Intervals as infinite unions While studying σ-algebras for probability, at some point I needed some knowledge from topology, dealing with unions of intervals.
I haven't studied topology so far but I do know that, for example, to write an open interval as unions of closed intervals we use: $$(x,y) = \bigcup_{n=1}^{\infty}[x+\frac{1}{n},y-\frac{1}{n}]$$
So my question is, if on the right-hand side we replaced the closed interval with an open one, like so: $ \bigcup_{n=1}^{\infty}(x+\frac{1}{n},y-\frac{1}{n})$, are we still capturing every element in $(x,y)$ this way too? If not then what is the meaning of this, if it has any?
 A: Yes we are. 
For every $n$ we have $(x+\frac1n,y-\frac1n)\subseteq(x,y)$ so that: $$\bigcup_{n=1}^{\infty}(x+\frac1n,y-\frac1n)\subseteq(x,y)$$
If $a\in(x,y)$ then for $k$ large enough we have $a\in(x+\frac1k,y-\frac1k)\subseteq\bigcup_{n=1}^{\infty}(x+\frac1n,y-\frac1n)$ so that: $$(x,y)\subseteq\bigcup_{n=1}^{\infty}(x+\frac1n,y-\frac1n)$$
A: Yes, $\bigcup_{n=1}^{\infty}{[x+\frac{1}{n}, y-\frac{1}{n}]}=\bigcup_{n=1}^{\infty}{(x+\frac{1}{n}, y-\frac{1}{n})}$. To see this, you can take both inclusions. 
The open sets are obviously included in the closed sets, i.e. $(x+\frac{1}{n}, y-\frac{1}{n})\subset[x+\frac{1}{n}, y-\frac{1}{n}]$. This gives the inclusion $RHS \subset LHS$.
It's also true that $[x+\frac{1}{n}, y-\frac{1}{n}] \subset (x+\frac{1}{n+1}, y-\frac{1}{n+1})$ for any $n\in\mathbb{N}$. From this the opposite inclusion follows, $LHS \subset RHS$. Since both sides are subsets of eachother, they must be the same set, i.e. $LHS=RHS$.
As to the meaning of it? There are several possible reasons it's been presented to you this way. It might be to show that the infinite union of closed sets needs not be closed, or to emphasize that there may be more than one way to write any given set. I'm not sure there's any inherent meaning to it other than the interval $(x,y)$ that it defines.
