1. ## Proof - sets

Hey guys, after spending a few hours on this question and not getting anywhere, I think I need some help.

Question

$\bigcap_{n \in N} A_n$ denotes the intersection of the sets $A_n$ and $n \in N$ and $\bigcup_{n \in N} A_n$ denotes the union of the sets $A_n$ and $n \in N$. ( $N$ is the set of Natural Numbers)

Prove that $(0, \frac{4}{3} ] = \bigcup_{n \in N} \left[ \frac{1}{n+2} , 1 + \frac{1}{n+2} \right]$

and prove that $\{1\} = \bigcap_{n \in N} \left[1, 1+ \frac{1}{n} \right]$

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We have been learning about the Archimedean property, so I've been trying to mess around with that. I mean the second one is fairly obvious. 1 will be included in the set for any n. And therefore it is the only number than is always in the set of natural numbers and the set [1, 1+ 1/n]...but I don't know how to write this formally.

Regarding the second one, again, I can see that 1/(n+2) tends to 0 as n tends to infinity. Perhaps I need to prove this via the archimedean property? But I don't think that'd be enough. 4/3 is the highest that 1+ 1/n+2 (n=1) could possibly be if n belongs to the set of natural numbers...but I'm just stating the obvious here.

2. Because I do not know the level of rigor demanded in your course, I can only give general directions.

For #1. We know that $\frac{4}{3} \in U_1 = \left[ {\frac{1}{3},\frac{4}{3}} \right]$ and $\forall n\left[ {1 + \frac{1}{{n + 2}} \le \frac{4}{3}} \right]
$
so that takes care of the right endpoint being included. Next suppose that $0 < t < 1$ then $\left( {\exists J} \right)\left[ {\frac{1}{{J + 2}} < t} \right]\quad \Rightarrow t \in \left[ {\frac{1}{{J + 2}},1 + \frac{1}{{J + 2}}} \right]$. That should suffice for the left endpoint. You may need to point out that $\forall n\left[ {0 \notin \left[ {\frac{1}{{n + 2}},1 + \frac{1}{{n + 2}}} \right]} \right]$.

3. Thanks for the reply, Plato!

Originally Posted by Plato
Because I do not know the level of rigor demanded in your course, I can only give general directions.

For #1. We know that $\frac{4}{3} \in U_1 = \left[ {\frac{1}{3},\frac{4}{3}} \right]$ and $\forall n\left[ {1 + \frac{1}{{n + 2}} \le \frac{4}{3}} \right]
$
Ah, clever - I understand this bit.

Originally Posted by Plato
Next suppose that $0 < t < 1$ then $\left( {\exists J} \right)\left[ {\frac{1}{{J + 2}} < t} \right]\quad \Rightarrow t \in \left[ {\frac{1}{{J + 2}},1 + \frac{1}{{J + 2}}} \right]$. That should suffice for the left endpoint.
I'm probably being very dull here, but why does that suffice? I don't understand much of this. I see why t is in that interval, but what does this mean about 0? Why does this show that 0 is the 'lowest' possible point? If it's not too much trouble, can you please explain this a bit more?

Originally Posted by Plato
You may need to point out that $\forall n\left[ {0 \notin \left[ {\frac{1}{{n + 2}},1 + \frac{1}{{n + 2}}} \right]} \right]$.
Not a problem, I think I should be able to prove this without any difficulty.

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Using the 'trick' you used for the first bit of this question, I will attempt a formal proof of #2 and post it. Thanks.

4. Originally Posted by Joel24
I'm probably being very dull here, but why does that suffice? I don't understand much of this. I see why t is in that interval, but what does this mean about 0? Why does this show that 0 is the 'lowest' possible point? If it's not too much trouble, can you please explain this a bit more?
That shows that for any positive number less that 1 belongs to one of those intervals.

5. Originally Posted by Plato
That shows that for any positive number less that 1 belongs to one of those intervals.
*D'oh* Of course!

I think I've completed the second proof relatively ok, so this thread can be closed.

6. To show that two sets A and B are equal, you must show that $A\subseteq B$ and $B\subseteq A$. So to show that $\left(0,\frac{4}{3}\right]=\bigcup_{n\,\in\,N} \left[ \frac{1}{n+2} , 1 + \frac{1}{n+2} \right]$, show both (i) and (ii) below:

$\color{white}.\quad.$ (i) $\left(0,\frac{4}{3}\right]\ \subseteq\ \bigcup_{n\,\in\,N} \left[ \frac{1}{n+2} , 1 + \frac{1}{n+2} \right]$

Let $x\in\left(0,\frac{4}{3}\right]$. First, note that $\forall\,n\in\mathbb{N}$, $1+\frac{1}{n+2}\leq1+\frac{1}{3}=\frac{4}{3}$

If $0, then choose an $N\in\mathbb{N}$ such that $\frac{1}{N+2}\leq x$. (This always exists, by the Archimedean property of $\mathbb{N}$.) Then $\frac{1}{N+2}\leq x\leq1\leq1+\frac{1}{N+2}$, i.e. $x\in\left[\frac{1}{N+2},1+\frac{1}{N+2}\right]$.

If $1, then $\frac{1}{3}<1, i.e. $x\in\left[\frac{1}{1+2},1+\frac{1}{1+2}\right]$.

So $x\in\left[\frac{1}{n+2},1+\frac{1}{n+2}\right]$ for some $n\in\mathbb{N}$, i.e. $x\in\bigcup_{n\,\in\,N} \left[ \frac{1}{n+2} , 1 + \frac{1}{n+2} \right]$. This proves (i).

$\color{white}.\quad.$ (ii) $\bigcup_{n\,\in\,N} \left[ \frac{1}{n+2} , 1 + \frac{1}{n+2} \right]\subseteq\left(0,\frac{4}{3}\right]$.

Let $x\in\bigcup_{n\,\in\,N} \left[ \frac{1}{n+2} , 1 + \frac{1}{n+2} \right]$. Thus $x\in\left[\frac{1}{n+2},1+\frac{1}{n+2}\right]$ for some $n\in\mathbb{N}$

$\Rightarrow\ 0<\frac{1}{n+2}\leq x\leq1+\frac{1}{n+2}\leq\frac{4}{3}$

$\Rightarrow\ x\in\left(0,\frac{4}{3}\right]$. This proves (ii).

(i) & (ii) $\Rightarrow\ \left(0,\frac{4}{3}\right]=\bigcup_{n\,\in\,N} \left[ \frac{1}{n+2} , 1 + \frac{1}{n+2} \right]$