Do $C_c(I)$ and $C(I)$ coincide for an open interval $I$ and are the functions in $C_c(I)$ even uniformly continuous?

Question

Most probably this question boils down to my confusion about the correct definition of $C_c(\Omega)$, but:

If $\Omega$ is a topological space and $\operatorname{supp}f:=\overline{\{x\in\Omega:f(x)\ne0\}}$ for any $f:\Omega\to\mathbb R$, then $C_c(\Omega)$ is the set of those $f$ for which $\operatorname{supp}f$ is a compact subset of $\Omega$.

Now, if $\Omega$ happens to be a subspace of another topological space, there is some room for confusion.

Question 1:

Consider the simple example of an open interval $I=(0,T)$, $T>0$. I wondered whether the spaces $C_c(I)$, $C_0(I)$ (vanishing at infinity) and $C(I)$ (continuous) actually coincide. It's clear to me that this is true, for $I$ replaced by $\overline I$, but I'm unsure whether it holds for $I$ itself as well.

I think the answer is no, but I would like to know whether if $f\in C_c(I)$, can we infer that $\operatorname{supp}f$ is the intersection of $I$ with a compact subset of $\mathbb R$?

Question 2:

Moreover, I would like to know whether a function in $C_c(I)$ is even uniformly continuous.

If I'm not mistaken, we should have the following result: If $\Omega$ is a locally compact metric space $\Omega$, then any function in $C_0(\Omega)$ is uniformly continuous. Since $C_c(\Omega)\subseteq C_0(\Omega)$, the same holds true for any function in $C_c(\Omega)$.

Now, if I'm not terribly wrong, any subset of a locally compact metric space is again a locally compact metric space (endowed with the restriction of the inherited metric) and since $\mathbb R$ is obviously locally compact this should be applicable for $\Omega=I$.

Answer 1

$C_c(I)$ is the set of all functions whose supports are compact subsets of $I.$ In particular, even though the support of a function like $f(x) = \chi_{(0,1)}(x)$, defined on $I = (0, 1)$ might seem to be $[0, 1],$ in the space $I$ the support is the set $(0,1),$ since $I$ can't see the two endpoints.

All functions in $C_c(I)$ are uniformly continuous. There are many ways of seeing this; the quickest is that any function with compact domain is uniformly continuous, and a compactly supported function behaves very much like a function with compact domain. You're also right that your other argument implies this, although applying it is overkill because the proof goes "for any $\epsilon,$ my fiunction is smaller than $\epsilon/2$ outside of some compact set and uniformly continuous on that compact set, so it's $\epsilon$-uniformly contninuous if I have some local compactness to make sure the boundary isn't too bad." In the end it is much trickier than showing $C_c(I)$ has this property.

Answer 2

Hint 1: Given $T > 0$, what are the compact subsets of $(0,T)$? (Here the topology of $(0,T)$ is the relative topology inherited from $\mathbb{R}$ or, in other words, restrict the usual absolute value metric to $(0,T)$.) This seems to be one of the sources of confusion. In particular, the answer is not "the intersection of a compact subset of $\mathbb{R}$ with $(0,T)$."

Hint 2: Given a locally compact metric space $\Omega$, what is the definition of $C_{0}(\Omega)$? Is it possible to be more specific when $\Omega = (0,T)$? $(0,T)$ is a very simple locally compact metric space due to the ordering it inherits from the real line, hence the definition of $C_{0}(\Omega)$ can be made much less abstract than in general.

As for the rest, yes, if $\Omega$ is a locally compact metric space, then every function in $C_{0}(\Omega)$ is uniformly continuous.

It is not true that any subset of a locally compact topological space (or even metric space) is locally compact. For example, $\mathbb{Q}$ is not locally compact. See here for a characterization of subsets that are locally compact.