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I'm studying Gemignani's Elementary Topology. I need to determine that the open interval $(0,1)$ and the closed interval $[0,1]$ are homeomorphic or not when considered as subspaces of the real numbers with the absolute value topology.

My claim is that they're not. Suppose there was a homeomorphism $f$ from $(0,1)$ onto $[0,1]$. Then $f(c)=0$ for some $c\in (0,1)$. Now either $f([c,1))$ or $f((0,c])$ must contain $1$ otherwise $f$ won't be onto. Assume wlog that $f((0,c])$ contains $1$ then it follows by the intermediate value theorem that $f((0,c])=[0,1]$. But then $f$ cannot be one one, thus, a contradiction.

Are there better ways to prove that there is no homeomorphism without using the notions of compactness and connectedness? Also the author asks to find a "topological property" which one of the spaces has but which the other doesn't if the spaces are not homeomorphic. I'm not sure how to answer this using the proof that I gave.

Edit: Perhaps I got my answer to the last paragraph. If $X,\tau$ is a topological space and $Y,Z$ are subspace of $X$ which are homeomorphic and if every open subset of $Y$ is an open subset of $X$ then this must be true for $Z$ as well. If not, there must be an open subset $U$ of $Z$ which is not open in $X$ then the image/inverse image of $U$ under the homeomorphism must not be open in $Y$.

In this case, every open subset of $(0,1)$ is open in $\mathbb{R}$ but $[0,1]$ is open in the subspace topology of $[0,1]$ but is not open in $\mathbb{R}$. Please someone correct me if I'm wrong.

ashK
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    Your argument looks fine. The "topological property" comment in the book is probably a reference to either connectedness or compactness, so I am not sure why you want to avoid mentioning those. Note that the intermediate value theorem is really using the notion of connectedness even if it isn't phrased using that word. – Nate Eldredge Apr 21 '20 at 12:36
  • @NateEldredge I am trying to avoid those because the author hasn't really discussed those topics yet. – ashK Apr 21 '20 at 12:39
  • See also https://math.stackexchange.com/a/42310/589 – lhf Apr 21 '20 at 13:12

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A property that $[0,1]$ has that $(0,1)$ has not not, is compactness, indeed. It's not one you use in your argument, though.

$(0,1)$ has the property "every point of $X$ is a cutpoint" (where a cutpoint (of a connected space) is a point $x$ such that $X\setminus \{x\}$ is not connected).

$[0,1]$ does not have this property, as $0$ and $1$ are not cutpoints, so for $[0,1]$ we can use : $X$ has two non-cutpoints. For $[0,1)$ we can use "X has one non-cutpoint", and so we can distinguish them too. Connectedness-related properties seem the most natural way to go here.

Henno Brandsma
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