Can a closed set in $\Bbb R$ be written in terms of open sets Is it possible to write any non-empty closed set in $\Bbb R$ as a combination of unions / intersection of open sets. Note that I don't demand just one union / intersection. I am happy with any combination (finite) of unions / intersection, but elements in each union/intersection must be open set. Any finite set, closed interval, cantor set can be written like this. 
 A: How about this:
$$\bigcap_{n\in\mathbb N}\left(-\frac{1}{n}, 1+\frac{1}{n}\right).$$
Can you see what this set is?
In fact, if you are not limited to countable unions, you can really go wild. Since every set is the union of its elements, you can also do this: take an arbitrary set $A$. Then for every $a\notin A$, the set $\mathbb R \setminus \{a\}$ is open. Then the set
$$B=\bigcap_{a\notin A} \left(\mathbb R \setminus \{a\}\right).$$
If you doubt that the sets are equal, look at the complement of $B$. It equals
$$B^c=\bigcup_{a\notin A}\{a\} = A^c,$$
meaning $A=B$.
A: Since any union of open sets is open, any non-open set cannot be written as a union of open sets.
On the other hand any closed subset $C$ can be written as intersection of open subsets, namely as the intersection of all open subsets containing $C$ (since $\Bbb R$ is such a subset, the collection is not empty). To see that $C$ equals this intersection, note first that it is trivially contained in the intersection. Now for any point$~x$ in the complement of $C$, there is an open neighbourhood of$~x$ contained in the complement of$~C$ (because $C$ is closed), and it is easily seen to contain a closed neighbourhood $N$ of$~x$. Now the complement of $N$ is an open set containing $C$, therefore being used in the intersection of open subsets, and it shows that $x$ is not in that intersection: the intersection is contained in (and hence equal to) $C$. One could limit $N$ to be an interval with rational bounds, showing that the $C$ is also the intesection of a countable familty of open subsets.
A: For any set $A\subset \mathbb R$ define $B(A, r)=\left\{x\in\mathbb R:\text{dist(x,A) < r}\right\}$, where $\,\text{dist(x,A)} = \inf_{y\in A}\rho(x,y)$ with metric $\rho$. $B(A,r)$ is open for any $A$ and $r$. Now for every closed set $F\subset\mathbb R$
$$F = \bigcap_{n\in\mathbb N}B(F, \frac 1 n)$$
So every closed set in $\mathbb R$ is indeed an intersection of countably many open sets (so it is a $G_\delta$ set).
A: All the answers seems to be considering countable operations, but you ask for finite unions and intersections as I read it. Then, the answer is no. To see this, just note that finite intersections of open sets are open and the union of any number of open sets is also open. Thus, if you are only allowed to use finite intersections and unions, you can only ever hope to construct open sets from open sets.
A: Let $[a,b]\subset\mathbb{R}$ be a closed set then you can write $[a,b]=\cap_n(a-\frac{1}{n},b+\frac{1}{n})$ as $n\to\infty$.
A: A topological space is called a $G_\delta$ space if every closed set can be written as a countable intersection of open sets. So here's a crisp, precise version of the question:


*

*Is $\mathbb R$ a $G_\delta$ space?


The answer is yes; see the other comments.
