Supremum, max, infimum, min of a set

Question

$S=\bigcap_{n \in \mathbb N}\left(3, \frac{7n+1}{n}\right)$

I still haven't full grasped how to solve these nested interval intersections problem as I have issues visualizing what is happening as $n$ increases.

For example, $n=1$ gives $\left(3, 8\right)$, which is the largest interval of this set and I believe this becomes a subset of every other set in the interval, or would it be $(3,7)$, the limit of the interval, that becomes a subset of every set in the interval?

Under the first assumption I get the following:

The interval: $(3,8]$

$\text{inf}S=3$

$\text{min}S=$DNE

$\text{sup}S=8$

$\text{max}S=8$

Answer 1

You have it slightly backwards. $n = 1$ yields $(3,8)$ which is the largest such interval. But this means every interval is a subset of it not the other way around. (Obviously $(3,8) \not \subset (3, 7 1/2)$.

For all $n$, $(7n + 1)/n = 7 + 1/n > 7$ so $(3,7]$ is smaller than all the intervals so $(3,7]$ is a subset of all $(3, (7n+1)/n)$ so $(3,7] \subset \cap \{(3,(7n + 1)/n)\}$.

The thing to note is for all numbers larger than $7$--- let's call such a number $7 + \epsilon$; $\epsilon$ can be any positive non-zero number--- There is a $7 < 7 + 1/n = (7n + 1)/n < 7 + \epsilon$. So $(3,7 + \epsilon) \not \subset \cap \{(3,(7n + 1)/n)\}$.

So, intuitively and visually, we may think of $(3,7]$ as the largest possible subset of $\cap \{(3,(7n + 1)/n)\}$. The "largest possible subset" of something is itself.

So $\cap \{(3,(7n + 1)/n)\} = (3,7]$.

So how to visualize it? Well the set of $(3, (7n + 1)/n)$ form a set of nested interval each on smaller than the previous one. There are an infinite number of them so there is no smallest one. But the get steadily smaller and tend toward a lower limit of $(3,7]$. That limit is the final interval that is the intersection of all of them.

What throws beginners off seems to be that as all the finite intersections-- $\cap_{i \le n}\{(3, (7n + 1)/7)\} = (3, 7 + 1/n)$ -- is open, then the infinite intersection should be open too.

This is not the case. $7 \in (3, (7n + 1)/n)$ for all $n$ so $7 \in \cap \{(3,(7n + 1)/n)\}$. So $(3, 7] \subset \cap \{(3,(7n + 1)/n)\}$.

This does boil done to $\inf$ vs. $\min$. Let $W = \{7n + 1)/n\}$. $\min W$ is the smallest real number in $W$. $\inf W$ is the largest number that is bigger that is at least as small as every real number in $W$. If $\min W$ exists, and there is a smallest member of $W$ then that number is the biggest number that is at least as small as every real number in $W$, so if $\min W$ exists then $\min W = \inf W$ and the intersection will be $(3, \min W)$.

But if $\min W$ doesn't exist, then $\inf W \not \in W$ and $\inf W < z \in W$ for all $z \in W$. So $(3 \inf W] \subset$ of the intersection. But $\inf W$ was the smallest such number. So $(3, z) \not \subset$ of the intersection for any $z > \inf W$. Therefore the intersect would be $(3, \inf W]$.

The latter is the case for $W = \{7n + 1)/n\}$. $\min W$ does not exist. $\inf W = 7$ but $7 \not \in W$. So the intersection is $(3, \inf W] = (3,7]$.

I know it's confusing but think it through and draw pictures. Those open intervals shave down everything above 7 but never actually shave the 7. So the 7 is "permanent", but everything above it is shaved off. Repeat and draw and repeat and draw until it clicks.

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SOrtof a picture:

(3, 7|... < 7 + 1/10000000 < 7 + 1/9999999 < ... 7 + 1/n.... 7 + 1/4 < 7 + 1/3 < 7 + 1/2 < 8)

THe intersection is the smallest of the (3, 7 < 7 + 1/n) but there is no smallest. So it keeps "shaving" off everything even slightly above 7. BUT THE 7 STAYS.

Answer 2

Define $A_n:=(3,\frac{7n+1}{n})$.

Thus we have: \begin{align} S=\bigcap_{n \in \mathbb{N}} A_n \end{align}

What are the $A_n$'s?

\begin{align} A_1=(3,8) \end{align}

\begin{align} A_2=(3,\frac{15}{2}) \end{align}

\begin{align} A_3=(3, \frac{22}{3}) \end{align}

If you look carefully, you will see that $A_{n+1}$ is smaller than $A_n$. So, $A_{n+1} \subset A_n$.

What you have done is what you should do if you take the union of alle sets $A_n, n\in \mathbb{N}$, but you have to take the intersection of them. So, you can't have the element "8" in $S$ if $8\notin A_2$. Every element of $S$ must be in $A_n , \forall n\in \mathbb{N}$.

Because $ A_{n+1} \subset A_n$, you can take limit $n$ to infinity of $A_n$: $\lim_{n\rightarrow \infty} A_n = (3,7] $ (Why is it an closed interval?)

So $S=(3,7] $. The intuition is important! You can now formally prove this by proving $S\subset\bigcap_{n \in \mathbb{N}} A_n$, and $\bigcap_{n \in \mathbb{N}} A_n\subset S$.

proof first part

"$\subset$":

Pick $x\in S$. So, $x$ satisfies the following inequality: $3<x<=7$. For all $n \in \mathbb{N}$ we have: $x<7+\frac{1}{n}$. So, $x\in A_n \forall n\in \mathbb{N}$. This implies $S\subset\bigcap_{n \in \mathbb{N}} A_n$.

Can you do the second part?

If you finish your proof, you will find what the sup, max etc is.

Answer 3

Here's how to visualize and gain insight into this kind of problem. The first thing to do is plug in actual numbers until you get a feel for problem. It's like warming up your brain.

For $n = 1, 2, 3, \dots$ you get $\frac{8}{1}$, which is $8$. Then $\frac{15}{2}$ which is $7 \frac{1}{2}$. Then $\frac{22}{3} = 7 \frac{1}{3}$, and so forth.

Then you see what's going on, and you look back at the formula and you realize that $\frac{7n+1}{n} = 7 + \frac{1}{n}$. Ah, now it's obvious. Each set is a a little bit larger than $(0,7)$ but by an amount that goes to zero. Then you remember your calculus and you see that the limit as $n \rightarrow \infty$ is $7$ and the nature of the sequence of intervals is obvious. It helps a lot to make diagrams and pictures if you are visual. Even most professional mathematicians get their insights from looking at examples and drawing pictures.

Next, what is the intersection of all these intervals? How could a point be in all of them? Well if $x \in (0, 7)$ then certainly $0 < x < 7 + \frac{1}{n}$ for all $n$, so $x$ is in the intersection.

On the other hand if $x \leq 0$, then $x$ isn't not in any of the intervals, let alone their intersection.

And if $x > 7$, then $|x - 7|$ is some positive real number; and then there is some $N \in \mathbb N$ with $0 < \frac{1}{N} < |x - 7|$. Therefore $x \notin (0, 7 + \frac{1}{N})$, so it can't be in the intersection.

I hope I was able to give a sense of how you approach a problem like this. Plug in simple values till you see what's going on. That often sparks understanding of the symbols, which leads to insight, which leads to the formal proof.