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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 $B_1$ ?

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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