I encountered an interesting statement:

Let $X$ and $Y$ be two topological spaces. Although there exist continuous bijections $f:X\to Y$ and $g:Y\to X$, they are still not necessarily homeomorphic.

One example is given as follows:

Set $X=\{n\in\mathbb{Z}\mid n<0\}\cup\bigcup_{k=0}^{\infty}{[2k,2k+1)}$ and $Y=X\cup\{1\}$. Define $f:X\to Y$ and $g:Y\to X$ as follows: \begin{equation*} f(x)=\begin{cases} x+1, & x\leq -2; \\ 1, & x=-1; \\ x, & x\geq 0. \end{cases}\quad{\rm and}\quad g(x)=\begin{cases} x, & x<0; \\ x/2, & x\in[0,1]; \\ (x-1)/2, & x\in[2,3); \\ x-2, & x\geq 4. \end{cases} \end{equation*} Apparently, both $f$ and $g$ are continuous bijections.

But why $X$ and $Y$ are not homeomorphic?

My reasoning is that $[0,1]$ is a connected component of $Y$, so it should be mapped to a connected component of $X$. However, there is no connected component of $X$ that is both compact and has the same cardinality as $[0,1]$, so $X$ and $Y$ are not homeomorphic.

I hope my previous reasoning is not wrong, but I wonder if there is any simpler reasoning. Any suggestion is highly welcomed.

  • 2
    $\begingroup$ Your argument is correct. You could also argue that $[0,1]$ is a connected component of $Y$ with two non-cut points, while $X$ has no such component. $\endgroup$ Jun 19, 2020 at 18:36
  • $\begingroup$ @BrianM.Scott This argument is nice. Thank you. $\endgroup$ Jun 19, 2020 at 18:52
  • $\begingroup$ In Euclidean subspace topology, $(a,1]$ is open in Y but not in X where $a>0$. $f^{-1}(a,1] = (a,1]$, thus f is not continuous? $\endgroup$
    – Nawaj
    Jun 19, 2020 at 19:13
  • $\begingroup$ @Nawaj No. If you assume $a\in(0,1)$, $f^{-1}((a,1])=\{-1\}\cup(a,1)$. This set is open in $X$. $\endgroup$ Jun 19, 2020 at 19:46

1 Answer 1


Suppose that there exists an homeomrphism $g:Y\rightarrow X$, $g([0,1])$ is an interval, implies that $g([0,1])\subset [2k,2k+1)$. Let $f$ be the inverse of $g$, $g([2k,2k+1)$ is an interval which contains $[0,1]$.

Suppose that $f([2k,2k+1)$ is not $[0,1]$, it implies that there exists $x\in f([2k,2k+1)$ which is in $\{n,n\in\mathbb{Z},n<0\}\cup_{p>0}[2p,2p+1)$. The segment $[0,x]$ is not contained in $Y$ this is in contradiction with the fact that $f([2k,2k+1)$ is connected. We deduce that $f([2k,2k+1))=[0,1]$ this is impossible since $[0,1]$ is compact but not $[2k,2k+1)$.

  • $\begingroup$ Thank you. Your reasoning is very clear, but I think you just explained why $[0,1]$ is mapped to $[2k,2k+1)$ for some $k$, which is very closed to my idea as the connected components in $X$ are either singletons of negative integer or half open intervals. $\endgroup$ Jun 19, 2020 at 18:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .