Definition $1$ : A function $f:X \to Y$ is continuous iff for every open set $E \subset Y$ , $f^{-1}(E)$ is an open set in $X$ .
Definition $2$ : A function $f: X \to Y$ is continuous iff for every point $x \in X$ and any open set $E \subset Y$ containing $f(x)$ , there exist an open set $F \subset X$ containing $x$ and $f(F)\subset E$
My attempt:
$(1)$ If $f$ is continuous by definition $1$ , then it is continous by difinition $2$ . Set $F=f^{-1}(E)$ then we can get the desired conclusion .
$(2)$ We accept axiom of choice , then if $f$ is continuous by definition $2$ , it is also continuous by definition $1$ . To see this , for every open set $E \subset Y$ , we can write $$E=\bigcup_{i\in I}y_i$$ And $$f^{-1}(E)=\bigcup_{i\in I} f^{-1}(y_i)$$ WLOG , we can assume for each $y_i$ we can find $G_i=\bigcup_{j\in J}x_j \subset X$ such that $f^{-1}(y_i)=G_i$ , and $$f^{-1}(E)=\bigcup_{i \in I} G_i=\bigcup_{k\in K}x_k$$ since $E$ is an open set contains each $f(x_k)$ , by definition of continuous we can find open set $K_k \subset X$ such that $K_k$ contains $x_k$ and $f(K_k)\subset E$. We have $$\bigcup_{k \in K}K_k \subset f^{-1}(E) $$ Also , since $x_k\in K_k$ we have $$f^{-1}(E)=\bigcup_{k\in K}x_k \subset \bigcup_{k \in K}K_k$$ Then we finished the proof .
My question :
① In my proof of $(2)$ , I have used axiom of choice to associate each $x_k$ with an open set $K_k$ . Can we have some other proof without using axiom of choice ?
② I know that two definition above did not require axiom of choice , so if we proved two definition are the same in ZFC , can we assert that they are also the same in ZF ?
Here's a proof without the axiom of choice, for the implication from Definition 2 to Definition 1. Suppose $f$ is continuous according to Definition 2, and consider any open set $E\subseteq Y$. I need to show that $f^{-1}(E)$ is open in $X$. Let $U$ be the union of all the open subsets of $f^{-1}(E)$. Then $U$ is open (being a union of open sets), so if I show $f^{-1}(E)=U$, I'll be done. Furthermore, the definition of $U$ immediately gives $U\subseteq f^{-1}(E)$, so all I need to do is prove $f^{-1}(E)\subseteq U$.
Consider any point $x\in f^{-1}(E)$. So $f(x)\in E$ and, since $f$ is continuous according to Definition 2, there's an open $F$ in $X$ with $x\in F$ and $f(F)\subseteq E$. The last of these properties of $F$ means that $F\subseteq f^{-1}(E)$. Since $F$ is open, it's one of the sets whose union I used as my definition of $U$, so $F\subseteq U$. Finally, since $x\in F$, I've shown $x\in U$. This completes the proof that every element $x$ of $f^{-1}(E)$ is also in $U$, as required.