point compactification of $\mathbb{R}^2$ being $S^2$. This can possibly be shown by analyzing its homology. $\endgroup$
    – freakish
    Commented Jun 1, 2018 at 8:41

4 Answers 4

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Since you mention the Jordan Curve Theorem, I will suppose that the machinery behind one of the modern proofs of the Jordan Curve Theorem --- singular homology --- is allowable. Anyway, this makes the problem an exercise in a first year algebraic topology course.

Let me denote $Y_n = X_n \times \mathbb{R}$. For any point $P = (p,t) \in Y_n$, it's pretty easy to compute the local singular homology group $H_2(Y_n,Y_n-P)$: it is isomorphic to $\mathbb{Z}^{n-1}$ if $p$ is the valence $n$ point of $X_n$; otherwise it is isomorphic to $\mathbb{Z}$.

But a homeomorphism preserves local homology groups: if $f : A \to B$ is a homeomorphism and if $a \in A$ and $b \in B$ then the induced relative homology homomorphism $f_* : H_i(A,A-a) \to H_i(B,B-b)$ is an isomorphism.

If $m \ne n$ and $m \ne 2$ then $Y_m$ has a point $P$ such that $H_2(Y_m,Y_m-P) = \mathbb{Z}^{m-1}$ which is not isomorphic to $H_2(Y_n,Y_n-Q)$ no matter what $Q$ is. Thus $Y_m$ and $Y_n$ are not homeomorphic.

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  • $\begingroup$ You had algebraic topology, with singular homologies, in the first year, as an undergrad? $\endgroup$
    – tomasz
    Commented Jun 2, 2018 at 21:03
  • $\begingroup$ Nope. I had it my first year in graduate school. $\endgroup$
    – Lee Mosher
    Commented Jun 2, 2018 at 21:24
  • $\begingroup$ Although I had a little taste of it in my senior year of college. $\endgroup$
    – Lee Mosher
    Commented Jun 2, 2018 at 21:24
  • $\begingroup$ Ok, that makes more sense. I would be curious where in the world they teach first year students singular homologies. $\endgroup$
    – tomasz
    Commented Jun 3, 2018 at 6:11
  • $\begingroup$ Yeah, when I grew up one didn't say "first year college", one said "freshman". $\endgroup$
    – Lee Mosher
    Commented Jun 3, 2018 at 16:08
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I could easily be wrong, but I don't think you need anything as fancy as JCT.

For a locally compact Hausdorff space $Y$ let $\omega Y = (Y \cup \{\infty\}, \infty)$, where $Y \cup \{\infty\}$ denotes the one point compactification. Then we have $$ \omega(X_n\times \mathbb{R}^d) \cong \omega X_n \wedge \omega(\mathbb{R}^d) \cong \omega X_n \wedge S^d \cong \Sigma^d \omega X_n $$ where $\cong$ means basepoint-preserving homeomorphism. Since $\omega X_n$ is homotopy-equivalent to a wedge sum of $n-1$ circles, its $d$-th reduced suspension should be homotopy-equivalent to a wedge sum of $n-1$ copies of $S^{d+1}$. Thus it should be possible to recover $n$ from $\pi_{d+1}(\omega(X_n \times \mathbb{R}^d))$. Since $\omega X_n$ is easy to triangulate, simplicial homology also seems a viable approach.

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You can use the Jordan Curve Theorem.

Note that a circle consisting of two halves on two different sheets of $X_n \times \mathbb{R}$ is a simple closed curve with a connected complement for $n>2$. This proves that $X_n \times \mathbb{R}$ can't be homeomorphic to $\mathbb{R}^2$.

This also works in higher dimensions, by applying the Jordan Brouwer separation theorem to a hypersphere constructed half on one sheet and half on another. The complement stays connected as long as there is a sheet other than the two containing the image of the hypersphere.

The theorems do a lot of heavy lifting, but it makes the rest of the proof trivial.

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You can use the idea that the continuous image of a connected set is connected.

We will take a look at the sets $X_{n}$ that you first described

Remove a small ball around the point where you have connected the half open intervals in U, let's call that point the origin in both U and V where a homeomorphism f:U -> V. With U consisting of m half open interval gluings and v consisting of k half open interval gluings are two such examples of the line segment gluings with $m,k>2$ described in your question and $m\neq k$. Take a Neighborhood around the origin and its image and look at U\N and V\f(N). For a small enogh neighborhood U\N has m connected components, and V\f(N) has either k connected components or 2 connected components depending on whether the origin in U maps to the origin in V or not. We can provide a contradiction in either case. If V\f(N) has fewer connected components then consider the map $f^{-1}$. Since the map is a bijection there must be a connected component in V\f(N) whose image is in two or more connected components in U\N. This contradicts the idea that the continuous image of a connected set is connected.

If U\N has fewer connected components then consider the map $f$. It is the same idea with the map $f$. Since the map is a bijection there must be a connected component in U\N whose image is in two or more connected components in V\f(N). This contradicts the idea that the continuous image of a connected set is connected.

With the hyperplane version you can use a more sophisticated version of the same idea.

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  • $\begingroup$ I agree that in the case of simple $n$-ods it's elementary, by counting the number of components the vertex cuts the space into. But I don't see how to easily generalize this. How do you prove that no copy of $\mathbb{R}^{n-1}$ disconnects $\mathbb{R}^n$ into at least three pieces without algebraic techniques? $\endgroup$ Commented Jan 30, 2021 at 18:06
  • $\begingroup$ @JohnSamples Your comment cuts right to the crux of the problem. Namely you agree that if I can replicate the argument that V\f(N) has either $k$ connected components or $2$ connected components, then the result will follow in the general case. However, the proof of that (fact, or necessary lema, etc.) where $f(N)$ is in one of the half planes or 'fins' bears a great similarity to a Jordan curve type result. Fortunately, I have many known 'elementary' analogue proofs of the Jordan curve theorem to draw from. Rather than cite them, I could bring over their main idea with a few minor tweaks. $\endgroup$ Commented Jan 31, 2021 at 0:58
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    $\begingroup$ @JohnSamples Since doing that feels like it might not fall inside the spirit of the problem, I will rather inform you of that fact and give you some references where you can see the originals. Elementary Analysis Proof: maths.ed.ac.uk/~v1ranick/jordan/tverberg.pdf A proof using graph theory: "The Jordan–Schönflies theorem and the classification of surfaces", American Mathematical Monthly, 99 (2): 116–130, doi:10.2307/2324180 $\endgroup$ Commented Jan 31, 2021 at 1:00
  • $\begingroup$ Thanks, hadn't seen this one before! $\endgroup$ Commented Feb 2, 2021 at 3:27

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