Proof of Inf(S) = 0 for Set S in Haggarty's Analysis Book

[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

proof by contradiction:
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?

 

[LCKurtz]

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

proof by contradiction:
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?

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.

 

[pasmith]

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

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.)

 

[phospho]

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.

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

 

[phospho]

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.)

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?

 

[LCKurtz]

1.
the archimedian postulate states that if we have a real number x, then any natural number n will be greater or equal to x.

No, it doesn't say any such thing.

 

[DrClaude]

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

$1/2^m + 1/3^n + 1/5^p \geq A$

can mean that

$2^m + 3^n + 5^p \geq A(2^m3^n5^p)$

 

[phospho]

No, it doesn't say any such thing.

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

 

[pasmith]

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

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 &gt; 0 for all n \in \mathbb{N}, and that if a > 0 and b > 0 then a + b > 0?

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"

"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 &gt; \pi, and there is no contradiction.

 

[LCKurtz]

1.
the archimedian postulate states that if we have a real number x, then any natural number n will be greater or equal to x.

No, it doesn't say any such thing.

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"

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.

 

[phospho]

"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 &gt; \pi, and there is no contradiction.

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

 

[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)

 

[LCKurtz]

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)

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.

 

[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?

 

[LCKurtz]

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 $\frac{1}{3} \epsilon > 1/3^n$

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.

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?

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.