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?
 A: 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.
A: $$\bigcap_{n\in\mathbb N}\{x\in\mathbb R^2\mid \|x\|\lt1+1/n\}=\{x\in\mathbb R^2\mid \|x\|\leqslant1\}$$
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.
A: 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.
A: 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: 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.
A: 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
$$\bigcap_{n\in \mathbb{N}} \left\{ x \in \mathbb{R}^n \text{ such that } \|x\| < 1+\frac{1}{n} \right\}= \{ x\in \mathbb{R}^n \text{ such that } \|x\|\leq 1\}$$ 
A: To answer the question from a different perspective, with a different example -
Baby Rudin Example 2.25 also addresses this exact thing. If $G_{n}$ = the open segment $(\frac{-1}{n}, \frac{1}{n})$ for $n = 1, 2, 3, ...$, Now each $G_{n}$ is an open subset of $\bf{R}$.
Consider now $G = \cap_{n=1}^{n=\inf}G_{n}$.
$G$ then is a single point $0$ (i.e. a closed set), and hence cannot be an open subset of $\bf{R}$
