0
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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, $B_r(f)\subset N^\complement$.

$\endgroup$
4
  • $\begingroup$ I presume then that $\mathring{N} = \emptyset $, but I am not able to construct a counterfunction. $\endgroup$
    – Kabouter9
    Jun 9 '20 at 15:59
  • $\begingroup$ Not at all. If $\mu(x)=x-\frac12$ then $B_{1/2}(\mu)\subset N$. $\endgroup$ Jun 9 '20 at 16:04
  • $\begingroup$ Then I have honestly no idea what the interior should be... $\endgroup$
    – Kabouter9
    Jun 9 '20 at 16:22
  • $\begingroup$ It consists of those functions $f$ for which there are $x,y\in[0,1]$ such that $f(x)<0<f(y)$. $\endgroup$ Jun 9 '20 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.