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.
Best Answer
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 $M=\psi^{-1}(\{0\})$.