Calculate the following union (proof needed)
$\bigcup_{0 < \epsilon \leq 1} [a-\epsilon, b+\epsilon), a<b \in \mathbb{R}$
and the intersection (with proof)
$\bigcap_{\epsilon > 0} [a-\epsilon,b+\epsilon], a < b \in \mathbb{R}$
I have tried and I do not know. Usually, in these type of exercises I know what the answer should be and then try to prove it, but I do not even the what the union and intersection would be. I am having difficulties with the epsilon
On
Let $a,b \in \mathbb{R}$ be fixed. Then $$\bigcup_{0 < \varepsilon \leq 1} [a-\varepsilon, b+ \varepsilon) = [a-1,b+1).$$ To see this, denote $f_{\varepsilon_i} = [a-\varepsilon_i, b+ \varepsilon_i)$. Then for $\varepsilon_2 > \varepsilon_1$, we clearly have that $f_{\varepsilon_1} \subset f_{\varepsilon_2}$ since $[a-\varepsilon_1, b+ \varepsilon_1) \subset [a-\varepsilon_2, b+ \varepsilon_2)$. Thus each interval in the union contains all the previous ones so it simply becomes the last one in the union.
Similarly, $$\bigcap_{\varepsilon > 0} [a-\varepsilon, b+ \varepsilon] = [a,b].$$ The reasoning is very similar, just note that each term in the intersection is contained in all the previous ones so the intersection is just the 'smallest one'. In this case, there is no smallest one, but we can easily take the limit as $\varepsilon \rightarrow 0$ and we get the desired result.
Claim 1: $\bigcup_{0 < \epsilon \leq 1} [a-\epsilon, b+\epsilon) =[a-1, b+1)$.
"$\subset$" is obvious, since for every $0 < \epsilon \leq 1$ the set $[a-\epsilon, b+\epsilon) $ is contained in $[a-1, b+1)$.
"$\supset$": also obvious, since the union contains $[a-1, b+1)$ as $\epsilon$ runs for $0 < \epsilon \leq 1$.
Claim 2: $\bigcap_{\epsilon > 0} [a-\epsilon,b+\epsilon]= [a,b]$
"$\subset$": Let $x \not \in [a,b]$. Assume, $x <a$. Then we can find $\epsilon_x >0$ with $a-x > \epsilon_x$. Therefore $x \not \in [a-\epsilon_x,b+\epsilon_x]$. Case $x >b$: similar.
"$\supset$": since $[a-\epsilon,b+\epsilon] \supset [a,b]$