# Thread: Prove the following equality of sets

1. ## Prove the following equality of sets

Hi,I'm a bit struggling with this question from my Analysis practice.
Here is the question:

Prove the following equality of sets:
(Large Union)for n=0 to infinite = (0,1/(n+1)]=(0,1]

Some people suggested that when try to prove two sets equal to each other, try to prove first one include in the second one,and second one include in the first one.So I tried,and I can prove (0,1/(n+1)] is in (0,1], but I cannot prove the other way around.

2. If so, note that, $\forall n \in \mathbb{N}$, $(0, \frac{1}{n + 1}] \subset (0,1]$. Let $x \in \cup_{n \in \mathbb{N}} (0, \frac{1}{n+1}]$. Then, for some $m \in \mathbb{N}$, we know that $x \in (0, \frac{1}{m+1}]$. But then, $x \in (0,1]$.

Then, you need to show that $(0,1] \subset \cup _{n \in \mathbb{N}} (0, \frac{1}{n+1}]$.

3. Originally Posted by Math Major
Is n a positive integer?

If so, note that, $\forall n \in \mathbb{Z}^+$, $(0, \frac{1}{n + 1}) \subset (0,1]$. Then, you need to show that $(0,1] \subset \cup _{n \in \mathbb{Z^+} (0, \frac{1}{n+1})$.

Thank you. But this is what I got so far. I can prove $(0, \frac{1}{n + 1}) \subset (0,1]$. However, I cannot find the way to show $(0,1] \subset \cup _{n \in \mathbb{Z^+} (0, \frac{1}{n+1})$. Could you please give me more suggestions? Thanks a lot.

4. Sure. Let $x \in (0,1]$. (edit, actually, just take m = 0) $0 < x \le \frac{1}{m+1}$. But then, $x \in (0, \frac{1}{m+1}]$, and it follows that $x \in \cup_{n \in \mathbb{N}} (0, \frac{1}{n+1}]$.

Sorry about the frequent edits! Sorry, fixing some typos.

5. Um, sorry, that should actually be n is a natural number including zero, not the positive integers. I'll go through, edit, and fix it. Sorry about the excessive confusion this might have caused. Tonight is just not my night. -_-;

Okay, I think I've fixed most of the typos. Man, I just shouldn't use the quick response button before I have a chance to read what I wrote! Sorry about that. But now I think I've fixed it so it's comprehensible.

6. Another way to do this is to note that if $A_n=\left( 0,\frac{1}{n+1} \right]$ with $n=0,1,2,...$ then $A_m\subset A_k$ whenever $k\leq m$ so the union of the $A_n$ is simply $A_0=(0,1]$