I have a question that goes as follows:
Consider the $\sup$ metric $$d(f,g) = \sup_{x\in[a,b]} |f(x)-g(x)|$$ Let $\mathcal{C}[0,1]$ be the space of continuous real-valued functions on $[0,1]$ with the sup metric and $A\subseteq[0,1]$. Show that the following subset is closed $\mathcal{C}[0,1]$: $$\{f\in\mathcal{C}[0,1]: f(a) = 0 \text{ for all } a \in A\} $$
I write below my (most probably illogical) proof of the question being asked. I have added remarks where the gaps of logic I think exist
Proof: Let us denote $$E = \{f\in\mathcal{C}[0,1] : f(a)=0 \text{ for all } a\in A \} $$ To show that $E$ is closed, we must show that $\mathcal{C} [0,1] \setminus E$ is open in $\mathcal{C} [0,1]$. This is equivalent to saying $$\mathcal{C} [0,1] \setminus E =\{f\in\mathcal{C} [0,1] : f(a)\ne 0 \text{ for all } a\in A \} $$ is open in $\mathcal{C} [0,1]$.
To show that this open, we must show that for every $f \in \mathcal{C} [0,1] \setminus E$ there exists a $\epsilon_f>0$ such that $$B_{\epsilon_f}(f) \subseteq \mathcal{C} [0,1] \setminus E$$
The ball in question can be expressed as $$B_{\epsilon_f}(f) = \{g \in\mathcal{C} [0,1] : d(f,g)<\epsilon_f \}$$
Let, $g,h \in \mathcal{C} [0,1] \setminus E$, then express $$\sup_{x\in[0,1]} |f(x)-g(x)| = \alpha$$
For $h \in B_{\epsilon_f}(f)$, there exists another ball centered at $h$, namely $B_{\epsilon_h}(h)$, with $\epsilon_h >0$. Choose $\epsilon_h = \epsilon_f-\alpha$.(*) Then we can say
$$d(f,g) = d(g,f) \le d(g,h) + d(h,f) < \epsilon_h + \alpha = \epsilon_f - \alpha + \alpha = \epsilon_f $$
Hence, since we can always find an $\epsilon_f$ for which $B_{\epsilon_f}(f)$ exists and $f \in \mathcal{C} [0,1] \setminus E$, then $\mathcal{C} [0,1] \setminus E$ is open, and therefore $E$ is closed as required. $\blacksquare$
Remarks:
- In the proof I have used no reasoning with respect to the fact that $f(a)=0$ or $f(a)\ne0$.
- I have not used any properties of continuous functions on the interval $[0,1]$
- (*) I am claming that $\epsilon_h >0$, however, it seems to be the case that $\epsilon_f-\alpha$ can take a value $<0$.
I just don't think my approach to proving this is necessarily correct nor intuitive with respect to what is being asked.
Please keep in mind this is with respect to an introductory course on Topology and Metric Spaces, so there is a "basic" level of knowledge that I can work with.
I would appreciate if someone could help explain what exactly the gaps in my proof consist of or how I can approach this to solidify my proof.
Thanks
As Pebeto correctly wrote in his comment your definition of the complement of $E$ was simply wrong, and the problems started from there. The definition should be like this:
$F:=C[0,1]\setminus E=\{f\in C[0,1]: \exists a\in A, f(a)\ne 0\}$
From here it should be simple. Let $f\in F$. Then there is a point $a\in A$ such that $f(a)\ne 0$. Let $r=f(a)$ and look at the ball $B_{|r|}(f)$. Take any function $g\in B_{|r|}(f)$. From the definition of your metric it follows that for each $x\in [0,1]$ we have $|f(x)-g(x)|<|r|$. But if we suppose that $g(a)=0$ then we get $|f(a)-g(a)|=|f(a)|=|r|$ which is a contradiction. Hence we showed $g(a)\ne 0$ which implies $g\in F$. So $B_{|r|}(f)\subseteq F$.