# $(\mathbb R, \tau_1) \not \cong (\mathbb R, \tau_2)$, where $\tau_1=\{(-n,n),n\in \Bbb N\}$ and $\tau_2=\{(-r,r),r\in \Bbb R\}$

In an exercice I'm asked to prove that $$(\mathbb R, \tau_1)$$, with $$\tau_1=\{\mathbb R,\emptyset\} \cup \{(-n,n),n\in \Bbb N\}$$ and $$(\mathbb R, \tau_2)$$, with $$\tau_2=\{\mathbb R,\emptyset\} \cup \{(-r,r),r\in \Bbb R\}$$ are not homeomorphic.

The textbook that I'm following said we can prove that 2 spaces are not homeomorphic if we find a topological property that is not preserved. At this point this is the list of all the topological properties that I've learned:

• $$T_0,T_1, T_2, T_3$$ and Regular spaces
• Satisfying the second axiom of countability
• Separable
• Discrete space
• Trivial space
• finite-closed topology
• countable-closed topology
• connectedness

I checked this properties and I found that, both spaces are not $$T_0,T_1, T_2, T_3$$ or Regular. Both do Satisfying the second axiom of countability and both are separable. Neither of the spaces are Discrete, trivial, finite-closed or countable-closed and both are connected.

Did I made any mistake and did I checked some property wrong? Or am I correct and there's another way of proving this?

If they were homeomorphic, the homeomorphism would give rise to a bijection between $$\tau_1$$ and $$\tau_2$$. This, however, is impossible, because $$\tau_1$$ is countable, and $$\tau_2$$ is uncountable.

You can also use the $$T_0$$ property, though not directly. For each $$x\in\Bbb R\setminus\{0\}$$ and $$i\in\{1,2\}$$ let $$A_i(x)$$ be the set of $$y\in\Bbb R\setminus\{x\}$$ such that no $$U\in\tau_i$$ contains exactly one of $$x$$ and $$y$$. It’s not hard to see that $$A_2(x)=\{-x\}$$, while $$A_1(x)$$ is infinite. If you’re interested, you can try showing that if $$h$$ were a homeomorphism from $$\langle\Bbb R,\tau_2\rangle$$ to $$\langle\Bbb R,\tau_1\rangle$$,

$$\left|A_1\big(h(x)\big)\right|=|A_2(x)|$$

would have to hold for every $$x\in\Bbb R\setminus\{0\}$$, and it cannot.

• Why would that imply that there is a bijection between $\tau_1$ and $\tau_2$? I've never heard of this Sep 2, 2020 at 20:54
• I do realize that that implies that there exists a function $f:\tau_1 \to \tau_2$, but how to we know that it is a bijection? Sep 2, 2020 at 20:56
• @EduardoMagalhães: Have you tried to prove the statement? There’s only one reasonable thing to try, and it works. Let $h:\Bbb R\to\Bbb R$ be the homeomorphism, and define $f:\tau_1\to\tau_2:U\mapsto h[U]$. Sep 2, 2020 at 21:00

Suppose that $$f : (\mathbb R, \tau_1) \to (\mathbb R, \tau_2)$$ is an homeomorphism.

$$f[(-1,1)]$$ has to be an open subset of $$(\mathbb R, \tau_2)$$, i.e. is equal to $$(-r_1, r_1)$$ for $$r_1>0$$.

But then $$S=f^{-1}[(-r_1/2, r_1/2)]$$ should be an open subset of $$(\mathbb R, \tau_1)$$ strictly included in $$(-1,1)$$ which can't be as $$S$$ is not empty.