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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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    You are using $S$ to denote both the set ad the boundary. You should be more careful with your notation. What have you tried? Your question shows little to no research effort of your own. Also, you should do your best to write your questions using $\LaTeX$ (See, e.g., http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference).2012-11-07
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    Do you know the definitions of boundary, closure, and interior? Have you tried to apply those definitions to these problems? What progress did you make? Where did you get stuck?2012-11-07
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    Closure includes its boundary points thus [0,1] Interior does not include boundary points thus (0,1) but I do not know how to explain The boundary is ∂S={0,1}2012-11-07
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    One definition of the boundary is the intersection of the closure of the set and the closure of the complement.2012-11-07
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    @copper.hat: If the OP is defining the closure using the boundary of $S$, then one cannot define the boundary of using the closure.2012-11-07
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    @JavaMan: I missed that, albeit it is (imho) a clumsy definition of closure.2012-11-07
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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 $x \notin \partial S, \ \forall x \neq 0,1.$