I need to tell if this space/set is compact in $C[0,1]$

$x_n(t) = t^n, n ∈ N$

Following Arzelà–Ascoli theorem, the set is compact when it has Uniform boundedness and Equicontinuity, is it correct?

So Uniform boundedness, in my case:

$Ǝ M: ∀φ(t)∈Q=>|φ(t)|≤M,∀t∈[0,1]$


But I'm really confused whether it's correct or not, or how to solve this overall

  • 1
    $\begingroup$ Do you have to use that theorem in order to prove that that space is not compact? $\endgroup$ – José Carlos Santos May 10 at 20:10
  • $\begingroup$ @JoséCarlosSantos Actually I don't, I just thought it's the easiest way $\endgroup$ – Alexander May 10 at 20:20
  • 1
    $\begingroup$ What function does $x_n$ converge to????? Is it in $C[0,1]$. $\endgroup$ – copper.hat May 10 at 20:22
  • $\begingroup$ @copper.hat I honestly have no idea, the question is all I have.. $\endgroup$ – Alexander May 10 at 20:26
  • $\begingroup$ What is $\lim_{n \to \infty} t^n$ for $t \in [0,1]$??? Just draw a few graphs for $n=1,2,3,...$. $\endgroup$ – copper.hat May 10 at 20:59

The space is not compact because every subsequence of the sequence $(x_n)_{n\in\mathbb N}$ converges pointwise to the function$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}0&\text{ if }x<1\\1&\text{ otherwise,}\end{cases}\end{array}$$which is discontinuous. But each $x_n$ is continuous. Therefore, the convergence cannot be uniform. In other words, every subsequence of the sequence $(x_n)_{n\in\mathbb N}$ diverges with respect to the $\sup$ metric.


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