# Prove that this space is compact/not compact by using Arzelà–Ascoli theorem

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:

$$Ǝ M: ∀φ(t)∈Q=>|φ(t)|≤M,∀t∈[0,1]$$

$$|X_n(t)|=|t^n|≤1$$

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

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