Inf(S) = 0 proof

1. Oct 12, 2013

phospho

This is from the haggarty book on analysis

take the set $S := \{ 2^{-m} + 3^{-n} + 5^{-p} | m,n,p \in \mathbb{N} \}$

claim $inf(S) = 0$

suppose $A > 0$ where A is the largest lower bound for S

then $1/2^m + 1/3^n + 1/5^p \geq A$
i.e. $2^m + 3^n + 5^p \geq A(2^m3^n5^p)$
let $X = 2^m + 3^n + 5^p$ and $Y = 2^m3^n5^p$ where $Y,X \in \mathbb{N}$
then $X \geq AY$ hence $Y \leq \frac{X}{A}$ which is a contradiction to the archimedean postulate which states for any real x a natural number n >= x hence the lower bound is 0

is this proof OK or have I gone wrong somewhere?

2. Oct 12, 2013

LCKurtz

I don't understand your contradiction. You made a hypothesis and just wrote it several different ways ending in $Y \leq \frac{X}{A}$. You haven't explained how it contradicts the Archimedian postulate. You also haven't observed that the glb is greater or equal 0.

3. Oct 12, 2013

pasmith

I would prove it directly: show that for all $\epsilon > 0$ there exist $m \in \mathbb{N}$ such that $2^{-m} < \frac13\epsilon$, $n \in \mathbb{N}$ such that $3^{-n} < \frac13\epsilon$, and $p \in \mathbb{N}$ such that $5^{-p} < \frac13\epsilon$, and then add the inequalities together. (This assumes you've shown that 0 is a lower bound for S.)

4. Oct 12, 2013

phospho

1.
well we know the X and Y are natural numbers, and A is any number greater than 0, the archimedian postulate states that if we have a real number x, then any natural number n will be greater or equal to x. Here we have Y (which is a natural number), which we showed to be less than or equal to a natural number over a real number A>0, which is a contradiction to the archimedian postulate.

2.
we know the glb is more than or equal to 0 as m,n,p are natural numbers, hence as m,n,p get large 1/2^m + 1/3^n + 1/5^p will become small, but most certainly not less than 0 (it's a summation of rational numbers).

Should I add 2. to the start of the proof, and 1. to the end to show it's a contradiction? Would that be more clear?

thank you

5. Oct 12, 2013

phospho

How would I prove 0 is a lower bound for S? It seems obvious but have no idea how to prove it

e.g. can I just say 0 is a lower bound as S is never negative?

6. Oct 12, 2013

LCKurtz

No, it doesn't say any such thing.

7. Oct 12, 2013

Staff: Mentor

I'll let the experts help you with the proof, but I'd like to understand how this
can mean that

8. Oct 12, 2013

phospho

The book I'm reading states this exactly:

"Another way of phrasing the Archimedean postulate is to say that, given any real number x, there exists an integer n with n ≥ x"

here we have the integer Y, and the real number X/A but Y ≤ X/A

9. Oct 12, 2013

pasmith

The relevant ordered field axioms are:

- For all a and b, if a < b then for all c > 0, ac < bc.
- For all a and b, if a < b then for all c, a + c < b + c.

Can you prove from those axioms that if a > 0 then $a^n > 0$ for all $n \in \mathbb{N}$, and that if a > 0 and b > 0 then a + b > 0?

"There exists" means that at least one integer is greater than or equal to x, not that all integers are greater than equal to x. For example, $\pi$ is a real number, and there are three strictly positive integers which are less than $\pi$. On the other hand $4 > \pi$, and there is no contradiction.

Last edited: Oct 12, 2013
10. Oct 12, 2013

LCKurtz

Do you not understand the difference between the statements "any natural number n will be..." and "there exists a natural number n with..."? The distinctions matter when you are constructing a proof.

11. Oct 12, 2013

phospho

so the proof is not valid?

I really don't understand

if the archimedean postulate states there exists an integer greater than or equal to a real number, and I just showed the an integer is less than or equal to a real number then it's not a contradiction?

I will show you one other solution in the book where they used the same method:

Q: http://gyazo.com/af55675c5d3919c8af02a4dffbff69e7
A: http://gyazo.com/e019c33314ce8a189b798ba2661fd437

they also go to use the same method for the inf

12. Oct 12, 2013

phospho

Also, isn't it a contradiction as I reached the statement for all Y <= X/A which is not true as there is a Y > X/A? (by the archimedean postulate)

13. Oct 12, 2013

LCKurtz

DrClaude pointed out to you in post #7 that you have an algebra error in your argument that the rest of us overlooked.

Regardless of that mistake, I can not make heads or tails of your reasoning, even if the algebra had been correct. So, no, I don't consider it a correct proof.

Finally, pasmith showed you in post #3 the obvious way to work the problem. I suggest you think about that.

14. Oct 13, 2013

phospho

OK going about it a pasmith's way:

I previously proved in the question before that for any $a \geq 2$ $a^n > n$. This question also hints to use for any "$\epsilon > 0 \exists n \in \mathbb{N}$ s.t. $1/n < \epsilon$. So with this we get $n > 1/\epsilon$ so take for 3^n $3^n > n > 1/\epsilon$ so $\epsilon > 1/3^n$ which would be the same as saying $\frac{1}{3} \epsilon > 1/3^n$ and by similar arguments you'd get the same for 2^-m and 5^-p and adding this you get $\epsilon > \frac{1}{2^m} + \frac{1}{3^n} + \frac{1}{5^p}$

but I don't understand how this shows the inf is 0?

15. Oct 13, 2013

LCKurtz

Why would $\epsilon > \frac 1 {3^n}$ be the same as saying $\frac{1}{3} \epsilon > 1/3^n$? They don't look the same to me.

You need to read post #3 again and actually try what he says.

With respect to not seeing how this shows the inf is 0, what are the two things you need to show that inf(S) = 0? It would be good to have them carefully in mind if you are trying to prove it.