Let $X=(C[0,1],||\cdot||_\infty)$ and $A=\{f\in X | f(0) \neq 0\}$. How can we decide whether $A$ is open or closed in $X$?

Question

Let $X=(C[0,1],||\cdot||_\infty)$ and $A=\{f\in X | f(0) \neq 0\}$. How can we decide whether $A$ is open or closed in $X$?

I am having difficulty trying to approach such problems.

We have $A^{c}=\{f\in X | f(0)=0\}$

take any sequence $f_n$ in $A^{c}$ that converges to $f_0$

$\Rightarrow ||f_n(x)-f_0 (x)||_\infty \to 0 \; \;\text{in} \; R$

$\Rightarrow max_{x \in [0,1]} \{|(f_n - f_0)(x)|\} = 0$

$\Rightarrow f_n (x) = f_0 (x) ]; \forall \; x \in [0,1] $

$\Rightarrow f_0(0) = f_n(0) =0 $

so $A^{c}$ is closed and so $A$ is open

Is this argument fine?

Answer 1

From $\|f_n-f_0\|\to0$ you cannot deduce that $$ \max_{x\in[0,1]}\{|f_n(x)-f_0(x)|\}=0 $$ However, you can deduce that, for every $\varepsilon>0$, there exists $N$ such that, for $n>N$, $\max_{x\in[0,1]}\{|f_n(x)-f_0(x)|\}<\varepsilon$.

In particular, you can deduce that $|f_n(0)-f_0(0)|<\varepsilon$, for $n>N$. Since $f_n(0)=0$, you get $|f_0(0)|<\varepsilon$ and, since $\varepsilon$ is arbitrary, $f_0(0)=0$.

Answer 2

Yes, that proves that $A^\complement$ is closed and therefore that $A$ is open.

It remains to be proved that $A$ is not closed. That's easy. If it was closed, $A^\complement$ would be open. Take the null function $f$, which belongs to $A^\complement$. If $\varepsilon>0$, let $g$ be the constant function which always take the value $\frac\varepsilon2$. Then $g\in B_\varepsilon(f)$, but $g\notin A^\complement$. Therefore, $A^\complement$ is not open, since it contains no open ball centered at $f$.