# Condition on convergent filter bases imply a function is continuous

I'm trying to understand an implication in a theorem. Suppose you're given topological spaces $(X,\mathcal{T})$ and $(Y,\mathcal{U})$ and $f\colon X\to Y$. Then if for every convergent filter base $\mathcal{F}\to x$ in $X$, $f[[\mathcal{F}]]\to f(x)$ in $Y$, then $f$ is continuous.

$f[[\mathcal{F}]]$ is the set of all direct images of $f$ on the sets $A$ in $\mathcal{F}$. The proof says take any $U\in\mathcal{U}$ and $x\in f^{-1}(U)$. The filter $\mathcal{F}$ of all neighborhoods of $x$ converges to $x$, so $f[[\mathcal{F}]]\to f(x)$. For some neighborhood $V$ of $x$, $f[V]\subset U$, so $V\subset f^{-1}(U)$, and $f^{-1}(U)\in\mathcal{T}$.

I don't understand the last part. Why would $V\subset f^{-1}(U)$ imply that $f^{-1}(U)$ is an open set in $X$?

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In general it does not true that $V\subseteq f^{-1}(U)$ implies that if $V$ is open then $f^{-1}(U)$ is open.
However, recall that a set $U$ is open if and only if for every $x\in U$ there is an open $V\subseteq U$ such that $x\in V$.
In the proof given in your question, we take an arbitrary $x\in f^{-1}(U)$. For every such $x$ set an open $V_x$ which is a subset of $f^{-1}(U)$ given by the proof.
Now we have that $V_x\subseteq f^{-1}(U)$ and therefore $\bigcup \{V_x\mid x\in f^{-1}(U)\}\subseteq f^{-1}(U)$. As well for every $x\in f^{-1}(U)$ we that $x\in V_x$, so the inclusion is actually equality.
Therefore $f^{-1}(U)$ is exactly the union of all these open sets, and thus open.