🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

We define the sequence $A_n:=(a,b-\frac{1}{n}]$ for each $n\in {\mathbb{N}}_1$ . We first show that the sequence is increasing, let $k\in {\mathbb{N}}_1$ in, and note that $k<k+1$, that is $\frac{1}{k}>\frac{1}{k+1}$, therefore $(a,b-\frac{1}{k}]\subseteq (a,b-\frac{1}{k+1}]$, so that $A_k\subseteq A_{k+1}$, therefore $\left(A_n\right)$ is increasing.

We now show that ${\bigcup }_{i=1}^{\infty }{A}_n=(a,b)$. Let $x\in {\bigcup }_{i=1}^{\infty }{A}_n$, by definition this means that there is some $j\in {\mathbb{N}}_1$ such that $x\in A_j$, in other words $x\in (a,b-\frac{1}{j}]$ since $b-\frac{1}{j}<b$ then we know that $(a,b-\frac{1}{j}]\subseteq (a,b)$ (by the above lemma), therefore we know that $x\in (a,b)$ .

We now work on the opposite inclusion, so assume that $x\in (a,b)$, note that $x<b$, so that $b-x>0$, and define $N:=\lfloor \frac{1}{b-x}\rfloor +1$, note that $N>0$ then we have the following: $\frac{1}{b-x}<N$ therefore by multiplying each side by $\frac{1}{N}$ and then $b-x$, we obtain that $\frac{1}{N}<b-x$.

Now notice that $\frac{1}{N}<b-x$ is equivalent to $-\frac{1}{N}>x-b$ which is the same as $b-\frac{1}{N}>x$ thus we can also say that $x<b-\frac{1}{N}<b$ and we we clearly see that $x\in (a,x]$ therefore we can conclude that $x\in (a,b-\frac{1}{N}]$, but $(a,b-\frac{1}{N}]$ is exactly $A_N$, so we've shown there is some $k\in {\mathbb{N}}_1$ such that $x\in A_k$ (namely $k=N$ ) therefore by definition $x\in {\bigcup }_{i=1}^{\infty }{A}_i$ as needed.