Let $M$ be a LCH space. Denote $C_0(M)$ the set of continuous functions that vanish at infinity. A function vanishes at infinity if for any fixed $\varepsilon >0$ there exists a compact subset $K$ s.t. $\sup_{x \in K^c}|f(x)|<\varepsilon$
I would like to show that if $f,g \in C_0(M)$ then $f-g \in C_0(M)$ I run into a complication; Here's my argument;
If $f,g \in C_0(M)$ then choose $K_g,K_f$ such that;
$$\sup_{x \in K_f^c}|f(x)|<\varepsilon/2$$
$$\sup_{x \in K_g^c}|g(x)|< \varepsilon/2$$
Then we have:
$$\sup_{x \in K_f^c \cap K_g^c}|f(x)-g(x)|\leq \sup_{x \in (K_f \cup K_g)^c}|f(x)|+|g(x)|<\varepsilon $$.
I'm concerned about the case when $(K_f \cup K_g)^c$ is empty. Is there a remedy to this or perhaps or a reason this is not possible?
Best Answer
Well, if $(K_f \cup K_g)^c$ is empty then $M$ is compact, so things get a bit trivial, but I'd say it's the $\sup$ that's giving you trouble, and me, I'd replace the
$$\forall \epsilon > 0 \;\;\; \exists K \text{ s.t. } \sup_{x \in K^c}|f(x)|<\epsilon$$
with
$$ \forall \epsilon > 0 \;\;\; \exists K \text{ s.t. } |f(x)|<\epsilon \;\;\;\;\forall x \notin K$$
which is equivalent for non-empty $K$ but doesn't raise questions when $K$ is empty.
You might also find that your text has a definition of what the $\sup$ of an empty set is, but I've never seen one that I really liked.