Continuous image of the intersection of decreasing sets in a compact space Suppose $B_{\epsilon}$ are closed subsets of a compact space and $B_{\epsilon} \supset B_{\epsilon'} \quad \forall \epsilon > \epsilon'$.  Furthermore, $B_0 = \bigcap_{\epsilon>0} B_{\epsilon}$.  For a continuous function $f$ can we conclude that
$$f(B_0) = \bigcap_{\epsilon>0} f(B_{\epsilon})?$$
I believe the answer to be yes.  It seems this should be a well-known property---I'm having trouble finding a reference.
 A: Suppose that $f\colon X\to Y$, then we assume that $X$ is compact and $f$ is continuous, however we need to assume that $Y$ is $T_1$ or that $X$ is Hausdorff.
First observe that since $B_\epsilon$ is closed it is compact too. Second we observe that the continuous image of a compact set is compact. (These two facts are true for all compact spaces, not only to Hausdorff spaces)
Since we also took the $B_\epsilon$ descending if the intersection is empty then there is $\epsilon$ such that $B_\epsilon=\varnothing$, and so $f(B_\epsilon)=\varnothing$ and the conclusion follows.
If the intersection is non-empty then we are in a situation where $f(B_\epsilon)$ form a decreasing chain of compact non-empty sets. The intersection cannot be empty, since by compactness we would have to have an empty set within the intersected family, which means $f(B_\epsilon)=\varnothing$, in contradiction to our assumption that we are in the case that $B_\epsilon\neq\varnothing$.
Furthermore, it is trivial that $f(B_0)\subseteq\bigcap f(B_\epsilon)$. So we only have to show the other direction. Suppose $y\in\bigcap f(B_\epsilon)$, then $y\in f(B_\epsilon)$ for all $\epsilon$, let $A=f^{-1}(y)$, this is a closed set, let $A_\epsilon=A\cap B_\epsilon$, this is a decreasing family of closed sets in a compact space, so by a similar argument as above we have that the intersection cannot be empty and must be a subset of $B_0$, therefore $y\in f(B_0)$ as wanted.

To see that we have to have $T_1$ in our assumptions, consider $\{0,1\}$ with the topology $\{\varnothing,\{1\},\{0,1\}\}$. Note that $1$ is dense and that $0$ is an accumulation point of $\{1\}$.
Consider the space $X=\mathbb N\cup\{\infty\}$ and the topology generated by initial segments of $\mathbb N$. So $U$ is open in $X$ if and only if $U=\varnothing$ or $U\cap\mathbb N=\{k\in\mathbb N\mid k<n\}$ for some $n\in\mathbb N$, and if $U\neq X$ then $\infty\notin U$.
Observe that $X$ is compact since if we cover $X$ by open sets we had to include $X$ itself in the covering, since the only open set which contains $\infty$ is $X$.
Now let $f\colon X\to\{0,1\}$ be defined as $f(n)=1$ for $n\in\mathbb N$ and $f(\infty)=0$. This is continuous since the preimage of $\{1\}$ is open, and the preimage of $\{0,1\}$ is open.
Let for $B_{\frac1n}=\{\infty\}\cup\{k\geq n\mid k\in\mathbb N\}$. Those are closed sets since their complement is an initial segment of $\mathbb N$.
whose intersection is not empty either. Observe that $f(B_n)=\{0,1\}$ for all $n$, however $\bigcap B_{\frac1n}=\{\infty\}$ and $f(\{\infty\})=\{0\}\neq\bigcap f(B_{\frac1n})$.
A: The following answer is a generalization of the one given by Asaf Karagila.
Theorem
Let $X$ and $Y$ be spaces, $I$ be an ordinal, $V : I \to \mathcal{P}(X)$ be decreasing, $V_i$ be closed, and $f : X \to Y$ have compact fibers. Then
$$f(\bigcap V_I) = \bigcap_{i \in I} f(V_i).$$
Proof
The $\subset$ direction holds by set theory; i.e. without any requirements. We will show the $\supset$ direction.
Let $y \in \bigcap_{i \in I} f[V_i]$, $C = f^{-1}[\{y\}]$, and $V'_i = V_i \cap C$. Then
\begin{aligned}
f[V'_i] & = f[V_i \cap C] \\
{} & = f[V_i \cap f^{-1}[\{y\}]] \\
{} & = f[V_i] \cap \{y\} \\
{} & = \{y\}.
\end{aligned}
Therefore $\bigcap_{i \in I} f[V'_i] = \{y\} \neq \emptyset$. By subspace topology, $V'_i$ is closed in $C$. By assumption, $C$ is compact. Since closed subsets of compact subsets are compact, $V'_i$ is compact. By the lemma below, $f[\bigcap V'_I] \neq \emptyset$. Therefore $f[\bigcap V'_I] = \{y\}$; i.e. $y \in f[\bigcap V_I]$. Therefore $f[\bigcap V_I] \supset \bigcap_{i \in I} f[V_i]$.
Special case
Suppose $X$ is compact, $Y$ is $T_1$, and $f$ is continuous. Then $f$ has compact fibers, since a closed subspace of a compact space is compact.
Lemma
Let $X, Y$ be spaces, $I$ be an ordinal, $V : I \to \mathcal{P}(X)$ be decreasing, $V_i$ be closed and compact, and $f : X \to Y$. Then
$$f[\bigcap V_I] = \emptyset \iff \bigcap_{i \in i} f[V_i] = \emptyset.$$
Proof
The direction $\impliedby$ holds by set theory without any restrictions. We will show $\implies$.
Suppose $f[\bigcap V_I] = \emptyset$. Then $\bigcap V_I = \emptyset$, and so $\bigcup_{i \in I} (V_0 \setminus V_i) = V_0$. By assumption, $V_i$ is closed. By subspace topology, $V_0 \setminus V_i$ is open in $V_0$. Since $V_0$ is compact, there exists finite $I' \subset I$ such that $\bigcup_{i \in I'} (V_0 \setminus V_i) = V_0$; i.e. $\bigcap V_{I'} = \emptyset$. Since $V$ is decreasing, $V_{\max(I')} = \bigcap V_{I'} = \emptyset$. Therefore $\bigcap_{i \in I} f[V_i] = \emptyset$.
Comments
The theorem does not generalize to locally compact fibers: a counter-example is given by $I = \mathbb{N}$, $V_i = \mathbb{R}^{\geq i}$, $f : \mathbb{R} \to \mathbb{R}$, and $f(x) = 0$. Then $f(\bigcap V_I) = \emptyset$ and $\bigcap_{i \in I} f(V_i) = \{0\}$.
