Proof that in the family of sets $A_t = (\frac{t}{2} ; \frac{t+1}{2})$ for $t \in (0;1)$, $\bigcup A_{t} = (0;1)$, $\bigcap A_{t} = \{\frac{1}{2}\}$ As in the title, I have to proof that for the family of sets given as: $A_t = (\frac{t}{2} ; \frac{t+1}{2})$ for $t \in (0;1)$:

*

*$\bigcup\limits_{t \in (0;1)} A_{t} = (0;1)$

*$\bigcap\limits_{t \in (0;1)} A_{t} = \{\frac{1}{2}\}$
I don't know how to do the proof form right to left. I mean proving that for every $x$ there is $t$ that represents a set that includes $x$. Usually I would use Archimedes' Axiom but out here I have no natural numbers. How then prove that for x close to 0 there is a set with a boundary that is even closer to 0? I need to do it "in the way of set theory" - I mean that I can not use boundary-value analysis nor function analysis.
 A: I suggest to prove the first equality with double inclusions:
since for every $t\in(0,1)$ one has $0<\frac t2<\frac{t+1}2<1$, it follows that $A_t\subset(0,1)$ for all such $t$, hence $\bigcup_{t\in(0,1)}A_t\subseteq(0,1)$.
Conversely take any $x\in(0,1)$ and search for some $t\in(0,1)$ such that $x\in A_t$, that is $\frac t2<x<\frac{t+1}2$; you'll see that $t=2x-\frac12$ will work. Hence $x\in\bigcup_{t\in(0,1)}A_t$ so you conclude that $(0,1)\subseteq
\bigcup_{t\in(0,1)}A_t$ and we are done with the first one.
Second one: we want to study $\bigcap_{t\in(0,1)}A_t$, that is we search for the elements $x$ belonging to $A_t$ for all $t\in(0,1)$:
\begin{align*}
x\in A_t\;\forall t\in(0,1)\;
&\Longleftrightarrow\frac t2<x<\frac{t+1}2\;\;\forall t\in(0,1)\\
\end{align*}
you can see that $x=1/2$ is satisfies this condition, that is
$$
\left\{\frac12\right\}\subseteq\bigcap_{t\in(0,1)}A_t$$
Moreover for every $x<1/2$ or $x>1/2$ you can find some $t\in(0,1)$ such that the last two equality are not simultaneously satisfied, from which the last inclusion is indeed an equality.
