Having defined metric balls and neighbourhoods, we now use them to define open sets and interior. An open set is a set that contains a small ball around each of its points. The interior of a set is the collection of all points where this happens. These ideas are the basis of topology in metric spaces and will be used in closure, boundary, subspaces, density, and continuity. :::definition[Open Set] Let $(X,d)$ be a metric space. A subset $G$ of $X$ is called an $open set$ if for every $x\in G$, there exists $r>0$ such that $$\begin{equation*} S(x,r)\subset G. \end{equation*}$$ Equivalently, every point of $G$ has a neighbourhood contained in $G$. ::: :::example[Open Intervals in the Real Line] In $\mathbb{R}$ with the usual metric, every bounded open interval is open because it is an open ball. The interval $(a,\infty)$ is also open. If $x\in(a,\infty)$, choose $r=x-a>0$. Then $S(x,r)\subset(a,\infty)$. ::: :::example[Open Sets in a Discrete Metric Space] Let $(X,d)$ be a discrete metric space. For every $x\in X$, $$\begin{equation*} S\left(x,\frac12\right)=\{x\}. \end{equation*}$$ If $A\subset X$ and $x\in A$, then $S(x,1/2)\subset A$. Hence every subset of a discrete metric space is open. ::: :::theorem[Algebra of Open Sets] Let $(X,d)$ be a metric space. Then $\varnothing$ and $X$ are open, the union of any family of open sets is open, and the intersection of any finite family of open sets is open. ::: :::proof Given that $(X,d)$ is a metric space. To prove the algebra of open sets. The empty set has no point at which openness can fail. For $X$, every ball centred at a point of $X$ lies in $X$. Hence $\varnothing$ and $X$ are open. Let $\{G_\alpha:\alpha\in\Lambda\}$ be open and put $G=\bigcup_{\alpha\in\Lambda}G_\alpha$. If $x\in G$, then $x\in G_{\alpha_0}$ for some $\alpha_0$. Since $G_{\alpha_0}$ is open, there exists $r>0$ such that $S(x,r)\subset G_{\alpha_0}\subset G$. Hence $G$ is open. Let $G_1,\ldots,G_n$ be open and put $H=\bigcap_{i=1}^{n}G_i$. If $x\in H$, then for each $i$ there is $r_i>0$ such that $S(x,r_i)\subset G_i$. Put $r=\min\{r_1,\ldots,r_n\}$. Then $r>0$ and $S(x,r)\subset H$. Hence $H$ is open. $\square$ ::: :::remark Infinite intersections of open sets need not be open. In $\mathbb{R}$, the sets $G_n=(-1/n,1/n)$ are open for every $n$, but $$\begin{equation*} \bigcap_{n=1}^{\infty}G_n=\{0\}, \end{equation*}$$ which is not open in the usual metric. ::: :::theorem[Open Sets as Unions of Balls] Let $(X,d)$ be a metric space. A subset $G$ of $X$ is open if and only if $G$ is a union of open balls. ::: :::proof Given that $G\subset X$. To prove the equivalence. Suppose $G$ is open. For each $x\in G$, choose $r_x>0$ such that $S(x,r_x)\subset G$. Then $$\begin{equation*} G=\bigcup_{x\in G}S(x,r_x). \end{equation*}$$ Conversely, suppose $G$ is a union of open balls. Since every open ball is open and arbitrary unions of open sets are open, $G$ is open. $\square$ ::: :::definition[Interior Point] Let $(X,d)$ be a metric space and let $A\subset X$. A point $x\in X$ is called an $interior point$ of $A$ if there exists $r>0$ such that $$\begin{equation*} S(x,r)\subset A. \end{equation*}$$ ::: :::definition[Interior of a Set] Let $(X,d)$ be a metric space and let $A\subset X$. The set of all interior points of $A$ is called the $interior$ of $A$ and is denoted by $A^\circ$. Thus $$\begin{equation*} A^\circ=\{x\in X:\text{there exists }r>0\text{ such that }S(x,r)\subset A\}. \end{equation*}$$ ::: :::example[Interior of a Closed Interval] In $\mathbb{R}$ with the usual metric, $$\begin{equation*} [0,1]^\circ=(0,1). \end{equation*}$$ The endpoints $0$ and $1$ are not interior points because every open interval around either endpoint contains points outside $[0,1]$. ::: :::theorem[Interior as the Largest Open Subset] Let $(X,d)$ be a metric space and let $A\subset X$. Then $A^\circ$ is open, $A$ is open if and only if $A=A^\circ$, and if $G$ is open with $G\subset A$, then $G\subset A^\circ$. ::: :::proof Given that $A\subset X$. To prove the three properties. Let $x\in A^\circ$. Then there exists $r>0$ such that $S(x,r)\subset A$. Since $S(x,r)$ is open, every point of $S(x,r)$ is an interior point of $A$. Hence $S(x,r)\subset A^\circ$, so $A^\circ$ is open. If $A$ is open, then every point of $A$ is an interior point, so $A\subset A^\circ$. Since $A^\circ\subset A$, we get $A=A^\circ$. Conversely, if $A=A^\circ$, then $A$ is open. If $G$ is open and $G\subset A$, then each $x\in G$ has a ball contained in $G$, hence contained in $A$. Thus $x\in A^\circ$ and $G\subset A^\circ$. $\square$ ::: :::theorem[Algebra of Interiors] Let $(X,d)$ be a metric space and let $A,B\subset X$. Then $A\subset B$ implies $A^\circ\subset B^\circ$, $$\begin{equation*} (A\cap B)^\circ=A^\circ\cap B^\circ, \end{equation*}$$ and $$\begin{equation*} A^\circ\cup B^\circ\subset(A\cup B)^\circ. \end{equation*}$$ ::: :::proof Given that $A,B\subset X$. To prove the stated relations. If $A\subset B$ and $x\in A^\circ$, then some ball centred at $x$ is contained in $A$, hence in $B$. Therefore $x\in B^\circ$. If $x\in(A\cap B)^\circ$, then some ball centred at $x$ is contained in both $A$ and $B$, so $x\in A^\circ\cap B^\circ$. Conversely, if $x\in A^\circ\cap B^\circ$, choose $r_1,r_2>0$ with $S(x,r_1)\subset A$ and $S(x,r_2)\subset B$. Put $r=\min\{r_1,r_2\}$. Then $S(x,r)\subset A\cap B$, so $x\in(A\cap B)^\circ$. The inclusion $A^\circ\cup B^\circ\subset(A\cup B)^\circ$ follows from monotonicity of interior. $\square$ ::: :::exercise [1] Prove that $\mathbb{R}$ is open in itself. [2] Find the interior of $[2,5]\cup\{8\}$ in $\mathbb{R}$. [3] Give an example where $(A\cup B)^\circ\neq A^\circ\cup B^\circ$. ::: :::answer [1] Every ball centred at a real number lies in $\mathbb{R}$. [2] The interior is $(2,5)$. [3] Let $A=[0,1]$ and $B=[1,2]$. Then $(A\cup B)^\circ=(0,2)$, while $A^\circ\cup B^\circ=(0,1)\cup(1,2)$. ::: :::faq[Frequently Asked Questions] Q: Is every open set a union of open balls? A: Yes. Each point contributes a ball contained in the open set. Q: Is the interior always open? A: Yes. It is the largest open subset contained in the set. Q: Why do finite intersections matter? A: The minimum of finitely many positive radii is still positive. ::: :::call-to-action[Continue to Limit Points] subtitle: Use open sets and interior to prepare for limit points, derived sets, and closure. button: Next Lesson | /functional-analysis/metric-spaces/topological-structure-of-metric-spaces/limit-points-and-closure button-ghost: Previous Lesson | /functional-analysis/metric-spaces/topological-structure-of-metric-spaces/metric-balls-and-neighbourhoods :::