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

## Theorem

Let $N \in \N$.

For all $k \in \left\{ {1, \ldots, N}\right\}$, let $T_k = \left({X_k, \tau_k}\right)$ 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 space topology on $X$.

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

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

Then $x_n$ converges to $x$ if and only if:

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

## Proof

### Necessary Condition

Let $x_n \to x$.

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

From Projection from Product Topology is Continuous it follows that $\operatorname{pr}_k$ is continuous.

By Continuous Mapping is Sequentially Continuous, $\operatorname{pr}_k$ is also sequentially continuous.

Hence $\operatorname{pr}_k \left({\mathcal x_n}\right) \to \operatorname{pr}_k \left({x}\right)$.

$\Box$

### 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 $x$.

$\displaystyle \mathcal B := \left\{ {U_{1} \times U_{2} \times \cdots \times U_{N} \quad \vert \quad \forall k \in \left\{ {1, \ldots, N}\right\} : U_k \in \tau_k} \right\}$

is an analytic basis for $\tau$.

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

$\displaystyle U = \bigcup_{ i \mathop \in I } \left({ U_{1,i} \times \cdots \times U_{N,i} }\right)$

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

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 $\operatorname{pr}_k \left({\mathcal x_n}\right) \to \operatorname{pr}_k \left({x}\right)$ it follows that:

$\displaystyle \forall k \in \left\{ {1, \ldots, N}\right\} \quad \exists M_k \in \N \quad \forall n \ge M_k: \operatorname{pr}_k \left({\mathcal x_n}\right) \in U_{k,i_0}$

Thus for all $n \ge M:=\operatorname{max} \left( {M_1, \ldots, M_N} \right)$ it holds that:

$\displaystyle x_n = \left( {\operatorname{pr}_1 \left({\mathcal x_n}\right), \ldots, \operatorname{pr}_N \left({\mathcal x_n}\right)} \right) \in U_{1,i_0} \times \cdots \times U_{N,i_0} \subset U$

Hence the result.

$\blacksquare$