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Prove: The set of interior points of A is the largest open set contained in A.

Definition of interior: Let $A$ be a set of real numbers. A point $p\in A$ is an interior point iff $p$ belongs to some open interval $S_{p}$ which is contained in A: $p\in S_{p}\subset A$

Definition of open: A set $A$ is open iff each of its points is an interior point

proof: Given $A$, we want to show the set of interior points of $A$ denoted int($A$) is the largest open set contained in $A$. Let $G$ be open and int($A$)$\subset G\subset A$. I want int($A$) $ = G$.

I am stuck here, if anyone can show me what to do next I would greatly appreciate it.

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  • $\begingroup$ Hint: show that an interior point of $G$ is an interior point of $A$. $\endgroup$ – Pedro M. Feb 20 '15 at 18:54
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    $\begingroup$ Aren't your definitions cyclic? Or is it known what an open interval is? $\endgroup$ – Mankind Feb 20 '15 at 19:07
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From the assumptions, you already know $int(A) \subset G$. You need only to show that $G \subset int(A)$. Let $a \in G$. Since $G$ is open, there exists an open set $V$ so that $a \in V \subset G \subset A$. This is precisely the definition of $a \in int(A)$ and the result follows.

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