Given is the set: $C([0,1]) = \{f: [0,1] \rightarrow \mathbb{R} \mid f \text{ is continuous} \}$ with the following metric: $d_\infty(f,g) = \sup\{ |f(x) – g(x)| \, | \, x \in \, [0,1] \}$.
Find the closure and interior of the following set: $N = \{ f: [0,1] \rightarrow \mathbb{R}\, | \, \exists x \in [0,1]: f(x) = 0 \}$.
My solution: $ \overline{N} = N$ and $\mathring{N} = N $.
Reason: The inclusion $ \mathring{N} \subseteq N$ is evident. Take now $f \in N$. Then there exist a $x_0 \in [0,1]$ such that $f(x_0) =0 $. Take now $\epsilon > 0$. We will show now that $B(f,\epsilon) \subseteq N$. Take $g \in B(f,\epsilon)$. We now have the following inequality: $|g(x_0)| = |f(x_0) – g(x_0)| \leq d_\infty(f,g) < \epsilon $. This shows that $g(x_0) = 0$ and that $g \in N$ because $\epsilon > 0$ was random.
For $\overline{N} \subseteq N$ I had an analogue reasoning.
My question is; am I right?
Thanks in advance!
Best Answer
The set $N$ is not an open set (and therefore $N\ne\mathring N$). For instance, the null function $\eta$ belongs to $N$. However, given $\varepsilon>0$, the constant function $\frac\varepsilon2$ belongs to $B_\varepsilon(\eta)$, but not to $N$. So, $B_\varepsilon(\eta)\varsubsetneq N$.
In fact, $\mathring N$ consists of those functions $f\in C([0,1])$ for which there are numbers $x,y\in[0,1]$ such that $f(x)<0<f(y)$.
But $N$ is closed (and therefore $N=\overline N$) because if $f\in N^\complement$, then $f$ has no zeros. Since $f$ is continuous and $[0,1]$ is compact, $\inf|f|>0$. Let $r=\inf|f|$. Then no function from $B_r(f)$ has a zero. In other words, $B_r(f)\subset N^\complement$.