How to conclusively determine the interior of a set
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.
$\endgroup$ 44 Answers
$\begingroup$Given $[0,3]\cup (3,5)$, we will prove that the interior contains $(0,5)$. Let $b\in (0,5)$. Three cases:
$b<3$. Then there is a neighborhood $(b-\epsilon, b+\epsilon)$ containing $b$ within $[0,3]$. We may find $\epsilon$ by taking $\min(b/2,(3-b)/2)$.
$b=3$. $(1,4)\subseteq [0,3]\cup(3,5)$ works.
$b>3$. Similar to case 1.
Now prove that no other points are in the interior.
$\endgroup$ 2 $\begingroup$The open sets in $\Bbb R$ are just unions of open intervals, and each interval must contain an irrational point.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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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