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My question is the following:

Do there exist two non-homeomorphic topological spaces $X$ and $Y$ such that there are embeddings $f : X \hookrightarrow Y$, $g : Y \hookrightarrow X$, with both $f(X)$ dense in $Y$ and $g(Y)$ dense in $X$?

It is easy to see that the answer is positive if one drops the requirement that the image be dense for one of the embeddings. For example, let $X = [0, 1[$, $Y = [0, 1]$, $f : X \rightarrow Y$ the inclusion, $g : Y \rightarrow X$ defined by $g(y) = \frac{1}{2} y$ for any $y \in Y$. It is clear that $f$ and $g$ are embeddings, and $f(X)$ is dense in $Y$, but $g(Y)$ is not dense in $X$.

However, I can't figure out an example of the stronger situation above, and I wouldn't know how to prove that in such a case the two spaces must in fact be homeomorphic.

Thanks in advance.

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2 Answers

up vote 11 down vote accepted
+50

Here is an example with a natural topology: $X$ is an open unit disk in the plane, and $Y$ is the same disk punctured at the origin. Here $Y$ is obviously embedded in $X$. If one removes one of the radii of $Y$ (i.e. interval connecting the origin to the boundary circle), one obtains an open set homeomorphic to the disk $X$.

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Update: Actually not an answer, just related idea.

Note that by composing $f$ and $g$ your question is stronger that being able to find non-trivial embedding of $X$ to itself with dense image or being able to find a nontrivial dense subspace homeomorphic to whole space. To this there exists at least this trivial example: Inclusion of (infinite) subset of the same cardinality as the whole space, with indiscrete topology so every non-empty subspace is dense.

Another example: Any non-trivial subspace (e.g. remove a point) of rational numbers without isolated points is homeomorphic to rational numbers.

Update: So spaces that contain non-trivial dense copy of itself include: any indiscrete space, $\mathbb{Q}$, $\mathbb{R^n}$ (by @user72694) and any topological sum of those.

Actually, $\mathbb{Q}$ is an anti-example to your question in the following sense: For $X = \mathbb{Q}$ there is no $Y, f, g$ such that your condition holds since $\mathbb{Q}$ is not densely embeddable to any its subspace with isolated points and any its subspace without isolated points is homeomorphic to it.

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Of course any answer to my question provides an example for your statement, but I can't see why the converse should hold. If I find such an embedding of X, what should I choose as Y? –  Luca Bressan Aug 11 '13 at 17:42
    
Just take $Y = X$ and the same embedding. –  user87690 Aug 11 '13 at 17:52
    
But then they would be homeomorphic, contrary to the assumption that they are not. –  Luca Bressan Aug 11 '13 at 17:55
    
@Luca: Oh, you're right. My bad. Just take my “answer” as related idea. –  user87690 Aug 11 '13 at 18:05
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