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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.

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    Here's another definition of a boundary: A point is a boundary point of a set iff every neighborhood contains points of the set and its complement. The boundary is the collection of boundary points.2012-11-07

1 Answers 1

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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 $x \notin \partial S, \ \forall x \neq 0,1.$