Equality between intersection of image of nested closed sets : $\bigcap_{n\in\mathbb{N}}f(F_{n}) = \bigcap_{n\in\mathbb{N}}\overline{f(F_{n})}$

Question

I'm trying to understand the following proof:

Theorem. Let $X$ and $Y$ be Hausdorff spaces. Let $f:X\rightarrow Y$ be a continuous mapping. Suppose that $\{F_{n}\}_{n\in\mathbb{N}}\subset X$ is a decreasing sequence of closed sets with compact intersection $K = \bigcap_{n\in\mathbb{N}}F_{n}.$ If for each open set $U$ in $X$, $K\subset U$ implies that there exists some $m\in\mathbb{N}$ such that $F_{m}\subset U,$ then $$f(K) = \bigcap_{n\in\mathbb{N}}f(F_{n}) = \bigcap_{n\in\mathbb{N}}\overline{f(F_{n})}.$$

Proof: For any $y\in\bigcap_{n\in\mathbb{N}}\overline{f(F_{n})}$, suppose that for each $x\in K$ there exists an open set $U_{x}$ containing $x$ such that $y\notin \overline{f(U_{x})}.$ The collection $\{U_{x}\}_{x\in K}$ is an open cover of $K.$ Then the union of the $U_{x}$ in the subcover gives an open set $U\supset K$ with $y\notin\overline{f(F_{n})}$ for $n$ large, a contradiction since $F_{n}\subset U$ for $n$ large. So take $x\in K$ with $y\in f(V)$ for every open $V$ containing $x.$ If $f(x)\neq y,$ then take disjoint open neighborhoods $W$ of $f(x)$ and $T$ for $y.$ Let $V = f^{-1}(W)$ to get a contradiction.

I understand every part of the proof except "So take $x\in K$ with $y\in f(V)$ for every open $V$ containing $x.$" Why $y\in\bigcap_{n\in\mathbb{N}}\overline{f(F_{n})}$ satisfies the previous affirmation? Because of the above $y\in\overline{f(V)}$ for each open set containg $x,$ but is not clear that must be $y\in f(V).$ Of course, for an element $y\in f(V)$ for each $V$ open set containg $x,$ the last part of the proof above give us the seeking contradiction finishing the argument.

Any kind of help is thanked in advance.

Answer 1

It has something to do with $Y$'s Hausdorffness. Let me fill in the details of the proof.

Since $K = \bigcap_{n\in\mathbb{N}}F_{n}$, we have $$f(K) =f(\bigcap_{n\in\mathbb{N}}F_{n}) \subseteq \bigcap_{n\in\mathbb{N}}f(F_{n}) \subseteq \bigcap_{n\in\mathbb{N}}\overline{f(F_{n})}$$ Therefore, we need to prove $$\bigcap_{n\in\mathbb{N}}\overline{f(F_{n})} \subseteq f(K)$$ Suppose that there exists a $y\in\bigcap_{n\in\mathbb{N}}\overline{f(F_{n})}$ such that $y \notin f(K)$, then for each $x \in K$, $f(x) \neq y$. Since $Y$ is Hausdorff, $f(x)$ and $y$ are separated by neighborhoods, thus there exists an open set $U_{x}$ containing $x$ such that $y\notin \overline{f(U_{x})}$ because $f$ is continuous. The collection $\{U_{x}\}_{x\in K}$ is an open cover of $K$, then the union of the $U_{x}$ in the finite subcover gives an open set $U=\bigcup_{i=1}^nU_{x_i}\supseteq K$ with $$y\notin \bigcup_{i=1}^n \overline{f(U_{x_i})}=\overline{\bigcup_{i=1}^n f(U_{x_i})}=\overline{f(\bigcup_{i=1}^n U_{x_i})} =\overline{f(U)} \supseteq \overline{f(F_{n})}$$ for $n$ large, a contradiction since $y\in \overline{f(F_{n})}$ for every $n \in \mathbb{N}$. This proves that $\bigcap_{n\in\mathbb{N}}\overline{f(F_{n})} \subseteq f(K)$. By the way, I don't think $X$ needs to be Hausdorff.