Show {x_n} is NOT a compact set in C([0,1])

**sweetadam** · May 4th 2012, 11:49 PM

This is the Question:

Let $x_n(t) = \begin{cases} nt & 0 \leq t \leq \frac{1}{n} \\ 2-nt & \frac{1}{n}<t\leq\frac{2}{n} \\0 & \frac{2}{n}<t\leq1 \end{cases}$

Show that $\{x_n\}$ is NOT compact set in $C([0,1])$

My idea is to show that $\mbox{\{x_n(t)\}}$ converge pointwise to $x(t)$ , and $x(t) \notin C([0,1])$

Where $x(t) = \lim_{n\to \infty}x_n(t) = \begin{cases}1&t=0\\0&0<t\leq1\end{cases}$

Is this enough to show $\{x_n\}$ is NOT compact set in $C([0,1])$ ???

And how do I state formally that $\mbox{\{x_n(t)\}}$ converge pointwise to $x(t)$ ???

**girdav** · May 5th 2012, 01:02 AM

In fact $\{x_n\}$ converges pointwise to the constant function equal to $0$ . If the set $\{x_n,n\in\Bbb N\}$ was compact, we would be able to extract a uniformly convergent sub-sequence $\{x_{n_k}\}$ . This one should converge uniformly to $0$ . Do you see why it is not possible?

**sweetadam** · May 5th 2012, 06:22 AM

Thank you for fast reply

I'm really new to these concepts, sorry if my answer is silly lol

from the $Def^\underline{n}$ of uniform convergence:

for any $x_m, m\in n_k,$ and any $\epsilon < 1:$

$\left|{x_m(\frac{1}{n}) - x(\frac{1}{n})\right|= 1-0 > \epsilon$

therefore no sub-sequence converge to 0 uniformly, hance $\{x_n,n\in\Bbb N\}$ is not compact

Is this right?

**girdav** · May 5th 2012, 07:21 AM

That right.

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