3

Let $X$ be a topological space and $f, g : X \to \Bbb{C}$ be two continuous, compactly-supported, complex-valued functions on $X$.

Support of $f$ is the closure of the set $\{x \in X : f(x) \neq 0\}$. Similarly for $g$. These are assumed to be compact sets.

I'm trying to prove that $C_c(X)$ the space of continuous, complex-valued, compactly-supported functions on $X$ is a vector space.

Rudin's "Real & Complex Analysis" book says that $\text{Support}(f + g) \subset \text{Support}(f) \cup \text{Support}(g) = $ union of two compact sets = compact set.

$$ \text{Support}(f + g) = \overline{\{ x \in X : f(x) \neq - g(x)\}} $$

I'm not seeing how this lies in that union.

Daniel Donnelly
  • 21,342

1 Answers1

3

We can decompose

$\{ x : f(x) + g(x) \neq 0 \} \subset \{ x : f(x) \neq 0 \} \cup \{ x : g(x) \neq 0 \} \cup \{ x : f(x) + g(x) \neq 0 \}$

But $f(x) + g(x) \neq 0$ implies $f(x) \neq 0$ or $g(x) \neq 0 $, so we immediately get

$\{ x : f(x) + g(x) \neq 0 \} \subset \{ x : f(x) \neq 0 \} \cup \{ x : g(x) \neq 0 \} $

Now you just need to consider the closures of either side. (e.g. here)

flawr
  • 16,533
  • 5
  • 41
  • 66