Showing a set of functions over a normed vector space is open. I am wanting to show that the following set $S$ is open an open set in $(C[0,1],\|\cdot \|_\infty)$.
$$S=\left \{f:\int^1_0 f(t) dt>1\right \}$$
I have tried to show the compliment is closed but others have now told me that it is simply easier to show it is open. I am not really sure of a suitable method to do this as I am told there are many ways. Any hints or solutions would be greatly appreciated thanks!
 A: Consider $\mu: (C[0,1], \lVert \cdot \rVert) \to \mathbb{R}$ given by $\mu(f) = \int_0^1 f(t) \, dt$. 
First note this functional is continuous:
$$\begin{align}\Bigg\lvert\int_0^1 f(t) - g(t) \, dt\Bigg\rvert &\leq \int_0^1 |f(t) - g(t)| \, dt \\
&\leq \sup_{x \in [0,1]} |f(t) - g(t)| \\
&\leq \lVert f-g \rVert\end{align}$$
Next observe $S = \mu^{-1}((1, \infty))$. This is the pre-image of an open set under a continuous mapping, hence $S$ is open.
A: To show that $S$ is open it suffices to show that for any $f \in S$ there exist an open ball $B(f, r)$ contained in $S$.
Take $f \in S$, and consider $$B\left(f, \underbrace{\int_0^1 f(t)\,dt - 1}_{>0}\right)$$
For any $g \in B\left(f, \int_0^1 f(t)\,dt - 1\right)$ we have:
\begin{align}\int_0^1 f(t)\,dt &= \int_0^1 f(t) - g(t)\,dt + \int_0^1 g(t)\,dt \\
&\le \int_0^1 \|f-g\|_\infty\,dt + \int_0^1 g(t)\,dt \\
&< \int_0^1 \left(\int_0^1 f(x)\,dx - 1\right)\,dt + \int_0^1 g(t)\,dt\\
& = \int_0^1 f(t)\,dt - 1 + \int_0^1 g(t)\,dt
\end{align}
Hence $\int_0^1 g(t)\,dt > 1$ so $g \in S$.
We conclude $B\left(f, \int_0^1 f(t)\,dt - 1\right) \subseteq S$ so $S$ is open.
