Determine if the following family, $\mathcal{F}$, of continuous functions is compact: $\{x^{n}\} \cup \{0\}$ in $[0,\frac{1}{2}]$
To do this I have to appeal to the Arzela Ascoli Theorem. In particular I have to establish that this family, $\mathcal{F}$, is:
- Bounded
- Closed
- Equicontinous
At the moment I am having trouble establishing if this family is closed or not. But I think it has more to do with how I'm using the idea of showing a set is closed in general.
So to show that this family $\mathcal{F}$ is closed I have to show that all of this sets limit points are in the set.
Observe that $f_{n}(x) = x^{n} \to 0$, when $n \to \infty$, when $x \in [0,\frac{1}{2}]$.
Proof: Suppose $f_{n}(x)$ converges uniformly to $f(x) \in C([0,\frac{1}{2}])$, where $f_{n} \in \mathcal{F}$.
Want to show: $f(x) = 0 \in \mathcal{F}$
This is where I'm stuck and feel I probably shouldn't be since I've done questions like this before involving real numbers.
I have to somehow show that $\|f_{n}(x) - f(x)\|_{\infty} \to 0$. I think....maybe I'm not even supposed to use the uniform norm in this scenario. As can be seen I'm stuck. SUggestions on how to proceed?
Alternative proof: In any topological space $\{x,x_1,x_2,...\}$ is compact if $x_n \to x$. This is because any open cover has a member containing $x$ and this member contains all but finitely many of the $x_n$'s. Hence there is a finite subcover. In the present case $|x^{n}| \leq (\frac 1 2)^{n} $ so $x^{n} \to 0$ uniformly.
Thus Arzela - Ascoli Theorem is an overkill for this question.