Finding a closed form expression for an infinite union

[Mr Davis 97]

Homework Statement

Show that $\displaystyle \bigcup_{n=2}^\infty \left[ \frac{1}{n} , \frac{n}{n+1} \right] = (0,1)$.

Homework Equations

The Attempt at a Solution

I'm not sure how to show this rigorously. It is sufficient to note that $\lim_{n\to\infty} \frac{1}{n} = 0$ and that $\lim_{n\to\infty}\frac{n}{n+1} = 1$? How can I verify that this union actually isn't $[0,1]$?

 

[Orodruin]

No, it is not sufficient to consider the limits. You need to show that for any $x \in (0,1)$ there is (at least) one set in the union that contains that $x$ and for any $x \notin (0,1)$ there is no set in the union that contains that $x$.

 

[RPinPA]

Note that the union is over finite values of n. x is an element of the union if and only if it falls in least one of the intervals for some finite n.

To show set equality you generally have to argue the subset relationship in both directions. So your proof has two parts.
1. Let x be in the union, i.e. in $\left [ 1/n, n/(n+1)\right ]$ for some $n$. Show it's in (0, 1). That means by definition the union is a subset of (0, 1).
Edit to add: To be more precise, that means the selected interval is a subset of (0, 1). But the choice of $n$ was arbitrary, so every such interval is a subset of (0, 1) and therefore so is the union.

2. Let x be in (0, 1). Show it falls in one of the intervals and therefore in the union. Therefore (0,1) is a subset of the union.

Final line of the proof, $A \subseteq B$ and $B \subseteq A$ therefore $A = B$

That's the general form of a proof that two sets are equal. The structure suggested by @Orodruin is equivalent.

You need to show that for any $x \in (0,1)$ there is (at least) one set in the union that contains that $x$

That's my part 2.

and for any $x \notin (0,1)$ there is no set in the union that contains that $x$.

That's the contrapositive of my part 1, and therefore equivalent.