inf, sup, max, min for $\bigcap_{n \in \mathbb N}\left[-\frac{1}{3^n},4+\frac{1}{2n}\right)$

Question

For the following set do I have the inf, sup, min, max correct?

$\bigcap_{n \in \mathbb N}\left[-\frac{1}{3^n},4+\frac{1}{2n}\right)$

$\text{inf}S=0$

$\text{sup}S= 4$

$\text{min}S=0$

$\text{max}S=$DNE

Additionally, this set is not compact, nor countable, right?

Answer 1

Hints:

• Is there any negative number in the set? Is $0$ in the set?
• Is there any number greater than $4$ in the set? Is $4$ in the set?
• Is there every number between $0$ and $4$ in the set?

What the intersection would be?

EDIT: In the learning steage you seem to be, I think it is not a good idea to trust in what you "see".

You should prove that there are no negative number in the set, and also that there is no number greater than $4$.

As an example, I'll prove the latter.

Take any number $x>4$. Let be $\epsilon=x-4>0$. Since $\lim (4+1/(2n))=4$, by definition of limit, there is some $k\in\Bbb N$ such that $|4+1/(2k)-4|<\epsilon$, that is, $1/(2k)<x-4$. Therefore, $x>4+1/(2k)$. Since $x$ is not in one set that we are intersecting, it is not in the intersection.

Nevertheless, note that $4$ is in every set, so it is in the intersection.

In general,

Let $a_n$ be an increasing sequence that converges to the real number $a$, $b_n$ be a decreasing sequence that converges to the real number $b$. Suppose that $a<b$. Then $$\bigcap_{n\in\Bbb N}(a_n,b_n)=[a,b]$$ The intersection is a closed interval, even if the intervals in the intersection are open, closed or open by one side and closed by the opposite.

Can you prove this?

Answer 2

Recall that, by definition, for any set $A$, collection of subsets $A_n \subset A$, and indexing set $J$,

$$\bigcap_{n \in J} A_n = \{x \in A \ | \ x \in A_n \text{ for all } n \in J\}$$

So, by defintion

$$\bigcap_{n \in \mathbb{N}} \bigg[\frac{-1}{3^n},4+\frac{1}{2n}\bigg) = \bigg\{x \in \mathbb{R} \ | \ x \in \bigg[\frac{-1}{3^n},4+\frac{1}{2n}\bigg) \text{ for all } n \in \mathbb{N}\bigg\}$$

From this, we are able to get

$$S=\bigcap_{n \in \mathbb{N}} \bigg[\frac{-1}{3^n},4+\frac{1}{2n}\bigg)=[0,4]$$

You can argue that $4 \in S$ because $4 \in \big[\frac{-1}{3^n},4+\frac{1}{2n}\big)$ for all $n \in \mathbb{N}$.