Where Does Arzela's Theorem Fail in this Classic Example?

Question

In demonstrating the fact that the closed unit ball in $C([0,1])$ under the sup-norm $\lVert f \rVert_{\infty}=\sup_{x\in[0,1]}f(x)$ is not sequentially compact, an example often cited (similarly to here) is something like that of the sequence of functions $(\space f_{n}\space)\subset C([0,1])$ defined as $0$ for $x\notin\left[\frac{1}{n+1},\frac{1}{n}\right]$ and otherwise giving the line segments between the points $\left(\frac{1}{n+1},0\right), \left(\frac{1}{2}(\frac{1}{n}+\frac{1}{n+1}),1\right)$, and $ \left(\frac{1}{n},0\right)$. Clearly, for $m\not=n, \space \lVert f_n-f_m\rVert_{\infty}=1$, since the supports have disjoint interiors, so there cannot exist a cauchy subsequence of $(\space f_n \space)$ under the given metric (even though the entire sequence converges point-wise to the $0$ function).

My question is regarding Arzela's Theorem, stated for my purposes as:

If $S$ is a compact metric space, any uniformly bounded and equicontinuous family of functions $ (\space f_n \space),\space f_n:S \to \Bbb{R}$, contains a uniformly convergent (to some function $\space f:S\to\Bbb{R}$) subsequence.

Supposing Arzela's Theorem applies, the $(\space f_n \space)$ defined above have a uniformly convergent subsequence, which is therefore convergent in $C([0,1])$ under the sup-norm, a contradiction to the conclusion that no Cauchy subsequence exists.

So, are the hypotheses of Arzela's Theorem not satisfied? $(\space f_n \space)$ is, of course, uniformly bounded by $1$. Uniform equicontinuity is the same as pointwise equicontinuity on compact sets, so we may restrict our attention to a given point $ x_0 \in [0,1] $, which lies in the support of at most two functions in $(\space f_n \space)$, so pointwise equicontinuity of the family reduces to equicontinuity of a finite number of continuous functions, which is immediate. Is my reasoning regarding the equicontinuity flawed, or is there some other reason Arzela's Theorem does not apply?

Answer 1

You have assumed that for a family $\mathcal F$ of functions to be equicontinuous at $x_0$, it suffices to consider only those $f\in\mathcal F$ such that $x\in\operatorname{supp}(f)$. This is not the case. Indeed, using this example, suppose $(f_n)$ is equicontinuous at $0$. Then there exists $\delta>0$ such that $$0\le x<\delta\quad\Rightarrow\quad|f_n(x)|<\tfrac12$$ for all $n\in\mathbb N$. (Note that $f_n(0)=0$.) Take $n>\frac1\delta$ and put $x=\frac12(\frac1n+\frac1{n+1})$. Clearly $0\le x<\delta$, and yet $f_n(x)=1>\frac12$. This is a contradiction.

The problem with your assumption is that even if $x_0$ is not in the support of $f$ for any $f\in\mathcal F$, it can get arbitrarily close, as in our case. Once you introduce your $\delta>0$, that becomes important.

Answer 2

This sequence is not equicontinuous; if it were, then there exists $\delta>0$ such that, for any $n\in\mathbb N$ and $x,y\in[0,1]$ with $|x-y|<\delta$, we have that $|f_n(x)-f_n(y)|<\frac{1}{2}$. If you apply this for $n>\frac{1}{\delta}$, $x=\frac{1}{n+1}$ and $y=\frac{1}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right)$, then you obtain that $$\frac{1}{2}>\left|f_n\left(\frac{1}{n+1}\right)-f_n\left(\frac{1}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right)\right)\right|=1,$$ which is a contradiction.