Why these two topological spaces are not homeomorphic

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.

• 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. Jun 19, 2020 at 18:36
• @BrianM.Scott This argument is nice. Thank you. Jun 19, 2020 at 18:52
• 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? Jun 19, 2020 at 19:13
• @Nawaj No. If you assume $a\in(0,1)$, $f^{-1}((a,1])=\{-1\}\cup(a,1)$. This set is open in $X$. Jun 19, 2020 at 19:46

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 $$[2k,2k+1)$$.
• 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. Jun 19, 2020 at 18:59