I was thinking about the following question:
If $p: X \rightarrow Y $ a continuous, closed, and surjective map with the property that for each $y \in Y$ we have $p^{-1}(\{y\})$ is compact. Prove that if $X$ is second countable then so is $Y$. Assume $\{B_i\}_{i=1}^{\infty}$ is the countable basis of $X$.
There is a hint for this question. as following:
For each finite set $J\subseteq \mathbb{N}$ make $U_J$ the union of sets of $p^{-1}(W)$, $W$ open in $Y$, and with the property that $p^{-1}(W)$ is in the $\cup_{i\in J}B_i$.
The only problem that I have with the proof is the set $\{U_J\}$ is countable. I think it should have the same cardinality of $P(\mathbb{Z}_{+})$, the power set of $\mathbb{Z}_{+}$.
I will be thankful for any helps.
The set of finite subsets of $\Bbb Z^+$ is countable, like $\Bbb Z^+$ itself is (the singletons are in obvious 1-1 correspondence with $\Bbb Z^+$, the two-point sets are an image of $(\Bbb Z^+)^2$, etc. and so for all sets of size $n$. All finite sets are just a union of these countably many countable sets, see Munkres 7.5 and 7.6).