Equivalence of Definitions of Continuous Mapping between Topological Spaces/Point

Theorem
Let $$X, Y$$ be topological spaces, $$x \in X$$ and $$f : X \rightarrow Y$$ a mapping. The following are equivalent:


 * 1) $$f$$ is continuous at $$x$$, i.e. for any neighborhood $$U \subseteq Y$$ of $$f(x)$$, $$f^{-1}(U)$$ is a neighborhood of $$x$$.
 * 2) For any filter $$\mathcal{F}$$ on $$X$$ that converges to $$x$$, the corresponding image filter $$f(\mathcal{F})$$ converges to $$f(x)$$.

(1) $$\Rightarrow$$ (2)
Assume $$f$$ is continuous at $$x$$ and $$\mathcal{F}$$ is a filter on $$X$$ that converges to $$x$$.

If $$U \subseteq Y$$ is a neighborhood of $$f(x)$$ then, since $$f$$ is continuous at $$x$$, $$f^{-1}(U)$$ is a neighborhood of $$x$$.

Since $$\mathcal{F}$$ converges to $$x$$, this implies that $$f^{-1}(U) \in \mathcal{F}$$.

By the definition of $$f(\mathcal{F})$$ it follows that $$U \in f(\mathcal{F})$$.

Thus $$f(\mathcal{F})$$ converges to $$f(x)$$.

(2) $$\Rightarrow$$ (1)
The set $$\mathcal{U}_x := \{ U \subseteq X:\, U \text{ is a neighborhood of } x \}$$ is a filter on $$X$$.

We set $$\mathcal{F} := \mathcal{U}_x$$.

Then $$\mathcal{F}$$ obviously converges to $$x$$.

By assumption this implies that $$f(\mathcal{F})$$ converges to $$f(x)$$.

Now let $$U \subseteq Y$$ a neighborhood of $$f(x)$$.

Then $$U \in f(\mathcal{F})$$ and by definition this implies $$f^{-1}(U) \in \mathcal{F} = \mathcal{U}_x$$.

Thus $$f^{-1}(U)$$ is a neighborhood of $$x$$.

Therefore $$f$$ is continuous at $$x$$.