Let $(X,\tau_1)$ and $(X,\tau_2)$ be topological spaces and $\tau_2$ finer than $\tau_1$. Prove if $\tau_2$ is second-countable then $\tau_1$ is also second-countable.

My try:
Let $B=\{B_n,n\in \mathbb{N}\}$ be countable basis for $\tau_2$. Let's prove that $B_1=B\cap \tau_1$ is countable basis for $\tau_1$. It's countable because it's subset of countable set.

And now I'm stuck, because I don't know how to prove it's basis for $\tau_1$. I tried to prove it like this: Let $U\in\tau_1\subset\tau_2$ that means there is $B'\subseteq B$ so $U=\cup B'$. What next? How to show it's also union of elements of $B_1$ ?

  • $\begingroup$ This doesn't make sense (indiscrete topology is counterexample my construction of $B_1$ is not good)... Maybe $B_1$ is set whose elements are finite unions of $B$? Would that be countable? Basis? $\endgroup$ – Meow May 14 '13 at 14:18
  • $\begingroup$ I find both counterexample and part from solution for one final exam online which states:"The second countability is preserved in coarser topologies" here- beginig of last page $\endgroup$ – Meow May 14 '13 at 14:42
  • $\begingroup$ That's problem from my final exam, and I find it on that link online too =D I'm confused. $\endgroup$ – Meow May 14 '13 at 14:43

Will a counterexample do? Let $\tau_2$ be the discrete topology on $\mathbb{N}$, and suppose $\tau_1$ is a non-first-countable (equivalently, non-second-countable) topology on $\mathbb{N}$. Since $\tau_2$ is second-countable and finer than $\tau_1$, this will be a counterexample to what you wanted to prove. So all we have to do is find a non-first-countable topology on $\mathbb{N}$. Here are two ways to do that.

I. Let $\mathcal{U}$ be a uniform ultrafilter on $\mathbb{N}$. Call a set $X\subseteq\mathbb{N}$ open if either $1\notin X$ or else $X\in\mathcal{U}$.

II. For $X\subseteq\mathbb{N}$ let $d(X)$ be the asymptotic density of $X$. Call a set X open if either $1\notin X$ or else $d(X)=1$.

In words: we make the topology discrete except at one point. For the neighborhood system of the special point we take some filter which is not countably generated.

  • $\begingroup$ I don't understand this example =D [I must confess... I tried :P] But I found counterexample with standar and cofinite topology on $\mathbb{R}$- standard is second countable and cofinite is not. Is that ok? $\endgroup$ – Meow May 14 '13 at 14:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.