Components Converge iff Convergence in Product Metric Space

Last updated Nov 1, 2022

Let $(M_{i}, d_{i}){i \in I}$ be Metric Spaces with Index Set $I$ and let $M = \prod\limits{i \in I}^{} M_{i}$ be the Product Metric Space with Supremum Metric. Let $(x^{(n)}){n=1}^{\infty} \subset M$ and let $x \in M$. Then $x^{(n)} \to x$ If and Only If $x^{(n)}{i} \to x_{i}$ uniformly for all $i \in I$.

($\Rightarrow$): Let $\epsilon > 0$. Since $x^{(n)} \to x$, there exists a $N \in \mathbb{N}$ so that for all $n \geq N$, $\sup\limits_{i \in I}d_{i}(x^{(n)}{i}, x{i}) < \epsilon$. Therefore, for all $i \in I$, $d_{i}(x^{(n)}{i}, x{i}) < \epsilon$. Thus $x^{(n)}{i} \to x{i}$ for all $i \in I$. $\checkmark$

$(\Leftarrow)$: Let $\epsilon > 0$. There exists $N \in \mathbb{N}$ so for all $n \geq N$ and $i \in I$, $d_{i}(x^{(n)}{i}, x{i}) < \epsilon$. Thus $\sup\limits_{i \in I}d_{i}(x^{(n)}{i}, x{i}) < \epsilon$ for all $n \geq N$. Therefore $x^{(n)} \to x$. $\checkmark$ $\blacksquare$