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Inf(S) = 0 proof

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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 [itex] inf(S) = 0 [/itex]

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?
 

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  • #2
LCKurtz
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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 [itex] inf(S) = 0 [/itex]

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.
 
  • #3
pasmith
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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 [itex] inf(S) = 0 [/itex]
I would prove it directly: show that for all [itex]\epsilon > 0[/itex] there exist [itex]m \in \mathbb{N}[/itex] such that [itex]2^{-m} < \frac13\epsilon[/itex], [itex]n \in \mathbb{N}[/itex] such that [itex]3^{-n} < \frac13\epsilon[/itex], and [itex]p \in \mathbb{N}[/itex] such that [itex]5^{-p} < \frac13\epsilon[/itex], and then add the inequalities together. (This assumes you've shown that 0 is a lower bound for S.)
 
  • #4
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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
 
  • #5
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I would prove it directly: show that for all [itex]\epsilon > 0[/itex] there exist [itex]m \in \mathbb{N}[/itex] such that [itex]2^{-m} < \frac13\epsilon[/itex], [itex]n \in \mathbb{N}[/itex] such that [itex]3^{-n} < \frac13\epsilon[/itex], and [itex]p \in \mathbb{N}[/itex] such that [itex]5^{-p} < \frac13\epsilon[/itex], 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?
 
  • #6
LCKurtz
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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.
 
  • #7
DrClaude
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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) ##
 
  • #8
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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
 
  • #9
pasmith
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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 [itex]a^n > 0[/itex] for all [itex]n \in \mathbb{N}[/itex], 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, [itex]\pi[/itex] is a real number, and there are three strictly positive integers which are less than [itex]\pi[/itex]. On the other hand [itex]4 > \pi[/itex], and there is no contradiction.
 
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  • #10
LCKurtz
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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.
 
  • #11
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"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, [itex]\pi[/itex] is a real number, and there are three strictly positive integers which are less than [itex]\pi[/itex]. On the other hand [itex]4 > \pi[/itex], 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
 
  • #12
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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
LCKurtz
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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.
 
  • #14
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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
LCKurtz
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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.
 

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