# Space of continuous functions vanishing at infinity

Let's denote with $C_0(X)$ the space of continuous functions $f$ on $X$ such that for every $\epsilon>0$ there exists a compact set $K_\epsilon\subset X$ satisfying $$\sup_{x\notin K_\epsilon }|f(x)|\leq\epsilon.$$ I have to prove that $C_0(C([0,1],\mathbb{R}))=\{0\}$.

Any suggestion? For me this is a consequence of the fact that compact sets in $C([0,1],\mathbb{R})$ have empty interior.

• How can $0$ be an element of $C_0(X)$ for $X=C([0,1],\mathbb R)$? $0$ is a real number, ant the elements of $C_0(X)$ are functions mapping $X$ (which is in itself a space of functions) to $\mathbb R$. – 5xum Jul 3 '14 at 8:05
• $0$ is quite a nice map from $C([0,1],\Bbb R)$ to $\Bbb R$, namely the constant function $0$. But which topology are you using on $C([0,1], \Bbb R)$? The metric topology inherited from the metric $d(f,g) = \int_0^1 \lvert f-g\rvert$? – Gaussler Jul 3 '14 at 8:10
• I'm using the supremum distance – max Jul 3 '14 at 8:12
• @Sue: your argument seems correct, what is wrong? – Siméon Jul 3 '14 at 8:21
• I just wanted to be sure – max Jul 3 '14 at 8:29

Suppose $0\neq h \in C_0(X)$, then assume $h \geq 0$ by replacing $h$ by $|h|$. Choose $x_0 \in X$ such that $h(x_0) > 0$. Then $\exists \epsilon > 0$ and an open set $U\subset X$ containing $x_0$ such that $$h(x) > \epsilon \quad\forall x\in U$$ Let $K$ be the compact set chosen such that $$\sup_{x\notin K} h(x) < \epsilon$$ so $U \subset K$. As you say, it now suffices to show that compact sets in $X$ have empty interior.
So suppose $K$ is a compact set with an interior point $x_0 \in K$. Replacing $K$ by $K-x_0$, we may assume that $x_0 = 0$. So $\exists \delta > 0$ such that $$\|x\| \leq \delta \Rightarrow x\in K$$ So $\overline{B(0, \delta)} \subset K$ must be a compact set in $X$. Hence, $\overline{B(0,1)}$ must be compact (since it is a homeomorphic image of $\overline{B(0,\delta)}$). But then consider the sequence $(x_n) \in X$ given by $$x_n(t) = t^n, 0\leq t\leq 1$$ It is easy to check that this does not have a convergent subsequence, and so $\overline{B(0,1)}$ is not compact.