Boundary of subset of continuous functions with supreme norm Let $X$ be a space of continuous functions on the set $[0,1]$ with values $[0,1]$ with the metric
\begin{equation}
d(f,g)=\sup_{x\in [0,1]} |f(x) - g(x)|.
\end{equation}
with induced topology.
Let $Y$ be a subset of functions $f$ with $f(0) \neq 0$ and $f(1) > 1/2$.
I want to find a boundary of $Y$
Ideas:
$\text{Boundary of } Y = \operatorname{closure}(Y)\smallsetminus\operatorname{interior}(Y).$
$\operatorname{Cl}(Y) = Y \cup Y'$, where $Y'$ is a set of limit points of $Y$. Because $Y$ is subset of $X$ by definition of induced topology $Y$ is open. It means that $X - Y$ is closed and contains some limit points of $Y$ (definition of closed set). So basically it means: $\operatorname{cl}(Y) = Y \cup (X-Y)'$
By definition of induced topology: $Y \subset X$, s.t for all $x \in Y$ exist $B \in$ B, s.t $x \in B \subseteq Y$.
Interior set is union of all  open balls which are completely contained in S. So interior of $Y$ is $Y$.
Comment: Can you please tell me if I am going in the right direction? I can be for sure completely false and don't blame me for that please. But if my ideas are correct - how should I proceed further?
Any corrections are welcome.
 A: Recall that $f \in \partial Y$ if and only if for every $r > 0$ we have $B(f, r) \cap Y \ne \emptyset$ and $B(f, r) \cap Y^c \ne \emptyset$.
If $f \in \left\{g \in C[0,1]: g(1) = \frac12\right\}$ then for every $r > 0$ the open ball $B(f, r)$ contains an element of $Y$:
Pick $c \in \langle 0, \frac12 + r\rangle \setminus \{-f(0)\}$ and notice that $g = f + c$ is in $Y$:
$$g(1) = f(1) + c > f(1) = \frac12$$
$$g(0) = f(0) + c \ne 0$$ 
Thus, $B(f, r) \cap Y \ne \emptyset$.
On the other hand, $f$ itself is in $Y^c$ so $B(f, r) \cap Y^c \ne \emptyset$.
Hence, $\left\{g \in C[0,1]: g(1) = \frac12\right\} \subseteq \partial Y$.

Now let $f \in \left\{g \in C[0,1]: g(1) > \frac12 \text{ and } g(0) =
 0\right\}$.
For every $r > 0$ we have $f \in B(f, r) \cap Y^c$ so $B(f, r) \cap Y^c \ne \emptyset$.
On the other hand, $f + \frac{r}2 \in B(f, r) \cap Y$ so $B(f, r) \cap Y \ne \emptyset$.
Hence, $\left\{g \in C[0,1]: g(1) > \frac12 \text{ and } g(0) =
 0\right\} \subseteq \partial Y$.

Now consider $f \notin \left\{g \in C[0,1]: g(1) = \frac12\right\} \cup \left\{g \in C[0,1]: g(1) > \frac12 \text{ and } g(0) =
 0\right\}$.
If $f(1) > \frac12$, then we have $f(0) \ne 0$ so for $r < \min\left\{f(1) - \frac12, |f(0)|\right\}$ we have $B(f, r) \subseteq Y$.
Similarly, if $f(1) < \frac12$, then for $r < \frac12 - f(1)$ we have $B(f, r) \subseteq Y^c$.
In any case, $f \notin \partial Y$.
Therefore:
$$\partial Y = \left\{g \in C[0,1]: g(1) = \frac12\right\} \cup \left\{g \in C[0,1]: g(1) > \frac12 \text{ and } g(0) =
 0\right\}$$
A: Let $f:[0,1]\to[0,1]$ be a continuous function satisfying $f(0)=0$ and $f(1)=1/2.$ Let $\varepsilon>0.$ Let $g(x) = \max\{1,f(x) + \varepsilon/2\}.$ Then $g\in Y$ and $d(f,g)<\varepsilon.$ Let $h(x)= \min\{f(x) - \varepsilon x/2,0\}.$
Then $h\notin Y$ and $d(f,h)<\varepsilon.$
So the $\varepsilon$-neighborhood of $f$ contains both points in $Y$ and points not in $Y.$ This is true no matter how small $\varepsilon>0$ is. Therefore $f$ is on the boundary of $Y.$
Next, suppose $f:[0,1]\to[0,1]$ is continuous and $f(1)<1/2.$ Let $\varepsilon= |f(1)- 1/2|/2.$ Then all functions $g$ within a distance $\varepsilon$ of $f$ fail to satisfy $g(1)>1/2,$ so $f$ is not on the boundary.
