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Are there compact subsets $A,B \subset \mathbb{R^2}$ with $A$ not homeomorphic to $B$ but $A \times [0,1]$ homeomorphic to $B \times [0,1]$?

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Yes. Consider the two sets below.

The desired sets

Edit: Jon, let me explain in more detail why the two sets (call them $A$ and $B$, respectively) are not homeomorphic. I find in practice that showing two sets to not be homeomorphic is a bit cumbersome. I didn't fill out all of the details for that reason, leaving them to you; let me know if you would like more explanation for anything.

Suppose, seeking a contradiction, that $h:A\to B$ is a homeomorphism. Let $\alpha:[0,1]\to B$ be a path between the two points where the crosses intersect the edge of the square and let $\beta:[0,1]\to B$ be a path between the two points where the sticks intersect the edge. Choose $\alpha$ and $\beta$ so that their images do not intersect, that is, so that $\alpha([0,1])\cap\beta([0,1])=\varnothing$. Now consider the paths $\alpha'=h^{-1}\circ\alpha:[0,1]\to A$ and $\beta'=h^{-1}\circ\beta:[0,1]\to A$. It is still true that $\alpha'$ is a path between the intersections of the crosses and the squares, and that $\beta$ is a path between the intersections of the sticks and the squares (why?). But the images of $\alpha'$ and $\beta'$ must intersect (why?), and that is a contradiction.

As for your second question -- why $A'\times[0,1]$ and $B'\times[0,1]$ are not homeomorphic -- let me give you some intuition. The sets $A\times[0,1]$ and $B\times[0,1]$ are homeomorphic because you can move the "branches" across the top of the cube contained in $A\times[0,1]$. But the space $A'\times[0,1]$ is $A\times[0,1]$ with the interior, top, and bottom of the cube removed -- there is no more "top" to move the branches across, and a homeomorphism won't work.

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Why are those sets not homeomorphic? – Barry Apr 21 '14 at 23:45
@Jin A homeomorphism would preserve the order of cross, line, cross, line around the outside, but the second subset doesn't have that order. – blue Apr 21 '14 at 23:53
@sea turtles Why would a homeomorphism preserve the order? – Barry Apr 22 '14 at 1:27
@Jin It should be intuitively obvious that no amount of warping can change the structure like that without also breaking it. But if you're interested in a sketch of a proof, here's a hint: consider parametrizing the boundary of the square... – blue Apr 22 '14 at 1:29
In three dimensions you can interchange the "branches" by moving them around, but in two dimensions it is impossible. – user134824 Apr 23 '14 at 0:42

This is an addendum to the nice example below.

It was proven by K. Borsuk (Theorem 16 in the paper "On the decomposition of a locally connected compactum into Cartesian product of a curve and a manifold", Fund. Math. 40, (1953) 140–159.) that if X, Y are locally connected compacta of dimension at most 1, and their products with interval are homeomorphic then X is homeomorphic to Y. In fact, instead of an interval, one can use any topological manifold with or without boundary.

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