Prove that a set is open or closed

Question

Let $M=\{f \in C[0,1] \mid f(0) = 0\}$ with $M \subset (C[0,1],\lVert \cdot \rVert_\infty)$. Is this set open or closed in the given normed vector space?

I think it is closed, but I'm not sure how to prove it. I tried working with the complement $\overline{M} = \{f \in C[0,1] \mid f(0) \neq 0\}$ and showing that this set is opened ($\forall x \in M \enspace \exists \varepsilon>0: B(x,\varepsilon) \subset M$) but I don't know how to proceed.

Answer 1

Your attempt can be made to show that the set is closed. If $f \in \overline{M}$, then $f(0)=f_0 \neq 0$. Take $\epsilon = \frac{|f_0|}2$. For each $g \in B(f,\epsilon)$ we have $\lVert f-g\rVert_\infty < \epsilon$, which implies by the definition of $\lVert\cdot\rVert_\infty$ that $\lvert f(0)-g(0)\rvert < \epsilon$. With $\lvert f(0)\rvert=\lvert f_0\rvert=2\epsilon$ this means $\lvert g(0)\rvert > \epsilon$, so $g \in \overline{M}$. This proves that $\overline{M}$ is open and hence $M$ is closed.

Answer 2

The set is closed, because the map$$\begin{array}{rccc}\psi\colon&\mathcal C[0,1]&\longrightarrow&\mathbb R\\&f&\mapsto&f(0)\end{array}$$is continuous (by the $\varepsilon-\delta$ definition; for each $\varepsilon>0$, just take $\delta=\varepsilon$) and $M=\psi^{-1}(\{0\})$.