Proving Infinite Intersection of (-1/n,1/n) To Be The Zero Set

Question

I am looking for a bit of guidance as I am examining some of the methods of proofs of infinite unions and the proper methods to doing so. I am hoping perhaps you will understand my question, I worry it is a bit difficult for me to convey via text but I am willing to try my best.

When looking at the infinite union of (-1/n,1/n) I believe that it is equal to the zero set. In proving that this I have so far gathered that I must prove that zero is a subset of the infinite intersection of (-1/n,1/n). I know that in doing so we can show that zero is an element of (-1/n,1/n) for all n. That clearly allows us to see that zero must also then be an element of the infinite intersection of (-1/n,1/n).

I am a bit confused as to where to progress from here. This is what I have come up with so far. I believe the next logical step would be to suppose there exists some value of x which is an element of the infinite intersection of (-1/n,1/n). This would lead us to know that x is an element of (-1/n,1/n) for all n. Now we need to prove the inequality -1/n<=x<=1/n not not hold for any value of x other than zero, is this correct? That would mean that when x greater than zero or x is negative we need to prove to be contradictions to the inequality? I guess I am confused as to how to go about that and am looking for some guidance. I know have used the Archmedian Axioms to prove that 1/n0 < x in the past for cases where x>0 but then i find myself caught up on proving the other side as x is negative to be a contradiction to the inequality.

Any insight is greatly appreciated as to where to go from here in proving my cases. I have attached below an image of my proof thus far up until where I find myself stuck

Work Thus Far

Answer 1

Consider this statement:

For any $n \in \mathbb N$ we have $\frac 1n > 0$.

So... have you seen that statement before? Do you believe it? Is it true?

I hope the answer for all of those is "yes".

So that means what.... It means

$-\frac 1n < 0 < \frac 1n$ for all $n\in \mathbb N$

Which means

$0 \in (-\frac 1n, \frac 1n)$ for all $n\in \mathbb N$.

Well, what does that tell you about the intersection $\bigcap_{n\in \mathbb N} (-\frac 1n, \frac 1n)$?

Well if $0 \in $ every one of the $(-\frac 1n, \frac 1n) $ then $0\in \bigcap_{n\in \mathbb N} (-\frac 1n, \frac 1n)$.

But what else is in the intersection?

What about this statement:

For any $r > 0$ then there is an $n\in \mathbb N$ so that $0 < \frac 1n < r$.

Have you seen that statement before? Do you believe it? Is it true?

Well, that might not be as obvious of the first statement but still the answer should be "Yes".

If that's not obvious: Pf: $r > 0$ so $\frac 1r > 0$. As the natural numbers are not bounded then there exists a natural number $n$ so that $n > \frac 1r > 0$. So $\frac 1n < r$.

So for any $r > 0$ we can always find and $n_r \in \mathbb N$ so that $0 < \frac 1{n_r} < r$.

That means

$-\frac 1{n_r }< \frac 1{n_r} < r$. So

$r \not\in (-\frac 1{n_r}, \frac 1{n_r})$.

So what does that tell us about $\bigcap_{n\in \mathbb N} (-\frac 1n, \frac 1n)$?

Well, $r \not \in (-\frac 1{n_r}, \frac 1{n_r})$ so $r \not \in $ every one of the $(-\frac 1n, \frac 1n)$. So $r$ is not in the intersection.

if $r > 0$ then $r\not \in \bigcap_{n\in \mathbb N} (-\frac 1n, \frac 1n)$.

....

Okay, what about $w < 0$? What can we say about that.

Well if $w < 0$ then $|w| > 0$ and there is an $n_w\in \mathbb N$ so that $0< \frac 1n < |w|$.

So multiply all that by $-1$ and we have $w = -|w| < -\frac 1{n_w} < 0 < \frac 1{n_w}$.

So $w\not \in (-\frac 1{n_w}, \frac 1{n_w})$.

So $w \not \in $ every one of the $(-\frac 1n , \frac 1n)$ and so $w \not \in \bigcap_{n\in \mathbb N} (-\frac 1n, \frac 1n)$

So we have:

1. $0 \in \bigcap_{n\in \mathbb N} (-\frac 1n, \frac 1n)$
2. if $r > 0$ then $r\not \in \bigcap_{n\in \mathbb N} (-\frac 1n, \frac 1n)$
3. if $w < 0$ then $w\not \in \bigcap_{n\in \mathbb N} (-\frac 1n, \frac 1n)$

So $0 \in \bigcap_{n\in \mathbb N} (-\frac 1n, \frac 1n)$ but nothing else is.

So $\bigcap_{n\in \mathbb N} (-\frac 1n, \frac 1n)=\{0\}$