# Why can the intersection of infinite open sets be closed?

I learned that the union of open sets is always open and the intersection of a finite set of open sets is open. However, the intersection of an infinite number of open sets can be closed. Apparently, the following example illustrates this.

In $E^2$, let $X$ be the infinite family of concentric open disks of radius $1 + 1/n$ for all $n \in \mathbb{Z^+}$. Why is $X$ a closed set? Can't I create a boundary set for $X$ that encloses all the elements in the interior?

• I thought that the question was about subsets of $\mathbb R^2$ the Euclidean plane? – Did Mar 25 '13 at 6:17

When you take the intersection you will have the set $$\bigcap_{n \in \mathbb{N}} \left(-1-\frac{1}{n}, 1+\frac{1}{n}\right)=[-1,1]$$ and this one is closed.

This gives the idea for $$\bigcap_{n\in \mathbb{N}} \left\{ x \in \mathbb{R}^n \text{ such that } \|x\| < 1+\frac{1}{n} \right\}= \{ x\in \mathbb{R}^n \text{ such that } \|x\|\leq 1\}$$

Look at the radii of your open disks: they’re the numbers $1+\frac1n$ for $n\in\Bbb Z^+$. In order for a point $p$ to be in the intersection of these open disks, its distance from the origin must be less than $1+\frac1n$ for each $n\in\Bbb Z^+$. But the infimum of these radii is $1$, so the intersection is precisely the closed disk $D=\{\langle x,y\rangle\in E^2:x^2+y^2\le 1\}$. And this is a closed set: if $p\notin D$, let $r$ be the distance from $p$ to the origin. Then $p>1$, and the open $(p-1)$-ball centred at $p$ is disjoint from $D$. Thus, $E^2\setminus D$ is open, and $D$ must be closed.

$$\bigcap_{n\in\mathbb N}\{x\in\mathbb R^2\mid \|x\|\lt1+1/n\}=\{x\in\mathbb R^2\mid \|x\|\leqslant1\}$$ The set on the RHS is the closed unit ball. Its boundary $\{x\in\mathbb R^2\mid \|x\|=1\}$ is the unit sphere.

Each ball $\{x\in\mathbb R^2\mid \|x\|\lt1+1/n\}$ is open. The closed unit ball $\{x\in\mathbb R^2\mid \|x\|\leqslant1\}$ is, well... closed. The unit sphere $\{x\in\mathbb R^2\mid \|x\|=1\}$ is closed.

Note that in any metric space the closed balls $\overline{B} ( x ; r ) = \{ y \in X : d ( x , y ) \leq r \}$ are closed sets, and for $r < r^\prime$ we have $$B ( x ; r ) \subseteq \overline{B} ( x ; r ) \subseteq B ( x ; r^\prime ).$$

So when you are talking about the intersection $$\bigcap_{n=1}^\infty B ( x ; 1 + \tfrac{1}{n} )$$ we can interleave these open balls with closed balls: $$\cdots \supseteq \overline{B} ( x ; 1 + \tfrac 1n ) \supseteq B ( x ; 1+\tfrac 1n ) \supseteq \overline{B} ( x ; 1 + \tfrac 1{n+1} ) \supseteq \cdots$$ and it is not too difficult to see that $$\bigcap_{n=1}^\infty B ( x ; 1 + \tfrac 1n ) = \bigcap_{n=1}^\infty \overline{B} ( x ; 1 + \tfrac 1n )$$ and the expression of the right-hand-side is an intersection of closed sets, and so the intersection must be closed.

The intersection will be the closed disk of radius $1$, i.e. having its boundary. Of course, you can remove the boundary, and that's open, but that's another set.

It is easy to see why an infinite intersection of open sets can be closed:

Given any point $x$ in any topological space, the intersection of all open sets containing $x$ is $\{ x \}$.

Moreover, if your topology has a countable basis of open sets, you can get any element as the intersection of open sets...

• A minor point: if $X$ is not T$_1$, the singleton $\{ x \}$ may not be the intersection of all open neighbourhoods of $x$. – user642796 Mar 24 '13 at 19:24
• yea I forgot to mention that, I am not a topologist so for me all spaces are haussdorff :) – N. S. Mar 24 '13 at 19:30