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

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

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

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

Denote by $\operatorname{pr}_i : X \to X_i$ the projection from $X$ onto $X_i$.

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

Necessary Condition
Let $\mathcal F$ converge to $x$.

Let $i \in I$.

From Projection from Product Topology is Continuous, $\operatorname{pr}_i : X \to X_i$ is continuous.

Thus $\operatorname{pr}_i \left({\mathcal F}\right)$ converges to $\operatorname{pr}_i \left({x}\right)$ as claimed.

Sufficient Condition
For all $i \in I$, let $\operatorname{pr}_i \left({\mathcal F}\right)$ converge to $\operatorname{pr}_i \left({x}\right)$.

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:
 * $\displaystyle U = \bigcap_{j \mathop \in J}\operatorname{pr}_j^{-1} \left({U_j}\right)$

where $U_j \subseteq X_j$ is an open set for all $j \in J$.

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

Since $\operatorname{pr}_j \left({\mathcal F}\right)$ converges to $x_j$ this implies $U_j \in \operatorname{pr}_j \left({\mathcal F}\right)$ for all $j \in J$.

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

Since $J$ is finite, it follows that:
 * $\displaystyle U = \bigcap_{j \mathop \in J} \operatorname{pr}_j^{-1} \left({U_j}\right) \in \mathcal F$

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.

Also see

 * Filter on Product Space Converges iff Projections Converge