Show $\bigcap_{p \in (1, \infty)} (\frac{1}{p}, 3p) \subseteq [1,3]$ 
$\bigcap_{p \in (1, \infty)} (\frac{1}{p}, 3p) \subseteq [1,3]$

My first intuition is to prove the contrapositive instead. That being: 
$x \notin [1,3] \Rightarrow x \notin \bigcap_{p \in (1, \infty)} (\frac{1}{p}, 3p)$
Meaning if we can find at least one $p$ where $x$ isn't a member of $(\frac{1}{p}, 3p)$, then the proof is complete.
Am I on the right track here?
 A: Note that for $p,q\in(1,\infty)$ with $p<q$, we have $$\left(\frac{1}{p},3p\right)\subset
\left(\frac{1}{q},3q\right)$$
because for $1<p<q$, $$\frac{1}{q}<\frac{1}{p}\qquad\text{and}\qquad 3p<3q$$
thus the collection $\left\{\left(\frac{1}{p},3p\right)\right\}_{p\in(1,\infty)}$ is a nested collection of intervals, therefore
$$ \bigcap_{p\in(1,\infty)}\left(\frac{1}{p},3p\right)=(1,3)\subset [1,3]$$
A: This question is about proving the following: 
If $a$ and $b$ are given real numbers, and $a<b+\epsilon $ for all $\epsilon >0$, then $a\leq b$. This can be prove by contrapositive as you suggested. 
Suppose $b<a$, and take $\epsilon =\frac{a-b}{2}$. Then by assumption $a< b+\epsilon= \frac{a+b}{2}$. Which means $a<b$, a contradiction.
A: Option:
$A_p:=(1/p,3p)$ , $p>1$.
$A_p \subset A_q$ for $p<q$.
1) $(1,3)\subset A_p$ , $p>1$.
$(1,3)\subset \cap_{p>1} A_p$; 
2)$\cap_{p>1}A_p \subset (1,3)$.
Let $x \in \cap_{p>1}A_p$ then 
$x$ in every $A_p$ , $p>1$, i.e.
$x$ in every $(1/p,3p)$ , $p>1$;
$1/p<x<3p$, for all $p>1$, i e.
$x \in (1,3)$.
3) $\cap_{p>1}A_p =(1,3)\subset [1,3]$.
