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.

  • $\begingroup$ Consider: is $2\in S$? $3\in S$? $\endgroup$ Oct 28 '19 at 20:56
  • $\begingroup$ @J.W.Tanner:Can you please elaborate your hint ? $\endgroup$
    – zeroflank
    Oct 28 '19 at 20:58
  • $\begingroup$ $y$ need not be greater than $2$ or less than $3$. $\endgroup$ Oct 28 '19 at 20:59
  • $\begingroup$ If $y\in S$ then $y>2-\frac1n\;\forall n\in \mathbb N$ -- that means $y\ge2$, which is slightly different from what you said $(y>2)$ $\endgroup$ Oct 28 '19 at 20:59

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.

  • $\begingroup$ You've only shown one implication. $\endgroup$ Oct 28 '19 at 21:01
  • $\begingroup$ Thank you for the answer, I have one doubt ,In the first inequality where you say, $2-1/n \le 2$ why are you using a equality because for any $n \in N$ we will always have a strict inequality . $\endgroup$
    – zeroflank
    Oct 28 '19 at 21:15
  • $\begingroup$ @zeroflank it does not matter since if somethig is bigger than something else then it will be also bigger equal.. i will edite though if you want. $\endgroup$ Oct 28 '19 at 21:17
  • 1
    $\begingroup$ @MariosGretsas: Thank you for writing such a detailed answer. $\endgroup$
    – zeroflank
    Oct 28 '19 at 21:19

You missed the boundary point, $2$ and $3$ because these points are included in all of your sets, therefore $$S = \displaystyle\bigcap_{n=1}^{\infty} \left[2-\dfrac{1}{n} , 3 + \dfrac{1}{n}\right]=[2,3]$$


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