Filter on Product Space Converges to Point iff Projections Converge to Projections of Point

Theorem
Let $$(X_i)_{i \in I}$$ be a family of non-empty topological spaces where $$I$$ is an arbitrary index set.

Denote with $$X := \prod_{i \in I} X_i$$ the corresponding product space.

Let $$\mathcal{F}$$ be a filter on $$X$$ and let $$x \in X$$.

Denote with $$\pi_i : X \to X_i$$ the projection from $$X$$ onto $$X_i$$.

Then $$\mathcal{F}$$ converges to $$x$$ iff for all $$i \in I$$ the image filter $$\pi_i \left({\mathcal F}\right)$$ converges to $$x_i := \pi_i(x)$$.

Proof

 * Assume first that $$\mathcal F$$ converges to $$x$$.

Let $$i \in I$$, then $$\pi_i : X \to X_i$$ is continuous.

Thus, $$\pi_i(\mathcal F)$$ converges to $$\pi_i(x)$$ as claimed.


 * Assume now that for all $$i \in I$$, $$\pi_i(\mathcal F)$$ converges to $$\pi_i(x)$$.

Let $$V \subseteq X$$ a neighborhood of $$x$$. We have to show that $$V \in \mathcal{F}$$.

Then there is a set $$U$$ from the natural basis of $$X$$ such that $$x \in U \subseteq V$$.

By the definition of the natural basis, there is a finite set $$J \subseteq I$$ such that $$U = \bigcap_{j \in J} U_j$$ where $$U_j \subseteq X_j$$ is an open set for all $$j \in J$$.

Thus $$U_j$$ is an open neigborhood of $$x_j$$ for all $$j \in J$$.

Since $$\pi_j(\mathcal{F})$$ converges to $$x_j$$ this implies $$U_j \in \pi_j(\mathcal{F})$$ for all $$j \in J$$.

This implies $$\pi_j^{-1}(U_j) \in \mathcal{F}$$ for all $$j \in J$$ by the definition of $$\pi_j(\mathcal{F})$$.

Since $$J$$ is finite, it follows that $$U = \bigcap_{j \in J} \pi_j^{-1}(U_j) \in \mathcal{F}$$.

Remember that $$U \subseteq V$$, $$\mathcal{F}$$ being a filter thus implies that also $$V \in \mathcal{F}$$.

Thus $$\mathcal{F}$$ converges to $$x$$ as claimed.