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My text provides the following two definitions and states that once combined they yield the Theorem. I do not understand. Can anyone explain it in a different way?

Def. Given a subset $A$ of a topological space, a point $x$ is said to be in $\operatorname{Int}(A)$ if $A$ is a neighborhood of $x$.

Def. Given a subset of $A$ of a topological space, if $x\in\operatorname{Int}(A)$, then $x \in U$ for some open set $U\subseteq A$.

Theorem: Given a subset $A$ of a topological space, \begin{equation} \operatorname{Int}(A) = \bigcup \{U \subseteq A : U \text{ is an open set}\} \end{equation}

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  • $\begingroup$ What is your definition of "neighbourhood"? Because to me, definition 2 defines what a non-open neighbourhood is. I.e. a set $A\subseteq X$ is a neighbourhood of $x$ if there is an open $U$ with $x\in U\subseteq A$. $\endgroup$ – Arthur Apr 11 '16 at 21:00
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A neighborhood of $x$ is a subset which contains an open set which contains $x$.
In other words: if $x\in G$ and $G$ is open, then $G$ is a neighborhood of $x$, and for any $H\supseteq G$, so is $H$.

For the theorem: we need to prove two containments:

  1. if $x\in Int(A)$, then $x$ is in the union of the right hand side, i.e. $x\in U$ for some some open $U\subseteq A$.
  2. if $x$ is in the union of open sets within $A$, then $x\in Int(A)$.

Basically, both are just the definition.


An alternative illustration: if $x\in A$ then $x$ is either in the interior of $A$ or on the boundary of $A$.
(The boundary of $A$ contains those points $x$ for which every neighborhood of $x$ intersects both $A$ and $A$ complement.)

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  • $\begingroup$ If x is in Int(A) and Int(A) is a subset of A, then x is in A. As defined U is a subset of A so x is in U? $\endgroup$ – NUG Apr 11 '16 at 21:41
  • $\begingroup$ $Int(A)$ is the largest open set inside $A$. For. 1, we need to find any such $U$ (open, contains $x$ and contained in $A$), But. let's take $U:=Int(A)$. $\endgroup$ – Berci Apr 11 '16 at 21:52

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