Set of bounded functions closed in B(X,R)

Question

NOTATION:$\mathbb{B}(X,\mathbb{R})$ the set of bounded functions with the supremum metric.

PROBLEM: Let $S\subset X$. Show that $Z_{S}=\{f:X\rightarrow\mathbb{R}$, $f$ bounded $\mid f(S)=0\}$ is a closed subset of $\mathbb{B}(X,\mathbb{R})$.

My Attempt: Let $\phi:\mathbb{B}(X,\mathbb{R})\rightarrow \mathbb{R}$ with $\phi(f)=f(S)$. If I can prove that $\phi$ is continuous, then $\phi^{-1}(\{0\})$ will be closed in $\mathbb{B}(X,\mathbb{R})$.

If we take $f,g\in \mathbb{B}(X,\mathbb{R})$, then $d_{\mathbb{R}}(\phi(f),\phi(g))=\mid\phi(f)-\phi(g)\mid \leq sup\{\mid f(z)-g(z)\mid\}=d_{\mathbb{B}}(f,g)$ for $z\in X$. Therefore, $\phi$ is continuous and since $\{0\}$ is closed, then $\phi^{-1}(\{0\})$ is closed in $\mathbb{B}(X,\mathbb{R})$.

QUESTION: Is this right? Cause I cant figure it out why it is not, but at the same time I feel like it was too easy and I didnt even used the fact that $S\subset X$ :/

Thanks for your answers <3

Answer 1

$f(S)$ is not a real number so your argument fails. For each $x \in S$ define $\phi_x (f)=f(x)$ and observe that $Z_S=\cap_{x\in S} \phi_x ^{-1} \{0\}$. This is an intersection of closed sets (because each $\phi_x$ is continuous) , hence closed.

Answer 2

Another way to see it, besides Kavi's fine answer, is to note that $B(S,\mathbb{R})$ is also a metric space in the supremum norm, and as $S \subseteq X$, there is a natural restriction map: $$r: B(X,\mathbb{R}) \to B(S, \mathbb{R}): r(f) = f|_S$$ which is well-defined as the restriction of a bounded function is still bounded. And clearly $d_{\sup}(f|_S, g|_S)= \sup\{|f|_S(x) - g|_S(x)|: x \in S\} \le \sup \{|f(x)-g(x)|: x \in X\}$ for all $f,g \in B(X, \mathbb{R})$ as we take the supremum of more numbers, and so this map $r$ is a weak contraction, and so is continuous.

The singleton $\{0_S\}$, where $0_S: S \to \mathbb{R}$ is the constantly $0$ function, is a closed subset of $B(S,\mathbb{R})$ and your set $Z_S$ equals $r^{-1}[\{0_S\}]$ by definition, so is closed by continuity of $r$.