# Sequence on Product Space Converges to Point iff Projections Converge to Projections of Point

## Theorem

Let $I$ be an arbitrary index set.

For all $i \in I$, let $T_i = \struct {X_i, \tau_i}$ be topological spaces.

Let $\ds X = \prod_{i \mathop \in I} X_i$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$.

Let $\tau$ be the product topology on $X$.

Denote by $\pr_i : X \to X_i$ the projection from $X$ onto $X_i$.

Let $\sequence {x_n}$ be a sequence on $X$ and let $x \in X$.

Then $\sequence {x_n}$ converges to $x$ if and only if:

for all $i \in I$ the sequence $\sequence {\map {\pr_i} {x_n} }$ converges to $\map {\pr_i} x$.

## Proof

### Necessary Condition

Let $x_n \to x$.

Let $i \in I$.

From Projection from Product Topology is Continuous it follows that $\pr_i$ is continuous.

By Continuous Mapping is Sequentially Continuous, $\pr_i$ is also sequentially continuous.

Hence $\map {\pr_i} {x_n} \to \map {\pr_i} x$.

$\Box$

### Sufficient Condition

Let $\map {\pr_i} {x_n} \to \map {\pr_i} x$ for all $i \in I$.

Let $U \in \tau$ be an open neighborhood of $x$.

Let $\BB$ be the collection of sets of the form $\ds \prod_{i \mathop \in I} U_i$, where:

for all $i \in I : U_i \in \tau_i$
for all but finitely many indices $i : U_i = X_i$

By Natural Basis of Product Topology and Synthetic Basis and Analytic Basis are Compatible it follows that $\BB$ is an analytic basis for $\tau$.

As $x \in U$ it follows that there exists $\ds \prod_{i \mathop \in I} U_i \in \BB$ such that:

$\ds x \in \prod_{i \mathop \in I} U_i \subseteq U$

Let $F := \set {i \in I : U_i \neq X_i}$.

By definition of $\BB$, $F$ is finite.

By our hypothesis $\map {\pr_i} {x_n} \to \map {\pr_i} x$ it follows that:

$\forall i \in F: \exists M_i \in \N : \forall n \ge M_i: \map {\pr_i} {x_n} \in U_{i}$

Furthermore, for $i \in I \setminus F$ we have $\map {\pr_i} {x_n} \in X_{i} = U_{i}$.

Thus for all $n \ge M := \max \set {M_i : i \in F}$ it holds that:

$x_n = \tuple {\map {\pr_i} {x_n} }_{i \mathop \in I} \in \prod_{i \mathop \in I} U_i \subseteq U$

Hence the result.

$\blacksquare$