# Closure and interior of set of functions.

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!

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 $$f(x)<0.
But $$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$$.
• I presume then that $\mathring{N} = \emptyset$, but I am not able to construct a counterfunction. Commented Jun 9, 2020 at 15:59
• Not at all. If $\mu(x)=x-\frac12$ then $B_{1/2}(\mu)\subset N$. Commented Jun 9, 2020 at 16:04
• It consists of those functions $f$ for which there are $x,y\in[0,1]$ such that $f(x)<0<f(y)$. Commented Jun 9, 2020 at 17:04