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

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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.

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

Let $\mathcal F$ be a filter on $X$.

Let $x \in X$.

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


Then $\mathcal F$ converges to $x$ if and only if:

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


Proof

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.

$\Box$


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 $j \in J$.

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

Since $J$ is finite, it follows that:

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

Recall that $U \subseteq V$.

As $\mathcal F$ is a filter, this implies that also $V \in \mathcal F$.

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

$\blacksquare$


Also see