I am searching for a two way embedding between two non-homeomorphic spaces. In other words, I want two non-homeomorphic spaces such that $X$ is embedded in $Y$ and $Y$ is embedded in $X$.

  • Recall given two spaces $(X, \mathfrak{T})$ and $(Y, \mathfrak{J}), A \subseteq Y$, if there is a homeomorphism $f: X \to A$ then we call $X$ embedded in $Y$

My initial attempt is to create a space with copies in each set like:

Let $X = (0,1) \cup \{2\}$, and $Y = (0,1)$

The two spaces are obviously not homeomorphic. Then take the identitify function $id(Y) = A \subset X$, then $Y$ is embedded in $X$. But this way $X$ is not embedded in $Y$, since we cannot map $\{2\}$ into $Y$ through a bijective function.

What modification can I make to the spaces to get the two way embedding without them being homeomorphic.

  • $\begingroup$ You certainly can map $(0,1) \cup \{2\}$ homeomorphically into $(0,1)$, e.g. by $t \to t/3$. $\endgroup$ – Robert Israel Jun 15 '16 at 0:10
  • $\begingroup$ You cannot do this with finite sets, because injection from $X$ to $Y$ and vice-versa will imply that $\#X=\#Y$. $\endgroup$ – Hmm. Jun 15 '16 at 0:11
  • $\begingroup$ @RobertIsrael Oh I didn't think of this...how can you see that it is a homeomorphism $\endgroup$ – Shamisen Expert Jun 15 '16 at 0:11
  • $\begingroup$ Let $C$ be middle-thirds Cantor set, and let $X=C\setminus\{1\}$. $C$ is compact, and $X$ isn’t, so the two spaces are not homeomorphic. The identity function embeds $X$ in $C$, and the function $f(x)=\frac{x}3$ embeds $C$ in $X$. $\endgroup$ – Brian M. Scott Jun 15 '16 at 20:33

You can consider $[0,1]$ and $\mathbb{R}$.

  • $\begingroup$ $\mathbb{R}$ is homeomorphic to $(-1,1)$, compose this with the canonical inclusion. $\endgroup$ – Hmm. Jun 15 '16 at 0:07

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