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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.

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