I am looking at the $2$ subspaces $X=L\cup(\{0\}\times\mathbb N)$ and $Y=L\cup(\{0\}\times\mathbb Z)$ where $L:=\{(\frac{1}{n},y):n\in \mathbb N, y\in \mathbb R\}$.
I wondering if these $2$ spaces are homeomorphic. The natural map is map $\mathbb N\to\mathbb Z$ by $1\to 0, 2\to1,3\to-1....$ and $L$ to $L$ by identity map. But this map is not even continuous as $\{(0,3)\}\cup (B_r(0,-1)\cap L)$ is not open.
Any help will be appreciated.
I understand $\mathbb{N}=\{1, 2, 3,\dots\}$ if $0$ should be included, then one can do the obvious modifications.
We are going to show that $X$ and $Y$ are not homeomorphic. We will prove this by contradiction. Assume there exists some homeomorphism $f: X \rightarrow Y$. As homeomorphisms map connected components to connected components, we find that there exists a bijection $h: \mathbb{N} \rightarrow \mathbb{N}$ and a family of homeomorphisms $g_n: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(1/n,x)=(1/h(n), g_n(x))$. Let me also introduce the function $\varphi: \mathbb{N} \rightarrow \mathbb{Z}$, such that $f(0,m)=(0,\varphi(m))$.
As $g_n$ are homeomorphism, they are monotone. Wlog assume that infinitely many of them are monotone increasing (the decreasing case is very much that same). We now want to show that $\varphi$ has its minimum at $\varphi(1)$ (in fact $\varphi$ is increasing). Note that this gives a contradiction as $(0,\varphi(1)-1)$ will not be in the image of $f$.
Let $m\in \mathbb{N}_{>1}$. By continuity there exists $\delta>0$ such that for all $ 1/n < \delta$ holds $$ \vert f(1/n,m) - f(0,m) \vert <1/4, \quad \vert f(1/n,1) - f(0,1) \vert <1/4 $$ However, this means $$ \vert g_n(m) - \varphi(m) \vert \leq \vert (1/h(n), g_n(m)) - (0, \varphi(m)) \vert = \vert f(1/n, m) - f(0,m) \vert < 1/4$$ and the same way $$ \vert g_n(1) - \varphi(1) \vert < 1/4. $$ However, as there are infinitely many $n$ such that $g_n$ is increasing, we can also pick such an $n$ with $1/n<\delta$ and we get $$ \varphi(m) = \varphi(m) - g_n(m) + g_n(m) > g_n(m) - 1/4 > g_n(1)-1/4 = g_n(1)-\varphi(1)+ \varphi(1)-1/4 > \varphi(1)-1/2. $$ As $\varphi$ takes values in the integers, this implies $\varphi(1)<\varphi(m)$.
Really, this is just a complicated way of saying that on the lines the map is monotone and by continuity we can push this to the boundary. However, there is no bijective order-preserving map $\mathbb{N} \rightarrow \mathbb{Z}$.