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?


2 Answers 2


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 $\langle\Bbb R,\tau_1\rangle$,


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

  • $\begingroup$ Why would that imply that there is a bijection between $\tau_1$ and $\tau_2$? I've never heard of this $\endgroup$ Sep 2, 2020 at 20:54
  • $\begingroup$ 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? $\endgroup$ Sep 2, 2020 at 20:56
  • $\begingroup$ @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 $f:\tau_1\to\tau_2:U\mapsto h[U]$. $\endgroup$ 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.


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