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In Dudley, Real Analysis and Probability (2nd ed.), Theorem 2.1.2 states:

Given topological spaces $(X,\mathcal{T})$ and $(Y,\mathcal{U})$ and a function $f:X\to Y$, if for every convergent filter base $\mathcal{F}\to x$ in $X$, $f[[\mathcal{F}]]\to f(x)$ in $Y$, then $f$ is continuous.

Proof. 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 $f^{-1}(U)\in\mathcal{T}$.

My question is: how does the last assertion follow? From $V\subset f^{-1}(U)$ we know that $f^{-1}(U)$ is a neighborhood of $x$ and therefore $f^{-1}(U)\in\mathcal{F}$.

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up vote 3 down vote accepted

Perhaps a slight re-wording of the argument make it more clear. To show that $f^{-1} [ U ]$ is open in $X$, it suffices to show for each $x \in f^{-1}[U]$ that there is an open neighbourhood $V$ of $x$ such that $V_x \subseteq f^{-1} [ U ]$. (Then $f^{-1} [ U ]$ is the union of the sets $V_x$, and is thus open.) The last little bit of the argument given is really a couple of steps compressed into one.

First, for a given $x \in f^{-1}[U]$ we have a neighbourhood $V$ of $x$ such that $f [ V ] \subseteq U$, and so $V \subseteq f^{-1} [ f[V] ] \subseteq f^{-1} [ U ]$. So we have found an appropriate neighbourhood for that particular element of $f^{-1} [ U ]$.

But this $x$ was just some arbitrary element of $f^{-1} [ U ]$, so we have actually shown that for all elements of $f^{-1} [ U ]$ there is an appropriate neighbourhood. And therefore $f^{-1} [ U ]$ is open, as desired.

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Thanks, I thought it was a more direct consequence. –  nokiddn Jun 7 '13 at 12:13

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