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I'm fairly confident I understand the meaning of the 'interior' of a set, but I can't figure out how to prove conclusively what said interior is for a given set. Consider this definition:

Let $S$ be a subset of $\mathbb{R}$. A point $x\in\mathbb{R}$ is an interior point of $S$ if there exists a neighborhood $N$ of $x$ such that $N\subseteq S$. The set of all interior points of $S$ is denoted by $\operatorname{int} S$ (or in some texts, $S^\circ$)

These examples were pulled from practice problems in my textbook, which only gives the final answer without showing why.

Given the set $[0,3]\cup(3,5)$, what is the interior? Intuitively I can just look at this and conclude that $[0,3]\cup(3,5)=[0,5)$, and the interior is everything but the boundary points -- namely $(0,5)$.

This may be conclusive enough for such a simple example, but what about the sets $S=\{1/n\mid n\in\mathbb{N}\}$ and $T=\{r\in\mathbb{Q}\mid 0<r<\sqrt{2}\}$? I know that $S^0=T^0=\emptyset$, as all the points in $S$ and $T$ are isolated points. But I don't know how to back up that claim.

Taking $S$, using the definition I pasted above, I feel like I should try to show that for every $x\in S$, there does not exist a neighborhood of $x$ that is a subset of $S$, but I don't know how to proceed with this.

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You would do well to abandon "confidence" and "intuitively" and instead use the definitions you are given. – vadim123 Nov 6 '13 at 1:08
Any open interval would contain irrational numbers. – a12345 Nov 6 '13 at 1:09
For your $[0,5)$ example, you can use the fact that the interior is precisely the complement of the boundary (in the closure of the set in question) to get your (correct) claim that the interior is (0,5). – zibadawa timmy Nov 6 '13 at 1:43
Your title made no sense, as you can only prove statements, and "the interior of a set" is not a statement. I changed the word for one that is more appropriate. – Andrés E. Caicedo Nov 6 '13 at 1:47

Given $[0,3]\cup (3,5)$, we will prove that the interior contains $(0,5)$. Let $b\in (0,5)$. Three cases:

  1. $b<3$. Then there is a neighborhood $(b-\epsilon, b+\epsilon)$ containing $b$ within $[0,3]$. We may find $\epsilon$ by taking $\min(b/2,(3-b)/2)$.

  2. $b=3$. $(1,4)\subseteq [0,3]\cup(3,5)$ works.

  3. $b>3$. Similar to case 1.

Now prove that no other points are in the interior.

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Shouldn't you also prove that points in $(-\infty,0] \cup [5, \infty)$ are not interior points for this to be complete? – MasterOfBinary Nov 6 '13 at 1:16
Good point, adjusted. – vadim123 Nov 6 '13 at 1:17

The open sets in $\Bbb R$ are just unions of open intervals, and each interval must contain an irrational point.

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Let's have a set $A\subset X$ where $X$ is a topological space. Then $intA=\cup${$B\subset A:B $open in $X$}. Also $x\in X$ is an interior point of $A$ if there is a neightborhood $N$ of $x$ such that $N\subset A$. $N$ is a neightborhood of $x$ <=> $x\in intA$

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For a set $ S = \{1/n | n \in \mathbb{N} \} $, as you mention you are aware that $ S $ is simply a collection of isolated points. Therefore, for any given element in $x \in S$, the interval $(x-\delta, x+\delta) \not \subset S $. Thus, the interior of S is the empty set. Similar logic follows for your other examples.

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