The boundary is $\partial S=\{0,1\}$.
The closure is $\textrm{cl}(S)=S \cup \partial S=[0,1]$.
The interior is $int S=S-\partial S=(0,1)$.
I need help explaining why this is the answer. Thank you.
The boundary is $\partial S=\{0,1\}$.
The closure is $\textrm{cl}(S)=S \cup \partial S=[0,1]$.
The interior is $int S=S-\partial S=(0,1)$.
I need help explaining why this is the answer. Thank you.
It seems that you need only the proof of $\partial S=\{0,1\}$.
Lets see why $0 \in \partial S$.
(I use the definition $x \in \partial S$ if $\forall \epsilon >0, \ (x-\epsilon,x+\epsilon) \cap S \neq \emptyset, \ \text{and} \ (x-\epsilon,x+\epsilon) \cap S^c \neq \emptyset$)
So let $\epsilon>0$. Then $-\frac{\epsilon}{2} \in (-\epsilon,\epsilon) \cap S^c$ and if $x= \min\{\frac{\epsilon}{2},\frac{1}{2} \} \Rightarrow x \in(-\epsilon,\epsilon) \cap S$ (or just say $0 \in(-\epsilon,\epsilon) \cap S$ ).
This proves that $0 \in \partial S.$
Now prove that $1 \in \partial S$ and that $x \notin \partial S, \ \forall x \neq 0,1.$