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I am looking for an example of an infinite metric space $X$ such that there exists a continuous bijection $f: X \to X$ which is not a homeomorphism. Please help. Thanks in advance.

Martin Sleziak
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  • It's easy to come up with continuous bijections $X \to Y$ which aren't homeomorphisms (i.e., not requiring the domain and codomain to be the same). I'm not sure about what you're asking for though... – Dustan Levenstein Mar 18 '16 at 11:43
  • @DustanLevenstein OP asked a completely different question – Anubhav Mukherjee Mar 18 '16 at 12:16
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    @Anubhav.K Your answer appears to work at a first glance, although it's not expressed in the language of metric spaces, so OP might find it a bit confusing. I'm not sure why you felt the need to respond to my comment though? I very clearly indicated that I wasn't sure about the answer to OP's question. – Dustan Levenstein Mar 18 '16 at 12:21
  • @DustanLevenstein I misjudge your comment, I thought you might confused the question...sorry – Anubhav Mukherjee Mar 18 '16 at 12:25
  • If you look at examples at some older questions (which did not ask for metric spaces, just for topological spaces), you will find out that some examples given there provide metrizables space. For example, this question or this question. – Martin Sleziak May 07 '16 at 10:01

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Consider $X= \sqcup_{n\in \mathbb Z} I_n$ (i.e disjoint union)where $I_n = [0,1) when \ \ n\leq 0$ and $I_n=S^1 when \ \ n\geq 1$.

Define $f: X\to X$ s.t $f|_{I_n} : I_n \to I_{n+1}$ is a homeomorphism for $n\neq 0$ ( since in that case $I_n,I_{n+1}$ are either both intervals or circles) and for $n=0$, $f|_{I_0} : I_0 \to I_1 = S^1$ is a continuous bijection (for example exponential map).

Then $f$ is a continuous bijection , but not a homeomorphism.

Anubhav Mukherjee
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Let $A=\left\{\frac1n:n\in\Bbb Z^+\right\}$, let $Y=\{0\}\cup A$, and let $X=\Bbb Z\times Y$. Define a metric $d$ on $X$ as follows:

$$d\big(\langle m,y_0\rangle,\langle n,y_2\rangle\big)=\begin{cases} 1,&\text{if }m\ne n\\ 1,&\text{if }m=n\le 0\text{ and }y_1\ne y_2\\ 0,&\text{if }m=n\le 0\text{ and }y_1=y_2\\ |y_1-y_2|,&\text{if }m=n>0 \end{cases}$$

If we let $Y_n=\{n\}\times Y$ for $n\in\Bbb Z$, this definition amounts to saying that $Y_n$ is a countably infinite space with the discrete metric if $n\le 0$, $Y_n$ is homeomorphic to $Y$ with the usual Euclidean metric if $n>0$, and if $p\in Y_m$, $q\in Y_n$, and $m\ne n$, then $d(p,q)=1$. I leave it to you to check that $d$ is a metric on $X$.

Now let $f:X\to X:\langle n,y\rangle\mapsto\langle n+1,y\rangle$. It’s not hard to verify that $f$ is a continuous bijection. To see that $f$ is not a homeomorphism, consider $f(\langle 0,0\rangle)$.

Brian M. Scott
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    This example has, in some sense, similar basic idea as the example from this answer (by Jim Belk; the other question asked for example of connected space rather than metrizable space). – Martin Sleziak May 07 '16 at 10:04
  • @Martin: I thought that there was a somewhat similar example around here somewhere, but I couldn’t remember enough to find it; thanks! – Brian M. Scott May 07 '16 at 10:13
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The Cantor set minus one point.

Indeed, in general let $u:U\to V$ is a non-open continuous bijection between topological spaces. Define $X=U\times\mathbf{N}^c\sqcup V\times\mathbf{N}$ (I assume $0\in\mathbf{N}$ and I mean the complement in $\mathbf{Z}$). Define $f(x,n)=(x,n+1)$ for $n\neq 0$ and $f(x,-1)=(u(x),0)$. Then $f$ is a non-open continuous permutation of $X$.

We can apply this to $V$ Cantor, $U$ Cantor minus one point. Indeed, fix a point $x\in U$ and let $U'=U\cup\{y\}$ be the 1-point compactification of $U$ and $V$ the quotient of $U'$ identifying $x$ and $y$. Then $V$ is a Cantor space, and the canonical map induced by inclusion $U\to V$ works.

Then $X$ is also homeomorphic to $U$. Indeed, first both $U\times\mathbf{N}$ and $V\times\mathbf{N}$ are homeomorphic to $U$, and hence so is their disjoint union $X$.

(I'm using that every metrizable, non-compact locally compact, non-empty, perfect and totally disconnected topological space $X$ is homeomorphic to $U$. Indeed, the 1-point compactification of $X$ then satisfies the same properties with non-compact replaced by compact, hence is a Cantor space, so $X$ is homeomorphic to a Cantor space minus a point, and this is unique, because the Cantor space is homogeneous under self-homeomorphisms, e.g., because it admits topological group structures.)

Added: For every $(X,f)$, $X$ topological space and $f$ bijective continuous non-open self map of $X$, we can obtain a another connected example taking the cone. So all examples here can be used to provide examples to the same question for restricted to connected spaces (2011 MSE post).

YCor
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