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

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

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

Let $\FF$ be a filter on $X$.

Let $x \in X$.

Let $\pr_i: X \to X_i$ denote the projection from $X$ onto $X_i$.

Then $\FF$ converges to $x$ if and only if:

for all $i \in I$ the image filter $\map {\pr_i} \FF$ converges to $x_i := \map {\pr_i} x$.


Necessary Condition

Let $\FF$ converge to $x$.

Let $i \in I$.

From Projection from Product Topology is Continuous, $\pr_i : X \to X_i$ is continuous.

Thus $\map {\pr_i} \FF$ converges to $\map {\pr_i} x$ as claimed.


Sufficient Condition

For all $i \in I$, let $\map {\pr_i} \FF$ converge to $\map {\pr_i} x$.

Let $V \subseteq X$ a neighborhood of $x$.

We have to show that $V \in \FF$.

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:

$\ds U = \bigcap_{j \mathop \in J} \map {\pr_j^{-1}} {U_j}$

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 $\map {\pr_j} \FF$ converges to $x_j$, this implies $U_j \in \map {\pr_j} \FF$ for all $j \in J$.

By the definition of $\map {\pr_j} \FF$, this implies $\map {\pr_j^{-1}} {U_j} \in \FF$ for all $j \in J$.

Since $J$ is finite, it follows that:

$\ds U = \bigcap_{j \mathop \in J} \map {\pr_j^{-1}} {U_j} \in \FF$

Recall that $U \subseteq V$.

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

Thus $\FF$ converges to $x$ as claimed.


Also see