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Let $$ X = \{(x,y) \in \mathbb{R}^2: 0 \le x \le 1\} \quad Y = \{(x,y) \in \mathbb{R}^2: 0 \le x\}. $$

Are $X$ and $Y$ homeomorphic?

I first thought no because their boundaries are not homeomorphic however the argument is flawed because the theorem I want to use is:

If $X \cong Y$ then for $A \subseteq X$ $$ \partial_X (A) \cong \partial_Y(f(A)). $$ So applying this theorem would yield $$ \partial_X(X) \cong \partial_Y(Y) \implies \emptyset \cong \emptyset $$ which is not helpful. Any hints?

jfab
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    Try looking at $X\setminus\mathrm{interior}(X)$ and $Y\setminus\mathrm{interior}(Y)$. Any homeomorphism should also be a homeomorphism on these sets, but these sets are not the same (One is path-connected, the other is not) – student91 Dec 21 '22 at 14:58
  • @student91 Again, $\operatorname{int}_X(X) = X$ because every set is open in itself. It's the same problem with the boundary case. Yo do not have an homeomorfism from $\mathbb{R}^2$ to itself. – jfab Dec 21 '22 at 16:41
  • @student91: What do you mean by "interior(X)"? The (topological) interior of $X$ as a subspace of $\mathbb{R}^2$? Or of $X$ itself? Or, as assumed in the answer of jfab, $X \setminus \delta X$, where $\delta X$ is the boundary of $X$ in the sense of algebraic topology? Please note that only the last option provides a valid proof. However, the background to this option is not as trivial as for the first two options. But perhaps you had this in mind, and my comment is irrelevant? – Ulli Dec 21 '22 at 21:03
  • Yes maybe my comment is a bit stupid. I was in the back of my head thinking of "interior" and "boundary" in terms of how one would do for manifolds with boundary, so "interior" point = one which has a nbh isomorphic to open disc, "boundary" point = one which has a nbh isomorphic to half-disc, but I guess this does not work here. – student91 Dec 21 '22 at 21:16
  • hmmm I guess it does work after all, $\mathrm{Int}(X):={x\in X\colon\exists f\colon D^2\to X,\ f^{-1}(x)\neq\varnothing}$ is a very sensible definition of interior that does help with this question. – student91 Dec 21 '22 at 21:27
  • Yes, sure, this will work. I just wanted to emphasise that it's not just considering interiors in the sense of a subspace of $\mathbb R^2$. And, hence, you need, at least a little bit, some machinery of algebraic topology. – Ulli Dec 21 '22 at 22:16

2 Answers2

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To proof it without using any algebraic topology, the same as here works:

Call a space $X$ compactly connected, if for each compact subset $A$ of $X$ there is compact subset $B$ of $X$, such that $A \subseteq B$ and $X \setminus B$ is connected.
$Y = [0,\infty) \times \mathbb{R}$ is compactly connected, since each compact subset is contained in a compact rectangle, such that one of its edges lies on the $y$-axis. The remainder consists of three (infinite, non-closed) "rectangles", where one intersects the other two. Hence it is connected.
$X = [0,1] \times \mathbb{R}$ is not compactly connected: Let $A := [0,1] \times \{0\}$. Then for any compact $B$ with $A \subset B$, $X \setminus B$ is not connected.

Ulli
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Since $$ X = [0,1] \times \mathbb{R} \cong [0,1] \times (0,1) \quad Y = [0,\infty) \times \mathbb{R} \cong [0,1) \times (0,1) $$ it suffices to show that the latter are not homeomorphic and this has been asked in Homeomorphisms in $\mathbb{R}^2$.

I believe the accepted answer is incorrect because of the same reason I mentioned, taking boundaries in $\mathbb{R}^2$ rather than on $X$ and on $Y$, however the comment to that answer appears to be correct (defining the "boundary" as the set of points so that $X \setminus \{p\}$ is simply connected).

jfab
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