Preliminary Definitions
Definition (Open Sets and Topological Space): Let $X$ be a set. A topology on $X$ is a subset $\mathcal{T}\subseteq \mathcal{P}(X)$ with the following properties:
- $\emptyset \in \mathcal{T}$ and $X\in\mathcal{T}$
- Unions of elements of $\mathcal{T}$ belong to $\mathcal{T}$.
- Intersectiosn of finitely many elements of $\mathcal{T}$ belong to $\mathcal{T}$.
The elements $T_{i}\in\mathcal{T}$ are called open sets and the pair $(X,\mathcal{T})$ is called a topological space.
Definition (Neighbourhood): Let $(X,\mathcal{T})$ be a topological space. A subset $U\subseteq X$ is called a neighbourhood of a point $x\in X$, if $U$ contains an open set $T_{x}\in\mathcal{T}$ containing $x$, i.e. $x\in T_{x}\subseteq U$.
Definition (Interior points and boundary points): Let $(X,\mathcal{T})$ be a topological space and $A\subseteq X$ a subset of $X$.
- A point $x\in X$ is called an interior point of $A$, if $A$ is a neighbourhood of $x\in X$. The set of all interior points of $A$ is called the interior of $A$, denoted as $A^{\circ}$.
- A point $x\in X$ is called a boundary point of $A$, if every neighbourhood of $x\in X$ contains points of $A$ and points of $X\setminus A$.
Question
Some texts state an open set $A\subseteq \mathcal{T}$ contains all interior points of $A$ but no boundary points of $A$ (which sounds kinda plausible since in the set of real numbers, for example, we call $(a,b)$ an open intervall and it obviously contains all interior points of that intervall but none of the two boundary points).
I tried to prove that statement using the definitions given above without success so far. Any ideas?
Best Answer
All sets contains its interior points by definition, because if $U$ is neighborhood of $x$ then $x\in U$
But if $A$ is open then all its points are interior points.
But interior point can't be boundary point, because if $x \in A^{\circ}$ then is neighborhood of $x$, but $A$ contains no points of $X\setminus A$, so $x$ not boundary for $A$.
Therefore $A$ contains no boundary points.