Infinite class of closed sets whose union is not closed 
Give an example of an infinite class of closed sets whose union is not closed.

 A: Can you express $(0,1)$ as an increasing union of closed sets? Maybe find a pair of sequences $a_n$ and $b_n$ with $a_n$ decreasing to $0$ and $b_n$ increasing to $1$? Then you can try taking $[a_n,b_n]$ and see if that works.
A: Here is a good example which clearly shows that the infinite union of closed sets may not be closed. consider the usual topology on $\mathbb{R}$, and let $\mathcal{C}$ be the collection of all closed sets of the form $( -\infty , \frac n{n+1} ]$ where $n \geq 1$.  Then $\bigcup \mathcal{C} = ( - \infty , 1 )$, which is open. So this union of infinitely many closed sets is open.
A: Rationals. $\mathbb{Q} = \cup_r \{r\}$ is a countable union of closed sets.
A: Every subset $S\subset X$ of a Hausdorff space is the union of its singleton subsets, which are closed : $$S=\bigcup_{s\in S} \lbrace s\rbrace $$
A: I think probably the most instructive example is considering $\displaystyle A_n=\left[\frac{1}{n},\infty\right)$.
A: The union of intervals of the form $\left[\frac{1}{n} ,1- \frac{1}{n}\right]$ will be one example:
$$
\bigcup_{n=2}^\infty \left[\frac{1}{n} ,1- \frac{1}{n}\right] = (0,1)
$$
The behaviour of the interval is already stated above.
A: As another example, let $X$ be any infinite set, and consider the cofinite topology on $X$ (ie all open sets are either the empty set or sets whose complement is finite).  Every proper closed subset of $X$ is finite.  So, fixing an element $x_0\in X$, we have the union closed sets equaling an open set:  $$X\setminus\{x_0\}=\bigcup\limits_{x\not=x_0} \{x\}$$
A: Example in R, take I= [0,1] with the usual topology inherited from R.
Then the union of all rationals in I
Is a union of singletons,  yet it's complement is not open since every ball around an irrational point contains rational points.
A: Consider the sets $[\frac{1}{n}, 2]$. Each of these sets is clearly closed. However, their union over all $\mathbb{N}$ is $(0, 2]$ which is neither open nor closed.
P.S: Note that the union cannot contain 0 because there is no $n \in \mathbb{N}$ such that $\frac{1}{n} = 0$.
A: Since singletons in R are closed in usual topology.
We can think about infinite class of singletons {x} where x belongs to (0,1] then there union will be (0,1] which is not closed in R.
