Intersection of a nested interval of $A_n=\left[3-{\frac{1}{\sqrt{n}},3+\frac{1}{3^n}}\right]$

Question

$A_n=\left[3-{\frac{1}{\sqrt{n}},3+\frac{1}{3^n}}\right]$ What is $\bigcap_{n=1}^{\infty}A_n$

Since every set becomes a subset of the next set, is it correct to say that the intersection of all sets $n$ is just:

$\left[2,\frac{10}{3}\right]$ since this is the only set in all of the sets?

@ZackNi is that because the limit of the fractional parts is $0$ creating an upper bound where everything before it becomes a subset? — nullByteMe, Jun 26 '16 at 11:54

score 1 · Accepted Answer · answered Jun 26 '16 at 11:59

1

$$3+\frac{1}{3^n}-(3+\frac{1}{3^{n-1}}) = \frac{-2}{3^{n}} < 0$$ for all $ n \ge 2$ (1)

$$3-{\frac{1}{\sqrt{n}}}-(3-{\frac{1}{\sqrt{n-1}}}) = \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n-1}\sqrt{n}} > 0 $$ for all $ n \ge 2$(2)

$$\lim_{x\to \infty}3+\frac{1}{3^n} = 3$$

$$\lim_{x\to \infty}3-{\frac{1}{\sqrt{n}}} = 3$$

Therefore By (1), $$3+\frac{1}{3^n} \ge 3$$

Therefore By (2), $$3\ge 3-{\frac{1}{\sqrt{n}}}$$

So the intersection set is $\{3\}$

answered Jun 26 '16 at 11:59

Zau

If this set were open instead of closed this wouldn't work because $3$ is not a member of any of the sets, correct? – nullByteMe Jun 26 '16 at 12:03
If the set is open it doesnt work really. The reason is that since all sets are open, the length of intersected interval shouldn't be 0 but when $n$ goes up to infinity the length of interval approach to 0 – Zau Jun 26 '16 at 12:09
The reason I ask is because my follow up question tells me to find a bounded but not closed nested interval where $\bigcap_{n=1}^{\infty} A_n=0$ and I figure I can use the same example but just open it. – nullByteMe Jun 26 '16 at 12:13
Yeah but is empty set bounded? – Zau Jun 26 '16 at 12:20
But it wouldn't be an empty set since there is no intersection, right? It would just be $0$? Or am I wrong? – nullByteMe Jun 26 '16 at 12:22
No it should be an empty set since (3,3) = {x| 3<x<3} = ∅ and empty set is well bounded because of this post:http://math.stackexchange.com/questions/532233/why-is-the-empty-set-bounded – Zau Jun 26 '16 at 12:25
Sorry, if I'm being unclear or obtuse. I just don't understand what you're saying... If I need to find $\bigcap_{n=1}^{\infty}A_n=\varnothing$ where the set is bounded but not closed, you're saying my set is unbounded? – nullByteMe Jun 26 '16 at 12:30
Let us continue this discussion in chat. – Zau Jun 26 '16 at 12:33

1 Answers1