Show that $[a,b] = \bigcap_{n=1}^{\infty} (a-1/n, b+1/n).$ 
Show that $[a,b] = \bigcap_{n=1}^{\infty} (a-1/n, b+1/n).$ 

Attempt: $a> a -1/n$ and $b < b+1/n$. This implies $[a,b] \subset (a-1/n, b+1/n)$. This implies $[a,b] \subset \bigcap_{n=1}^{\infty} (a-1/n, b+1/n) $.

Show that $(a,b) = \bigcup_{n=1}^{\infty} [a+1/n, b-1/n].$ 

Attempt: Similarly, $a< a+1/n$ and $b> b-1/n$. Therefore, $[a+1/n, b-1/n] \subset (a,b)$. This implies $\bigcup_{n=1}^{\infty} [a+1/n, b-1/n] \subset (a,b)$.
Is this correct? and could you give some hint for how to proceed the other side? 
Thank you in advance. 
 A: It is fine so far. For added correctness I would add "for any $n$" somewhere appropriate in both cases. For instance right before or right after the inequalities.
For the top one, I would continue by considering an $x\notin [a,b]$ and show that it is not in the intersection (i.e. that at least one of the sets you're intersecting doesn't contain $x$). For the bottom one I would continue by considering a $y\in(a,b)$ and show that it is in the union (i.e. that at least one of the sets you're unioning contains $y$). The Archimedean property will be of use in both cases.
A: Here is another approach that might help you.  I would try to avoid proving both directions separately, and treat this as a simplification problem, simplifying the right hand side to the left hand side.$%
\require{begingroup}
\begingroup
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
%$
For example, for the second problem, which $\;x\;$ are in the RHS set?  We calculate, for any $\;x\;$,
$$\calc
    x \in \bigcup_{n=1}^{\infty} [a+1/n, b-1/n].
\op\equiv\hint{definition of $\;\bigcup\;$}
    \langle \exists n : n \ge 1 : a+1/n \le x \le b-1/n \rangle
\op\equiv\hint{arithmetic -- the only real moment of inspiration}
    \langle \exists n : n \ge 1 : a+1/n \le x \le b-1/n \;\land\; a<x<b \rangle
\op\equiv\hint{...a simple calculation to isolate $\;n\;$...}
    \langle \exists n : n \ge 1 : n \ge \ldots \land n \ge \ldots \rangle \;\land\; a<x<b
\op\equiv\hint{choose a large enough $\;n\;$}
    a<x<b
\op\equiv\hint{definition of open interval}
    x \in (a,b)
\endcalc$$
That proves the second statement.
$%
\endgroup
%$
