Jtdylktuy

Let $a,binmathbb{R}$ with $a<b$ we must prove that $$(a,b)=bigcup_{n=1}^{+infty} bigg (a,b-frac{1}{n}bigg].$$
Obviously $$bigcup_{n=1}^{+infty}bigg(a,b-frac{1}{n}bigg]subseteq (a,b),$$
in fact, if $xinbigcup_n^{+infty}big(a,b-frac{1}{n}big]$ exists $n_0inmathbb{N}$ such that $xinbig(a,b-frac{1}{n_0}big]subseteq(a,b)$, then $xin(a,b)$.

On the other hand $$bigcup_{n=1}^{+infty}bigg(a,b-frac{1}{n}bigg]supseteq (a,b),$$ about that let $xin (a,b)$. From density of $mathbb{Q}$ in $mathbb{R}$ there are infinite values of $ninmathbb{N}$ such that $xle q_n<b$, therefore $xin (a,q_n]$ for infinite values of $ninmathbb{N}$.

At this point, how do we say that these $q_n$ are of the form $b-frac{1}{n}$?

Thanks!

Suppose $x in (a,b)$, then we have $b-x > 0$.

We can find $N in mathbb{N}, N>0$ such that $frac1N < b-x $ and hence $x < b-frac1N$.

Hence $x in (a, b-frac1N) subset (a,b-frac1N]$

Hence $x in bigcup_{n=1}^infty (a,b-frac1n]$.

Remark: To prove that something is in the union, we just have to show that it is inside one of the set in the union.