What does "closed" mean in Heine Borel for $C^0$?

Question

Heine Borel for $C^0$:

A set $\mathcal{E} \subseteq C^0([a,b], \mathbb{R})$ is compact if it is closed, bounded and equicontinuous.

I don't really understand what closed mean in the definition.

Consider the $f_n(x) = \sin(x+n),x \in [0,1]$, and let $\mathcal{E} = \{f_n(x)\} $ I want to show that this set is compact

1. We know $\mathcal{E}$ is bounded since $\sin(x+n)$ is bounded

2. We know $\mathcal{E}$ is equicontinuous, since $\sin(x+n)$ is bounded and its derivative $\cos(x+n)$ is also bounded, by intermediate value theorem, $\mathcal{E}$ is equicontinuous

3. Closed.....?

What does it mean by $\{\sin(x+n)\}$ to be closed? Can someone provide an example of a set $\{f_n\}$ that is closed and not closed?

Answer 1

So first, the Arzela-Ascoli theorem usually says something to the effect of "if $E$ is bounded (in the uniform metric) and equicontinuous, then $E$ is precompact". Precompact means "has compact closure". A closed, precompact set is compact and vice versa. I just mention this because this is what you are more likely to find if you go searching in books or online.

That aside, closed here means the same as it does in any topological space. However, it is probably easier to think about it in terms of metric spaces: $E$ is closed if any sequence in $E$ which converges in the uniform metric has its limit in $E$.

This matters to the theorem above because a set in a metric space which is not closed can't be compact. This is because you could have a sequence which is convergent to a point outside the set, and such a sequence can't have a convergent subsequence.

An example of a non-closed set in $C^0$ is $\{ f_n : n \in \mathbb{N} \}$ where $f_n(x)=1/n$ (regardless of $x$). Here the limit is $f(x)=0$ (again regardless of $x$) but this is not in the set. Another non-closed set is any open ball, e.g. $\{ f : \| f \|_\infty < 1 \}$.

Answer 2

Your example is a singleton set and as such is closed in our normed space $C_0([a,b],\Bbb R)$. (Note that the space in question is a space of functions, i.e., when we refer to a point of this space we really mean a continuous function).

A more general subset $S$ of our space is closed if its complement is open, that is for efery $f\notin S$ ther exists $r>0$ such that for all $g\in C^0([a,b],\Bbb R)$ with $d(f,g)<r$ we also have $g\notin S$. (Also recall that $d(f,g)=\|f-g\|_\infty=\max\{\,|f(x)-g(x)|\mid x\in[a,b]\,\}$.