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

## Theorem

Let $N \in \N$.

For all $k \in \set {1, \ldots, N}$, let $T_k = \struct{X_k, \tau_k}$ be topological spaces.

Let $\displaystyle X = \prod_{k \mathop = 1}^N X_k$ be the cartesian product of $X_1, \ldots, X_N$.

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

Denote by $\pr_k : X \to X_k$ the projection from $X$ onto $X_k$.

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 $k \in \set{1, \ldots, N}$ the sequence $\sequence {\map {\pr_k} {x_n}}$ converges to $\map {\pr_k} x$.

## Proof

### Necessary Condition

Let $x_n \to x$.

Let $k \in \set {1, \ldots, N}$.

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

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

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

$\Box$

### Sufficient Condition

Let $\operatorname{pr}_k \left({\mathcal x_n}\right) \to \operatorname{pr}_k \left({x}\right)$ for all $k \in \left\{ {1, \ldots, N}\right\}$.

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

By definition of the product topology and Synthetic Basis and Analytic Basis are Compatible it follows that

$\displaystyle \BB := \set {U_1 \times U_2 \times \cdots \times U_N \quad \vert \quad \forall k \in \set{1, \ldots, N} : U_k \in \tau_k}$

is an analytic basis for $\tau$.

Hence there exists an index set $I$ such that:

$\displaystyle U = \bigcup_{ i \mathop \in I } \paren{ U_{1,i} \times \cdots \times U_{N,i} }$

where $U_{k,i} \in \tau_k$ for all $i \in I, k \in \set{1, \ldots, N}$.

As $x \in U$ it follows that there exists $i_0 \in I$ such that

$\displaystyle x \in U_{1,i_0} \times \cdots \times U_{N,i_0}$

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

$\displaystyle \forall k \in \set{1, \ldots, N} \quad \exists M_k \in \N \quad \forall n \ge M_k: \map {\pr_k} {x_n} \in U_{k,i_0}$

Thus for all $n \ge M:=\max \set {M_1, \ldots, M_N}$ it holds that:

$\displaystyle x_n = \tuple{\map {\pr_1} {x_n}, \ldots, \map {\pr_N} {x_n}} \in U_{1,i_0} \times \cdots \times U_{N,i_0} \subset U$

Hence the result.

$\blacksquare$