Proof that $C(S)$ is connected for any subset of $\mathbb{R}$

Question

I am confused with how to start this proof

For any $S \subseteq \mathbb{R}$, prove that $C(S)$ is connected, where $$C(S) = \{f: \text{real-valued, bounded, continuous functions on } S\}$$

with the metric

$$d(f,g) = \sup\{|f(x) - g(x)| : x\in S\}$$

I was thinking about trying to prove that it is discontinuous

If $C(S)$ is discontinuous, then there are open subsets $U_1, U_2 \subseteq C(S)$ such that

$$(C(S) \cap U_1) \cap(C(S) \cap U_2) = \varnothing$$ $$(C(S) \cap U_1) \cup(C(S) \cap U_2) = C(S)$$ $$C(S) \cap U_1 \neq \varnothing$$ $$C(S) \cap U_2 \neq \varnothing$$

But I don't know how to proceed. Do I just pick some $f \in U_1$ and $g \in G$? Should I assume $U_1$ and $U_2$ are disjoint?

Answer 1

You can join any two points $f,g$ of $C(S)$ by the path $t \to tg+(1-t)f$ so $C(S)$ is (path) connected. In your approach you can define $V_1=\{t\in [0,1]: tg+(1-t)f \in U_1\}$ and $V_2=\{t\in [0,1]: tg+(1-t)f \in U_2\}$. You can then verify that this leads to the wrong conclusion that $[0,1]$ is not connected.