Metric spaces - prove the closedness of a subset of a continuous function One question in my book is to prove the continuity of a subset of $C[0,1]$ (continuous real functions on [0,1]) with metric $d(f,g) = sup_{x\in[0,1]} |f(x)-g(x)|$. However, I am not sure on how to continue. The subset in question is $\{f \in C[0,1] | f(a) = 0 ~ \forall a \in A \}$ where $A \subseteq [0,1]$.
I've tried the limit point approach. We have not had limit points' relations to sequences yet, so I'd like to avoid that. The definition of limit points is given as "a point $x$ in a metric space $X$ is a limit point of $A \subseteq$ X if $\forall \epsilon > 0~ (B_\epsilon(x)-\{x\})\cap A \not= \emptyset$". The subset A in the definition would be $\{f \in C[0,1] | f(a) = 0 ~ \forall a \in A \}$ in this question.
So then I let $f_0 \in C[0,1]$ be a limit point of $\{f \in C[0,1] | f(a) = 0 ~ \forall a \in A \}$ and $\epsilon > 0$. By definition of a limit point, I can define a function $f_1 \in (B_\epsilon(f_0)-\{f_0\}) \cap \{f \in C[0,1] | f(a) = 0 \forall a \in A\}$. This means that $f_1 \in B_\epsilon(f_0), f_1 \notin \{f_0\} (\text{which means} f_1 \not= f_0)$ and $f_1(a) = 0~\forall a \in A$.
$f_1 \in B_\epsilon(f_0)$ implies $d(f_1, f_0) < \epsilon$, hence $sup_{x\in[0,1]} |f_1(x) - f_0(x)| < \epsilon$ but this is where I get stuck. First of all, the values of x are taken over [0,1] and only something of the values of $f_1(x)$ where $x\in A$ is known and I do not see how to prove that $f_1$ is in the subset, which will prove that the subset is closed. I hope someone can help me out.
Another possibility might be to prove that $C[0,1] - \{f \in C[0,1] | f(a) = 0 ~ \forall a \in A \}$ is open, but I do not get any further using this approach either. Taking a subset of continuous functions instead of a set of numbers confuses me.
 A: First of all, for the love of Gauss call your $\{f\in C[0,1]\mid \forall a\in A: f(a)=0 \}$ something. I'll arbitrarily call it $S$.
It looks like it would indeed be easiest to show that $C[0,1]\setminus S$ is open. In order to do this, you let $g\notin S$ be arbitrary and seek to prove that there is a ball around $g$ that doesn't touch $S$.
By unfolding the definition, the assumption that $f\notin S$ means that there is some $a\in A$ with $g(a)\ne 0$. Now what can you say about, say, $B_{|g(a)|/2}(g)$?

However, to complete the approach you've already started, what you need to prove is that $f_0\in S$. Therefore let $a\in A$ be arbitrary; you then need to prove that $f_0(a)=0$.
Your notation $f_1$ is not good, because $f_1$ depends on $\epsilon$. It should probably be called $f_\epsilon$ instead. Then for each $\epsilon>0$ you know that $f_\epsilon\in S$ and $d(f_0,f_\epsilon)<\epsilon$. The latter condition means in particular that $|f_0(a)-f_\epsilon(a)|<\epsilon$, and since $f_\epsilon(a)\in S$ this is the same as $|f_0(a)|<\epsilon$. This holds for every $\epsilon>0$ ...
