# prove a lemma for a metric space

• Oct 17th 2007, 10:35 AM
hercules
prove a lemma for a metric space
I need help proving this

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

Thanks guys
• Oct 17th 2007, 01:12 PM
ThePerfectHacker
Quote:

Originally Posted by hercules
I need help proving this

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

Thanks guys

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

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