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Prove: The set of interior points of any set $A$, written int($A$), is an open set.

Let $p\in$ int($A$), then by definition $p$ must belong to some open interval $S_{p}\subset A$. Now since we know that the real line itself is open then $S_{p}\subset \mathbb{R}$. Now suppose we pick some $q\in A$. I want $q\in$ int($A$).

I am stuck here I am not sure if my logic is correct, any suggestions would be greatly appreciated.

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

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  • $\begingroup$ What's your definition of the interior of a set ? $\endgroup$ – Tryss Feb 20 '15 at 16:02
  • $\begingroup$ Are you working in a general topological space, or in the real line? $\endgroup$ – Stefan Hamcke Feb 20 '15 at 16:06
  • $\begingroup$ First, it does not follow that if $q \in A$ then $q \in int(A)$. Surely you meant something else. Also, not providing what you know about interiors and open sets will hinder progress. For instance, one definition of the interior of a set is the union of all open sets contained in $A$. Clearly this is open and needs no further investigation. $\endgroup$ – Sloan Feb 20 '15 at 16:06
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    $\begingroup$ 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: $\endgroup$ – Wolfy Feb 20 '15 at 16:08
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    $\begingroup$ If you are trying to do this proof for a metric space then you can use the definition of metric function. For a general topological space you can use the definition of base, I guess. $\endgroup$ – Samrat Mukhopadhyay Feb 20 '15 at 16:12
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Hint 1:

Any union of open sets is open. (This is axiomatic.)

Hint 2:

We would like to show that the interior is a union of open intervals. Can you see which intervals in particular? Use the definition of interior points here.

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  • $\begingroup$ I am not sure if I understand your hints, we want me to state that since any union of open sets is open then let the point $p$ which be containted in the union of open intervals. Using the definition stated above the union of open intervals is contained in $A$. I am not sure if I am following... $\endgroup$ – Wolfy Feb 20 '15 at 16:18
  • $\begingroup$ I am saying that you can write $int(A)$ as a union of open intervals. If you can do that, then thanks to Hint 1 it is automatic that $int(A)$ is open. $\endgroup$ – Gyu Eun Lee Feb 20 '15 at 16:33
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Given $A$, we want to show that the set of interior points $\operatorname{int}(A)$ is open, meaning that each point $p\in\operatorname{int}(A)$ is an interior point of $\operatorname{int}(A)$.

By definition, $p\in S_p\subseteq A$ for some open interval $S_p$. Now consider some $q\in S_p$. Since $q$ belongs to the open interval $S_p\subseteq A$, so $q$ is also an interior point of $A$. Therefore $S_p\subseteq\operatorname{int}(A)$, so $p\in S_p\subseteq \operatorname{int}(A)$ and $p$ is also an interior point of $\operatorname{int}(A)$.

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  • $\begingroup$ Clean, Thanks man $\endgroup$ – Wolfy Feb 20 '15 at 16:22
  • $\begingroup$ No, recall that the goal was to show that $p$ is interior to $\operatorname{int}(A)$. I wrote "also" because we started out assuming only that $p$ is an interior point of $A$. $\endgroup$ – Mario Carneiro Feb 20 '15 at 16:25
  • $\begingroup$ I see, well Thank you that is exactly what I was looking for $\endgroup$ – Wolfy Feb 20 '15 at 16:43
  • $\begingroup$ @MorganWeiss Don't forget to upvote and/or accept any answers you find useful (use the arrow buttons and checkmark next to the answers). $\endgroup$ – Mario Carneiro Feb 21 '15 at 1:42

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