help finding interior,exterior and boundary of subset in metric space So I have a metric space $(\mathscr{C}[a,b],d_{\infty})$
Where $d_{\infty}=\sup\limits_{x\in[a,b]}${$|f(x)-f(y)|$}
So i have two subsets:
$A$={$f\in\mathscr{C}[a,b], f(a)\in[0,1)$}
$B$={$f\in\mathscr{C}[a,b], \exists x \in [a,b], f(x)=0$}
So i need to find both subsets $A,B$ INTERIOR,EXTERIOR and BOUNDARY..
I am wondering how i can find in such a metric space, and with these subsets.
Since some functions are probably polaynomials, i think that they are in boundary, but am not sure.
Any help or hints woould be welcomed. 
 A: A complete solution (if you care to read it all) based on @Hellen's hints:
$\DeclareMathOperator{\Int}{Int}$
For A:
Let $S = \{f\in\mathscr{C}[a,b]: f(a)\in(0,1)\} \subseteq A$. We shall show $S = \Int A$. Take $f \in S$. Let's show that $B(f, \min\{f(a),1-f(a)\})\subseteq S$. For $g \in B(f, \min\{f(a),1-f(a)\})$ we have:
$$g(a) = f(a) + (g(a) - f(a)) \leq f(a) + d_\infty(f, g) < f(a) + 1 - f(a) = 1$$
$$f(a) - g(a) \leq d_\infty(f, g) < f(a) \implies g(a) > 0$$
Thus, $g(a) \in (0, 1)$ so $g\in S$. This shows that $S$ is open.
Now consider $f \in A$ such that $f(a) = 0$. Notice that for every $r > 0$ we can find $g_1, g_2 \in B(f, r)$ such that $g_1 \in A$, $g_2 \notin A$. Let $g_1 = f + \min\{\frac{r}{2}, 1\} \in B(f, r)$. Then $g_1(a) = \min\{\frac{r}{2}, 1\} \in [0, 1)$ so $g_1 \in A$. Let $g_2 = f - \frac{r}{2} \in B(f, r)$. We see that $g_2(a)=-\frac{r}{2} < 0$ so $g_2 \notin A$.
This shows that such $f$ are not in the interior of $A$ since there does not exist $B(f, r) \subseteq A$. From this follows that $S$ is the largest open set contained in $A$, hence $\Int A = S$.
Consider $f \in\mathscr{C}[a,b]$ such that $f(a) = 1$. Notice that for every $r > 0$ we can find $g_1, g_2 \in B(f, r)$ such that $g_1 \in A$, $g_2 \notin A$. Let $g_1 = f - \min\{\frac{r}{2}, 1\} \in B(f, r)$. Then $g_1(a) = 1 - \min\left\{\frac{r}{2}, 1\right\} \in [0, 1)$ so $g_1 \in A$. Let $g_2 = f + \frac{r}{2} \in B(f, r)$. We see that $g_2(a)=1 + \frac{r}{2} > 1$ so $g_2 \notin A$.
This shows that $\{f\in\mathscr{C}[a,b]: f(a)\in\{0,1\}\} \subseteq \partial A$. Let's show that in fact $\{f\in\mathscr{C}[a,b]: f(a)\in\{0,1\}\} = \partial A$.
Take $f\in\mathscr{C}[a,b]$ such that $f(a)< 0$.
We shall show that $B(f, -f(a))\subseteq \{h\in\mathscr{C}[a,b]: h(a)<0\}$. For $g \in B(f, -f(a))$ we have:
$$g(a)-f(a) \leq d_\infty(f, g) < -f(a) \implies g(a) < 0$$
Similarly, take $f\in\mathscr{C}[a,b]$ such that $f(a)> 1$.
We shall show that $B(f, f(a)-1)\subseteq \{h\in\mathscr{C}[a,b]: h(a)>1\}$. For $g \in B(f, f(a)-1)$ we have:
$$f(a)-g(a) \leq d_\infty(f, g) < f(a)-1 \implies g(a) > 1$$
Thus, $\{f\in\mathscr{C}[a,b]: f(a)\in\{0,1\}\} = \partial A$, since for any other $f\in\mathscr{C}[a,b]$ there does not exist $r > 0$ such that $B(f, r)$ intersects both $A$ and $A^c$.
The closure of $A$ is $\overline{A} = \Int A \cup \partial A = \{f\in\mathscr{C}[a,b]: f(a)\in [0,1]\}$. Exterior of $A$ is $\bigl(\overline{A}\bigr)^c = \{f\in\mathscr{C}[a,b]: f(a)\notin [0,1]\}$
For B:
Let $T = \{f\in\mathscr{C}[a,b]: \exists x_1, x_2 \in [a, b] \text{ such that } f(x_1) > 0, f(x_2) < 0\}$. Notice that $T \subseteq B$ because of continuity. We shall show that $T = \Int B$. Take $f \in T$. Let's prove that $B(f, \min\{f(x_1), -f(x_2)\}) \subseteq T$. For arbitrary $g \in B(f, \min\{f(x_1), -f(x_2)\})$ we have:
$$f(x_1)-g(x_1)<d_\infty(f, g) < f(x_1) \implies g(x_1) > 0$$
$$g(x_2)=f(x_2) + (g(x_2)-f(x_2)) \leq f(x_2) + d_\infty(f, g) < f(x_2) -
 f(x_2)= 0$$
Thus, $T$ is open.
Consider now $B \setminus T$, the set of functions which have at least one zero, but do not change sign. Let $f \in B \setminus T$, and WLOG assume that $f \geq 0$. We shall show that for every $r > 0$ there exist $g_1, g_2 \in B(f, r)$ such that $g_1 \in B$ and $g_2 \notin B$. $f$ is continuous on $[a, b]$ which is a compact set so $f$ attains its minimum and maximum: let $f(x_{min}) = 0$ and $f(x_{max}) = \max f([a, b])$.
Take $g_1 = f - \min\left\{\frac{r}{2}, f(x_{max})\right\} \in B(f, r)$ and notice that $g_1(x_{max}) > f(x_{max}) - f(x_{max}) = 0$ and $g_1(x_{min}) > f(x_{min}) = 0$. By continuity of $g_1$ there exists $x$ between $x_{min}$ and $x_{max}$ such that $g_1(x)=0$, so $g_1 \in B$.
Take $g_2 = f + \frac{r}{2} \in B(f, r)$ and notice that $g_2 > f \geq 0$, so $g_2$ has no zeroes. $g\notin B$ follows.
Since for $f \in B \setminus T$ there doesn't exist $r > 0$ such that $B(f, r) \subseteq B$, $B \setminus T$ cannot intersect the interior, so we have that $T$ is the largest open set contained in $B \implies \Int B = T$.
And finally, let's demonstrate that $\partial B = B\setminus T$. Take $f \in B^c = \{f\in\mathscr{C}[a,b]: f > 0 \text{ or } f < 0\}$. WLOG let $f > 0$. It is obvious that $B(f, f(x_{min})) \subseteq \{f\in\mathscr{C}[a,b]: f > 0\} \subseteq B^c$.
The closure of $B$ is $\overline{B} = \Int B\cup \partial B = B$, so $B$ is closed. Now, the exterior of $B$ is simply $B^c$.
