If X satisfies the first or second countability axiom then $F(X)$ satisfies the same condition.

Question

I have 2 questions related to the proof given here,

1. In I, why to propose $\{F(B_i) \cap F(X)\}_{i \in I}$ as a basis for $f(p)$ when $\{B_i\}_{i \in I}$ is a basis for $p$? That is what we want to prove so why to give it? We are supposed to construct the base.

2. In II, at the second-to-last line why this $\{B_i \cap F(X)\}_{i \in I}$? Should not be $\{F(B_i )\}_{i \in I}$?

Let $F: X \rightarrow Y$ be a continuous open map. Show that if X satisfies the first or second countability axiom then $F(X)$ satisfies the same condition.

Proof made by math_at_you user in this post Second Countability Proof.

We have an open, continuos map $F: X \rightarrow Y$. We'll use $I$ as a countable indexing set.

I. Suppose $X$ is first countable. Then, for each $p \in X$ there exists a countable collection of neighbourhoods of $p$, $\{B_i\}_{i \in I}$, such that for any neighbourhood $U_p$ of $p$, there's a $j \in I$ for which $B_j \subseteq U_p$.

Given that $F$ is an open map, we'll propose $\{F(B_i) \cap F(X)\}_{i \in I}$ as a basis for $f(p)$ when $\{B_i\}_{i \in I}$ is a basis for $p$. Since $F(B_i) = F(B_i) \cap F(X)$, $F(B_i)$ will be open in both $Y$ and $F(X)$ for any $i$. Take a point $f(p) \in F(X)$. Let $V_{f(p)} = V \cap F(X)$ be an arbitrary neighbourhood of $f(p)$ where $V$ is open in $Y$ (Also, since $F$ is an open map, $F(X)$ is open, so $V_{f(p)}$ is open in $Y$ too). Then, $F^{-1}(V_{f(p)})$ is a neighbourhood of $p$, because $F$ is continuous. As such, there's a $j \in I$ for which $p \in B_j \subseteq F^{-1}(V_{f(p)})$. Then $f(p) \in F(B_j) \subseteq F(F^{-1}(V_{f(p)})) = V_{f(p)} \cap F(X) = V_{f(p)}$; this proves our collection is a basis for $f(p)$, which proves $F(X)$ is first countable.

II. Suppose $X$ is second countable. Then there exists a countable basis for $X$, namely $\{B_i\}_{i \in I}$. Take $U$ open in $F(X)$. Then $U = V \cap F(X)$, where $V$ is open in $Y$, and $F^{-1}(U) = F^{-1}(V) \cap F^{-1}(F(X)) = F^{-1}(V)$, which is open because $F$ is continuous; therefore $F^{-1}(U)$ is open. Then, there exists $J \subseteq I$ such that $\bigcup_{i \in J} B_i = F^{-1}(U)$. From here it follows that $F(\bigcup_{i \in J} B_i) = F(F^{-1}(U))$. Since $F(\bigcup_{i \in J} B_i) = \bigcup_{i \in J} F(B_i)$ and $F(F^{-1}(U)) = U \cap F(X)$, we get $\bigcup_{i \in J} F(B_i) = U \cap F(X) = U$. Because $F$ is an open map, $F(B_i)$ is open in $Y$ for any $i$, and because $F(B_i) \cap F(X) = F(B_i)$, $F(B_i)$ is also open in $F(X)$. We have now proven that any open set $U$ in $F(X)$ is a union of the sets from $\{B_i \cap F(X)\}_{i \in I}$. Therefore, this collection is a basis for $F(X)$, which proves $F(X)$ is second countable.

Answer 1

Intersecting with F(X), though redundant, clarifies that $F(B_i)$ is open within the F(X) subspace. An alternative theorem would take F as surjective.

In II you found a mistake.