# How to find the set which is written as an intersection of infinite sets?

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.

• Consider: is $2\in S$? $3\in S$? Oct 28 '19 at 20:56
• @J.W.Tanner:Can you please elaborate your hint ? Oct 28 '19 at 20:58
• $y$ need not be greater than $2$ or less than $3$. Oct 28 '19 at 20:59
• 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 $(y>2)$ 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.

• You've only shown one implication. Oct 28 '19 at 21:01
• 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 . Oct 28 '19 at 21:15
• @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. Oct 28 '19 at 21:17
• @MariosGretsas: Thank you for writing such a detailed answer. 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 $$S = \displaystyle\bigcap_{n=1}^{\infty} \left[2-\dfrac{1}{n} , 3 + \dfrac{1}{n}\right]=[2,3]$$