Open Set for a continuous function I'm having a difficulty in understanding how to determine if a set of continuous functions with closed bounds is an open set or not with a norm of uniform convergence.
ex: {g ∈ C(Ω) | g(x) > 0, ∀x ∈ Ω}
By definition of Open Set:A subset S ⊆ E of a normed space E is open if for every x ∈ S there exist ε > 0 such that B(x, ε) ⊆ S .
I can visualise how this can be done for a subset of $\mathbb{R}^2$. Wouldn't there be an infinite number of functions, how do you check for the open ball in a continuous function.
 A: Let $O=\{g\in C(\Omega):g(x)>0\ \forall x\in\Omega\}$. Wether $O$ is open or not will depend on $\Omega$.
If $\Omega$ is compact, then $O$ is open. If $f\in O$, since $\Omega$ is compact, $f$ attains its minimum $m$, which will be strictly positive. Take $0<\epsilon<m$. If $g\in C(\Omega)$ and $|g(x)-f(x)|\le\epsilon$ for all $x\in\Omega$, then $g(x)\ge m-\epsilon>0$ for all $x\in\Omega$ and $g\in O$.
Consider now $\Omega=\mathbb{R}$ and take $f(x)=1/(1+x^2)$. For any $\epsilon>0$ there are functions $g\in C(\Omega)$ such that $|f(x)-g(x)|\le\epsilon$ and $g(x)<0$ for sufficiently large $x$. 
A: I will use the example you provided. Let $S = \{g \in C(\Omega) \mid \forall x \in \Omega : g(x) >0\}$. We want to show that $S$ is open, this can only be the case if $\Omega$ is compact, which I will assume. Now because we are working in a normed space that is impossible to visualize in any way, we can only use the definition - not intuition. 
Pick a $ g \in S$. We need to construct an $\varepsilon > 0 $ such that $B(g,\varepsilon) \subseteq S$. Since we are using the uniform norm on $C(\Omega)$, we know:
$$B(g,\varepsilon) = \{f \in C(\Omega) : \|f-g\|_\infty < \varepsilon\}$$
This is the tricky part, what $\varepsilon$ can we pick? There isn't really one method, you should just get a 'feel' for it. In this case, pick $\varepsilon = \inf\{g(x) : x\in \Omega\}$. Because $g(x)>0$ for all $x \in \Omega$ and $\Omega$ is compact, we know that $\varepsilon >0$. 
Now we must prove that $B(g,\varepsilon) \subseteq S$. Choose $f \in B(g,\varepsilon)$. Then we know per the definition of $B(g,\varepsilon)$ that for all $ x\in\Omega: |f(x) - g(x)| < \varepsilon$. This is equivalent to:
$$\forall x \in \Omega:\quad g(x) - \varepsilon < f(x) < g(x) + \varepsilon$$
Because we know that for all $x \in \Omega : g(x) \geq \varepsilon$, we can conclude that $f(x) > 0$ for all $x$. This is exactly what we needed for $f$ to be in $S$. And thus $B(g,\varepsilon) \subseteq S$, which completes the proof that $S$ is open.
