I need to find the following set
$S = \displaystyle\bigcap_{n=1}^{\infty} \left[2-\dfrac{1}{n} , 3 + \dfrac{1}{n}\right]$
Now using hints from these questions :Union and intersection of the family of sets $[-1,1-\frac1n]$; describe and prove! ,and Does $\bigcup_{n=1}^\infty \left(-\infty, 1-\frac{1}{n}\right) = (-\infty, 1)$?
I have tried to solve this problem on my own and would like to know if it is right or wrong.
So, I claim : The given set $S = (2,3)$
Suppose $y \in S$ then clearly $y \gt 2$ and $y \lt 3$ Hence $y \in (2,3)$
So , $S \subset (2,3)$
Now I am having difficulty in proving the reverse claim ie How can I prove that
$(2,3) \subset S$ . I think to produce an $n$ which satisfies the given inequality I need to use Archimedian Property , But I am not sure how to do that.
I have two questions at this point.
(i) Is my proof upto given point ie $S \subset (2,3)$ correct ?
(ii) How can I show the reverse claim ?
Thank you.
Best Answer
Note that a countable intersection of closed sets is closed.
Let $x \in [2,3]$
Then $\forall n \in \Bbb{N}$ we have that $$2-\frac{1}{n} <2\leq x \leq 3 < 3 + \frac{1}{n}$$
Thus $[2,3] \subseteq S$
Let $x \in S$ then $2-\frac{1}{n} \leq x \leq 3 + \frac{1}{n},\forall n\in \Bbb{N}$
The sequence $2+\frac{1}{n} \to 2$ and $\frac{1}{n}+3 \to 3$
also the constant sequence $a_n=x \to x$
So $2\leq x\leq 3$ since limits of sequences preserve inequalities.