# How to Show that $X \times \mathbb{R}$ is not Homeomorphic to $\mathbb{R}^2$

This is a geometrically obvious statement, but I only just realized that an explicit proof might not be entirely trivial. I was curious to know if anybody knows an elementary proof of this fact, and if not, maybe a slick one. The problem is actually a bit more general than the title suggests.

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 $\mathbb{R}$.

There is a theorem of Moore, simplified by Bing, that there is no uncountable collection of copies of $X_3$ contained in the plane, which distinguishes $\mathbb{R}^2 = X_2 \times \mathbb{R}$ from $X_3 \times \mathbb{R}$ (and by inclusion, from $X_n \times \mathbb{R}$ for all $n \geq 3$). The proof can then be modified inductively to show that there is no uncountable collection of $X_{n+1}$ in $X_n \times \mathbb{R}$ but the details are a bit complicated. However, the proofs of this theorem depend crucially on the Jordan Curve Theorem and its corollaries.

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.

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.

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 way to prove this, preferably in a way that generalizes to the higher dimensional analogues, that is as simple as it is obvious? Or is it equivalent to Jordan Curve Theorem, which would be really interesting?

• The title doesn't seem to correspond to the question. – freakish Jun 1 '18 at 7:13
• 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. – Mikhail Katz Jun 1 '18 at 7:43
• 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 $[0,1]$. – freakish Jun 1 '18 at 7:48
• 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. – freakish Jun 1 '18 at 8:01
• 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. – freakish Jun 1 '18 at 8:41

## 2 Answers

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.

• You had algebraic topology, with singular homologies, in the first year, as an undergrad? – tomasz Jun 2 '18 at 21:03
• Nope. I had it my first year in graduate school. – Lee Mosher Jun 2 '18 at 21:24
• Although I had a little taste of it in my senior year of college. – Lee Mosher Jun 2 '18 at 21:24
• Ok, that makes more sense. I would be curious where in the world they teach first year students singular homologies. – tomasz Jun 3 '18 at 6:11
• Yeah, when I grew up one didn't say "first year college", one said "freshman". – Lee Mosher Jun 3 '18 at 16:08

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.