Let $f:X\longrightarrow Y$ be a function, $A,A_1,A_2$ subsets of $X$ and $B,B_1,B_2$ be subsets of $Y$1
Prove that $f(\lim\sup A_n)\subset \lim\sup f(A_n)$.
$\textbf{Proof:}$ Assume $y\in f(\lim\sup A_n).$ Then $\exists x\in\lim\sup A_n$ such that $y=f(x)$
I know my ultimate goal here is to show that $y\in\lim\sup f(A_n)$ but im not sure how to go about proceeding to the next step. What does it mean for the existence of $x\in\lim\sup A_n$? Please help on proceeding to the next step.
By definition of set-theoretic limit
$$\limsup A_n=\bigcap_{n\ge1}\bigcup_{j\ge n}A_j$$
If $y\in f(\limsup A_n)$, then $\exists x$, $\forall n,\exists j\ge n, x\in A_j$ and $y=f(x)$.
We want to show $y\in\limsup f(A_n)$ i.e. $y\in\bigcap_{n\ge1}\bigcup_{j\ge n}f(A_j)$ i.e. $\forall n$, $\exists j\ge n$, $\exists x\in A_j$, $y=f(x)$.
From the logic the former (stronger) implies the latter.