A question about a closed set Let $X = C([0; 1])$. For all $f, g \in X$, we define the metric $d$ by
$d(f; g) = \sup_x |f(x) - g(x)|$. Show that $S := \{ f\in X : f(0) = 0 \}$ is closed in $(X; d)$.
I am trying to show that $X \setminus S$ is open but I don't know where to start showing that.
I wanna add something more, I have not much knowledge about analysis and I am just self taught of it, what I have learnt so far is just some basic topology and open/closed sets.
 A: Prove that the map $\varphi : X \to \mathbb{R}$ defined by $\varphi(f) = f(0)$ is continuous.
($\varphi$ is even Lipschitz-continuous)
Then notice that $S = \varphi^{-1}(\{0\})$.
A: I'd try to show that $S$ contains all its limit points. To this end let $f$ be a limit point of $S$. Let $f_n$ be a sequence in $S$ converging to $f$ in the sup norm. 
Now we show that $f$ is also in $S$:
By assumption, for $\varepsilon > 0$ you have that $\sup_{z \in [0,1]}|f_n(z) - f(z)| < \varepsilon$ for $n$ large enough. In particular, $|f_n(0) - f(0)| = | f(0)| < \varepsilon$ for $n$ large enough. Now let $\varepsilon \to 0$.
A: Here's a direct argument: Suppose $f\in X\setminus S$. Let $d=|f(0)|\ne 0$. Then the open ball of radius $d$ around $f$ is disjoint from $S$.
A: You can try to show that $X \backslash S$ is open by a contradiction argument. This proof will only assume the very basics of analysis and topology, namely open sets and metrics. 
Suppose $X\backslash S$ is not open. That means there is an $f \in X\backslash S$ such that $B_r(f) \cap S \neq \emptyset$ for all $r$. You probably know this, but $B_r(f) = \{g \in C[0,1] : d(f, g) < r\}$ and is called the "open ball of radius $r$". Now, let $f(0) = y \neq 0$. Since $B_r(f) \cap S \neq \emptyset$ for all $r$, we can take a sequence $\{r_n\}$ such that $r_n \to 0$ as $n \to \infty$. This gives rise to a sequence of $g_n \in S$ such that $d(f, g_n) < r_n$, which is equivalent to saying  that $\sup_x|f(x) - g_n(x)| < r_n$. Recall that $g_n(0) = 0$, which means that $|f(0) - g_n(0)| < r_n$. This could definitely be the case, but remember that $f(0)$ is a fixed number $y \neq 0$. Since $\{r_n\} \to 0$, we can choose one specific radius $r_N$ such that $r_N < y$. By our assumption, there exists a $g_N \in S$ such that $|f(0) - g_N(0)| < r_N < y = |f(0) - g_N(0)|$. This is a contradiction and so $X\backslash S$ is open.
