# Thread: prove a lemma for a metric space

1. ## prove a lemma for a metric space

I need help proving this

Lemma: A sequence $\displaystyle \left( \bold{ x^{(n)}} \right)$ in $\displaystyle \mathbb{R}^k$ converges if and only if for each $\displaystyle j = 1,2,...,k,$ the sequence $\displaystyle \left( x_j^{(n)}\right)$ converges in $\displaystyle \mathbb{R}$.

Thanks guys

2. Originally Posted by hercules
I need help proving this

Lemma: A sequence $\displaystyle \left( \bold{ x^{(n)}} \right)$ in $\displaystyle \mathbb{R}^k$ converges if and only if for each $\displaystyle j = 1,2,...,k,$ the sequence $\displaystyle \left( x_j^{(n)}\right)$ converges in $\displaystyle \mathbb{R}$.

Thanks guys
Say $\displaystyle |\bold{x}_k - \bold{y} | <\epsilon$.
Then it means,
$\displaystyle \sqrt{(x^{(1)}_k - y_1)^2+...+(x^{(n)} - y_k)^2} < \epsilon$
But that means,
$\displaystyle \max | x^{(i)}_k - y_i| < \epsilon$.
Thus, each component converges.

Conversely if each component converges it means,
$\displaystyle \max |x^{(i)}_k - y_i| < \epsilon$.
That means,
$\displaystyle \sqrt{(x^{(1)}_k - y_1)^2+...+(x^{(n)} - y_k)^2} \leq \sqrt{ n \max (x^{(i)}_k - y_i)^2} = \sqrt{n}\max | x^{(i)} - y_i| < \sqrt{n} \epsilon$.