Mixing
Proving an infimum using the archimedean property of R
$begingroup$
As the title states, I am working on proving an infimum with the archimedean property of $mathbb{R}$. The problem is stated as follows:
Use the archimedean property of $mathbb{R}$ to rigorously prove that $inf {frac{1}{n} : n in mathbb{N}} = 0$
A note: I realize the below is probably the wrong way to go about it, but it is an attempt - im not great at proofs yet...
I start out by stating that $A = {frac{1}{n} : n in mathbb{N}}$
I want to prove that $infA = 0$.
I feel like a proof by contradiction is a good way to go about it:
So, suppose not, suppose $infA neq 0$
Then let some $s$ be a lower bound for $A$.
Following the definition of an infimum, given any lower bound $b$, it follows $s ge b$.
Since by our premise $infA neq 0$, then $s gt 0$.
Now, following the archimedean property, since $s in mathbb{R}$ and $s gt 0$, we can find some $n in mathbb{N}$ such that $frac{1}{n} < s$.
It follows then that $1 < sn$. But the supremum of the set $A$ is 1. Thus we have reached a contradiction, and it must be that $inf A = 0$.
To be honest I feel intimidated by trying to prove this stuff, and could really use the support. I have ordered a book on proofs, but I feel like im struggling to really "grasp" what is going on beyond just putting stuff together. Aside from helping with the proof itself, I'd appreciate any advice on how to decide when to use proof by contradiction in the future. Thanks!
real-analysis proof-verification
asked Feb 2 at 5:32
CL40CL40
2136
$endgroup$
add a comment |
$begingroup$
As the title states, I am working on proving an infimum with the archimedean property of $mathbb{R}$. The problem is stated as follows:
Use the archimedean property of $mathbb{R}$ to rigorously prove that $inf {frac{1}{n} : n in mathbb{N}} = 0$
A note: I realize the below is probably the wrong way to go about it, but it is an attempt - im not great at proofs yet...
I start out by stating that $A = {frac{1}{n} : n in mathbb{N}}$
I want to prove that $infA = 0$.
I feel like a proof by contradiction is a good way to go about it:
So, suppose not, suppose $infA neq 0$
Then let some $s$ be a lower bound for $A$.
Following the definition of an infimum, given any lower bound $b$, it follows $s ge b$.
Since by our premise $infA neq 0$, then $s gt 0$.
Now, following the archimedean property, since $s in mathbb{R}$ and $s gt 0$, we can find some $n in mathbb{N}$ such that $frac{1}{n} < s$.
It follows then that $1 < sn$. But the supremum of the set $A$ is 1. Thus we have reached a contradiction, and it must be that $inf A = 0$.
To be honest I feel intimidated by trying to prove this stuff, and could really use the support. I have ordered a book on proofs, but I feel like im struggling to really "grasp" what is going on beyond just putting stuff together. Aside from helping with the proof itself, I'd appreciate any advice on how to decide when to use proof by contradiction in the future. Thanks!
real-analysis proof-verification
asked Feb 2 at 5:32
CL40CL40
2136
$endgroup$
1
$begingroup$
When you suppose $inf Ane 0$ how do you know inf A exists at all?
$endgroup$
– fleablood
Feb 2 at 6:40
1
$begingroup$
Yous say let $s $ be *some* lower bound then for any other lower bound you say $sge b $. Why? You never said $s $ was the greatest lower bound. Then you say $s>0$. Why? You've never said $0$ is a lower bound.
$endgroup$
– fleablood
Feb 2 at 6:49
add a comment |
$begingroup$
As the title states, I am working on proving an infimum with the archimedean property of $mathbb{R}$. The problem is stated as follows:
Use the archimedean property of $mathbb{R}$ to rigorously prove that $inf {frac{1}{n} : n in mathbb{N}} = 0$
A note: I realize the below is probably the wrong way to go about it, but it is an attempt - im not great at proofs yet...
I start out by stating that $A = {frac{1}{n} : n in mathbb{N}}$
I want to prove that $infA = 0$.
I feel like a proof by contradiction is a good way to go about it:
So, suppose not, suppose $infA neq 0$
Then let some $s$ be a lower bound for $A$.
Following the definition of an infimum, given any lower bound $b$, it follows $s ge b$.
Since by our premise $infA neq 0$, then $s gt 0$.
Now, following the archimedean property, since $s in mathbb{R}$ and $s gt 0$, we can find some $n in mathbb{N}$ such that $frac{1}{n} < s$.
It follows then that $1 < sn$. But the supremum of the set $A$ is 1. Thus we have reached a contradiction, and it must be that $inf A = 0$.
To be honest I feel intimidated by trying to prove this stuff, and could really use the support. I have ordered a book on proofs, but I feel like im struggling to really "grasp" what is going on beyond just putting stuff together. Aside from helping with the proof itself, I'd appreciate any advice on how to decide when to use proof by contradiction in the future. Thanks!
real-analysis proof-verification
asked Feb 2 at 5:32
CL40CL40
2136
$endgroup$
As the title states, I am working on proving an infimum with the archimedean property of $mathbb{R}$. The problem is stated as follows:
Use the archimedean property of $mathbb{R}$ to rigorously prove that $inf {frac{1}{n} : n in mathbb{N}} = 0$
A note: I realize the below is probably the wrong way to go about it, but it is an attempt - im not great at proofs yet...
I start out by stating that $A = {frac{1}{n} : n in mathbb{N}}$
I want to prove that $infA = 0$.
I feel like a proof by contradiction is a good way to go about it:
So, suppose not, suppose $infA neq 0$
Then let some $s$ be a lower bound for $A$.
Following the definition of an infimum, given any lower bound $b$, it follows $s ge b$.
Since by our premise $infA neq 0$, then $s gt 0$.
Now, following the archimedean property, since $s in mathbb{R}$ and $s gt 0$, we can find some $n in mathbb{N}$ such that $frac{1}{n} < s$.
It follows then that $1 < sn$. But the supremum of the set $A$ is 1. Thus we have reached a contradiction, and it must be that $inf A = 0$.
To be honest I feel intimidated by trying to prove this stuff, and could really use the support. I have ordered a book on proofs, but I feel like im struggling to really "grasp" what is going on beyond just putting stuff together. Aside from helping with the proof itself, I'd appreciate any advice on how to decide when to use proof by contradiction in the future. Thanks!
real-analysis proof-verification
real-analysis proof-verification
asked Feb 2 at 5:32
CL40CL40
2136
asked Feb 2 at 5:32
CL40CL40
2136
asked Feb 2 at 5:32
CL40CL40
2136
asked Feb 2 at 5:32
CL40CL40
2136
asked Feb 2 at 5:32
CL40CL40
2136
2136
1
$begingroup$
When you suppose $inf Ane 0$ how do you know inf A exists at all?
$endgroup$
– fleablood
Feb 2 at 6:40
1
$begingroup$
Yous say let $s $ be *some* lower bound then for any other lower bound you say $sge b $. Why? You never said $s $ was the greatest lower bound. Then you say $s>0$. Why? You've never said $0$ is a lower bound.
$endgroup$
– fleablood
Feb 2 at 6:49
add a comment |
1
$begingroup$
When you suppose $inf Ane 0$ how do you know inf A exists at all?
$endgroup$
– fleablood
Feb 2 at 6:40
1
$begingroup$
Yous say let $s $ be *some* lower bound then for any other lower bound you say $sge b $. Why? You never said $s $ was the greatest lower bound. Then you say $s>0$. Why? You've never said $0$ is a lower bound.
$endgroup$
– fleablood
Feb 2 at 6:49
1
1
$begingroup$
When you suppose $inf Ane 0$ how do you know inf A exists at all?
$endgroup$
– fleablood
Feb 2 at 6:40
$begingroup$
When you suppose $inf Ane 0$ how do you know inf A exists at all?
$endgroup$
– fleablood
Feb 2 at 6:40
1
1
$begingroup$
Yous say let $s $ be *some* lower bound then for any other lower bound you say $sge b $. Why? You never said $s $ was the greatest lower bound. Then you say $s>0$. Why? You've never said $0$ is a lower bound.
$endgroup$
– fleablood
Feb 2 at 6:49
$begingroup$
Yous say let $s $ be *some* lower bound then for any other lower bound you say $sge b $. Why? You never said $s $ was the greatest lower bound. Then you say $s>0$. Why? You've never said $0$ is a lower bound.
$endgroup$
– fleablood
Feb 2 at 6:49
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
To prove $inf{1/n:ninmathbb{N}}=0$, it must be shown that (i) $0$ is a lower bound and (ii) $0$ is the greatest lower bound.
Proof of (i): Fix $ninmathbb{N}$. Observe $0<1/n$. We are done.
Proof of (ii): Let $x$ be an arbitrary lower bound. It remains to show $xle 0$. By contradiction, suppose $0<x$. Then by Archimedes, there exists $N$ such that $1<Nx$. Therefore $1/N<x$. Therefore $x$ is not a lower bound---a contradiction. As a result, $0nless x$. Because $mathbb{R}$ is totally ordered, it follows that $xle 0$ as desired. We are done.
answered Feb 2 at 5:38
Alberto TakaseAlberto Takase
2,403719
$endgroup$
$begingroup$
Interesting, so you proceeded directly by the definition of an infimum and then used contradiction to prove the second part. How did you decide contradiction was the correct choice without brute forcing a direct proof first? I think I was confused. Your proof makes it clear I should proceed from what I know rather than jumping into a proof head first...its really difficult not to.
$endgroup$
– CL40
Feb 2 at 5:42
$begingroup$
The hint was to use Archimedes principle. Therefore a positive number was required in order to invoke it. It seemed natural given that one wants to show $xle 0$ in part (ii).
$endgroup$
– Alberto Takase
Feb 2 at 5:45
add a comment |
$begingroup$
In the context of this problem, the lemma says:
Let $0$ be a lower bound for $A$. Then $0 = $inf $A$ if and only if for every $ε > 0$, there exists $a in A$ such that $a < 0 + ε$.
To use this Lemma to prove $0 =$ inf $A$, you must show that $0$ satisfies two statements:
$(i)$ $0$ is a lower bound for $A$.
$(ii)$ For every $ε > 0$, there exists $a in A$ such that $a < ε$.
Explain that this is what you need to show and why. Then prove $(i)$ and $(ii)$ as two separate proofs.
answered Feb 2 at 5:51
Key FlexKey Flex
8,58171233
$endgroup$
$begingroup$
Similar the other answer you proceeded directly from the definition of an infimum. I am trying to build the intuition to just "follow the definitions", it's a lot harder than I thought.
$endgroup$
– CL40
Feb 2 at 6:05
1
$begingroup$
@CL40 You'll find yourself coming back to the basic definitions frequently. It's the best way to build intuition about problems.
$endgroup$
– Shubham Johri
Feb 2 at 6:46
add a comment |
$begingroup$
You get the logic, but you fall short in formalizing it rigorously.
$A$ is non-empty and bounded below by $0$, since $1/n>0forall ninBbb N$. So $inf A$ exists, let it be $s$.
Since $0$ is a lower bound, $sge0$, by the basic definition of infimum. But if $s>0,exists ninBbb N|1/n<s$ by the Archimedian property. Thus $snot>0$ because you can always find elements of $A$ smaller than $s$ in that case. Thus, $s=0$.
There is no need to allude to the supremum of $A$.
answered Feb 2 at 7:13
Shubham JohriShubham Johri
5,613818
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To prove $inf{1/n:ninmathbb{N}}=0$, it must be shown that (i) $0$ is a lower bound and (ii) $0$ is the greatest lower bound.
Proof of (i): Fix $ninmathbb{N}$. Observe $0<1/n$. We are done.
Proof of (ii): Let $x$ be an arbitrary lower bound. It remains to show $xle 0$. By contradiction, suppose $0<x$. Then by Archimedes, there exists $N$ such that $1<Nx$. Therefore $1/N<x$. Therefore $x$ is not a lower bound---a contradiction. As a result, $0nless x$. Because $mathbb{R}$ is totally ordered, it follows that $xle 0$ as desired. We are done.
answered Feb 2 at 5:38
Alberto TakaseAlberto Takase
2,403719
$endgroup$
$begingroup$
Interesting, so you proceeded directly by the definition of an infimum and then used contradiction to prove the second part. How did you decide contradiction was the correct choice without brute forcing a direct proof first? I think I was confused. Your proof makes it clear I should proceed from what I know rather than jumping into a proof head first...its really difficult not to.
$endgroup$
– CL40
Feb 2 at 5:42
$begingroup$
The hint was to use Archimedes principle. Therefore a positive number was required in order to invoke it. It seemed natural given that one wants to show $xle 0$ in part (ii).
$endgroup$
– Alberto Takase
Feb 2 at 5:45
add a comment |
$begingroup$
To prove $inf{1/n:ninmathbb{N}}=0$, it must be shown that (i) $0$ is a lower bound and (ii) $0$ is the greatest lower bound.
Proof of (i): Fix $ninmathbb{N}$. Observe $0<1/n$. We are done.
Proof of (ii): Let $x$ be an arbitrary lower bound. It remains to show $xle 0$. By contradiction, suppose $0<x$. Then by Archimedes, there exists $N$ such that $1<Nx$. Therefore $1/N<x$. Therefore $x$ is not a lower bound---a contradiction. As a result, $0nless x$. Because $mathbb{R}$ is totally ordered, it follows that $xle 0$ as desired. We are done.
answered Feb 2 at 5:38
Alberto TakaseAlberto Takase
2,403719
$endgroup$
$begingroup$
Interesting, so you proceeded directly by the definition of an infimum and then used contradiction to prove the second part. How did you decide contradiction was the correct choice without brute forcing a direct proof first? I think I was confused. Your proof makes it clear I should proceed from what I know rather than jumping into a proof head first...its really difficult not to.
$endgroup$
– CL40
Feb 2 at 5:42
$begingroup$
The hint was to use Archimedes principle. Therefore a positive number was required in order to invoke it. It seemed natural given that one wants to show $xle 0$ in part (ii).
$endgroup$
– Alberto Takase
Feb 2 at 5:45
add a comment |
$begingroup$
To prove $inf{1/n:ninmathbb{N}}=0$, it must be shown that (i) $0$ is a lower bound and (ii) $0$ is the greatest lower bound.
Proof of (i): Fix $ninmathbb{N}$. Observe $0<1/n$. We are done.
Proof of (ii): Let $x$ be an arbitrary lower bound. It remains to show $xle 0$. By contradiction, suppose $0<x$. Then by Archimedes, there exists $N$ such that $1<Nx$. Therefore $1/N<x$. Therefore $x$ is not a lower bound---a contradiction. As a result, $0nless x$. Because $mathbb{R}$ is totally ordered, it follows that $xle 0$ as desired. We are done.
answered Feb 2 at 5:38
Alberto TakaseAlberto Takase
2,403719
$endgroup$
To prove $inf{1/n:ninmathbb{N}}=0$, it must be shown that (i) $0$ is a lower bound and (ii) $0$ is the greatest lower bound.
Proof of (i): Fix $ninmathbb{N}$. Observe $0<1/n$. We are done.
Proof of (ii): Let $x$ be an arbitrary lower bound. It remains to show $xle 0$. By contradiction, suppose $0<x$. Then by Archimedes, there exists $N$ such that $1<Nx$. Therefore $1/N<x$. Therefore $x$ is not a lower bound---a contradiction. As a result, $0nless x$. Because $mathbb{R}$ is totally ordered, it follows that $xle 0$ as desired. We are done.
answered Feb 2 at 5:38
Alberto TakaseAlberto Takase
2,403719
answered Feb 2 at 5:38
Alberto TakaseAlberto Takase
2,403719
answered Feb 2 at 5:38
Alberto TakaseAlberto Takase
2,403719
answered Feb 2 at 5:38
Alberto TakaseAlberto Takase
2,403719
2,403719
$begingroup$
Interesting, so you proceeded directly by the definition of an infimum and then used contradiction to prove the second part. How did you decide contradiction was the correct choice without brute forcing a direct proof first? I think I was confused. Your proof makes it clear I should proceed from what I know rather than jumping into a proof head first...its really difficult not to.
$endgroup$
– CL40
Feb 2 at 5:42
$begingroup$
The hint was to use Archimedes principle. Therefore a positive number was required in order to invoke it. It seemed natural given that one wants to show $xle 0$ in part (ii).
$endgroup$
– Alberto Takase
Feb 2 at 5:45
add a comment |
$begingroup$
Interesting, so you proceeded directly by the definition of an infimum and then used contradiction to prove the second part. How did you decide contradiction was the correct choice without brute forcing a direct proof first? I think I was confused. Your proof makes it clear I should proceed from what I know rather than jumping into a proof head first...its really difficult not to.
$endgroup$
– CL40
Feb 2 at 5:42
$begingroup$
The hint was to use Archimedes principle. Therefore a positive number was required in order to invoke it. It seemed natural given that one wants to show $xle 0$ in part (ii).
$endgroup$
– Alberto Takase
Feb 2 at 5:45
$begingroup$
Interesting, so you proceeded directly by the definition of an infimum and then used contradiction to prove the second part. How did you decide contradiction was the correct choice without brute forcing a direct proof first? I think I was confused. Your proof makes it clear I should proceed from what I know rather than jumping into a proof head first...its really difficult not to.
$endgroup$
– CL40
Feb 2 at 5:42
$begingroup$
Interesting, so you proceeded directly by the definition of an infimum and then used contradiction to prove the second part. How did you decide contradiction was the correct choice without brute forcing a direct proof first? I think I was confused. Your proof makes it clear I should proceed from what I know rather than jumping into a proof head first...its really difficult not to.
$endgroup$
– CL40
Feb 2 at 5:42
$begingroup$
The hint was to use Archimedes principle. Therefore a positive number was required in order to invoke it. It seemed natural given that one wants to show $xle 0$ in part (ii).
$endgroup$
– Alberto Takase
Feb 2 at 5:45
$begingroup$
The hint was to use Archimedes principle. Therefore a positive number was required in order to invoke it. It seemed natural given that one wants to show $xle 0$ in part (ii).
$endgroup$
– Alberto Takase
Feb 2 at 5:45
add a comment |
$begingroup$
In the context of this problem, the lemma says:
Let $0$ be a lower bound for $A$. Then $0 = $inf $A$ if and only if for every $ε > 0$, there exists $a in A$ such that $a < 0 + ε$.
To use this Lemma to prove $0 =$ inf $A$, you must show that $0$ satisfies two statements:
$(i)$ $0$ is a lower bound for $A$.
$(ii)$ For every $ε > 0$, there exists $a in A$ such that $a < ε$.
Explain that this is what you need to show and why. Then prove $(i)$ and $(ii)$ as two separate proofs.
answered Feb 2 at 5:51
Key FlexKey Flex
8,58171233
$endgroup$
$begingroup$
Similar the other answer you proceeded directly from the definition of an infimum. I am trying to build the intuition to just "follow the definitions", it's a lot harder than I thought.
$endgroup$
– CL40
Feb 2 at 6:05
1
$begingroup$
@CL40 You'll find yourself coming back to the basic definitions frequently. It's the best way to build intuition about problems.
$endgroup$
– Shubham Johri
Feb 2 at 6:46
add a comment |
$begingroup$
In the context of this problem, the lemma says:
Let $0$ be a lower bound for $A$. Then $0 = $inf $A$ if and only if for every $ε > 0$, there exists $a in A$ such that $a < 0 + ε$.
To use this Lemma to prove $0 =$ inf $A$, you must show that $0$ satisfies two statements:
$(i)$ $0$ is a lower bound for $A$.
$(ii)$ For every $ε > 0$, there exists $a in A$ such that $a < ε$.
Explain that this is what you need to show and why. Then prove $(i)$ and $(ii)$ as two separate proofs.
answered Feb 2 at 5:51
Key FlexKey Flex
8,58171233
$endgroup$
$begingroup$
Similar the other answer you proceeded directly from the definition of an infimum. I am trying to build the intuition to just "follow the definitions", it's a lot harder than I thought.
$endgroup$
– CL40
Feb 2 at 6:05
1
$begingroup$
@CL40 You'll find yourself coming back to the basic definitions frequently. It's the best way to build intuition about problems.
$endgroup$
– Shubham Johri
Feb 2 at 6:46
add a comment |
$begingroup$
In the context of this problem, the lemma says:
Let $0$ be a lower bound for $A$. Then $0 = $inf $A$ if and only if for every $ε > 0$, there exists $a in A$ such that $a < 0 + ε$.
To use this Lemma to prove $0 =$ inf $A$, you must show that $0$ satisfies two statements:
$(i)$ $0$ is a lower bound for $A$.
$(ii)$ For every $ε > 0$, there exists $a in A$ such that $a < ε$.
Explain that this is what you need to show and why. Then prove $(i)$ and $(ii)$ as two separate proofs.
answered Feb 2 at 5:51
Key FlexKey Flex
8,58171233
$endgroup$
In the context of this problem, the lemma says:
Let $0$ be a lower bound for $A$. Then $0 = $inf $A$ if and only if for every $ε > 0$, there exists $a in A$ such that $a < 0 + ε$.
To use this Lemma to prove $0 =$ inf $A$, you must show that $0$ satisfies two statements:
$(i)$ $0$ is a lower bound for $A$.
$(ii)$ For every $ε > 0$, there exists $a in A$ such that $a < ε$.
Explain that this is what you need to show and why. Then prove $(i)$ and $(ii)$ as two separate proofs.
answered Feb 2 at 5:51
Key FlexKey Flex
8,58171233
answered Feb 2 at 5:51
Key FlexKey Flex
8,58171233
answered Feb 2 at 5:51
Key FlexKey Flex
8,58171233
answered Feb 2 at 5:51
Key FlexKey Flex
8,58171233
8,58171233
$begingroup$
Similar the other answer you proceeded directly from the definition of an infimum. I am trying to build the intuition to just "follow the definitions", it's a lot harder than I thought.
$endgroup$
– CL40
Feb 2 at 6:05
1
$begingroup$
@CL40 You'll find yourself coming back to the basic definitions frequently. It's the best way to build intuition about problems.
$endgroup$
– Shubham Johri
Feb 2 at 6:46
add a comment |
$begingroup$
Similar the other answer you proceeded directly from the definition of an infimum. I am trying to build the intuition to just "follow the definitions", it's a lot harder than I thought.
$endgroup$
– CL40
Feb 2 at 6:05
1
$begingroup$
@CL40 You'll find yourself coming back to the basic definitions frequently. It's the best way to build intuition about problems.
$endgroup$
– Shubham Johri
Feb 2 at 6:46
$begingroup$
Similar the other answer you proceeded directly from the definition of an infimum. I am trying to build the intuition to just "follow the definitions", it's a lot harder than I thought.
$endgroup$
– CL40
Feb 2 at 6:05
$begingroup$
Similar the other answer you proceeded directly from the definition of an infimum. I am trying to build the intuition to just "follow the definitions", it's a lot harder than I thought.
$endgroup$
– CL40
Feb 2 at 6:05
1
1
$begingroup$
@CL40 You'll find yourself coming back to the basic definitions frequently. It's the best way to build intuition about problems.
$endgroup$
– Shubham Johri
Feb 2 at 6:46
$begingroup$
@CL40 You'll find yourself coming back to the basic definitions frequently. It's the best way to build intuition about problems.
$endgroup$
– Shubham Johri
Feb 2 at 6:46
add a comment |
$begingroup$
You get the logic, but you fall short in formalizing it rigorously.
$A$ is non-empty and bounded below by $0$, since $1/n>0forall ninBbb N$. So $inf A$ exists, let it be $s$.
Since $0$ is a lower bound, $sge0$, by the basic definition of infimum. But if $s>0,exists ninBbb N|1/n<s$ by the Archimedian property. Thus $snot>0$ because you can always find elements of $A$ smaller than $s$ in that case. Thus, $s=0$.
There is no need to allude to the supremum of $A$.
answered Feb 2 at 7:13
Shubham JohriShubham Johri
5,613818
$endgroup$
add a comment |
$begingroup$
You get the logic, but you fall short in formalizing it rigorously.
$A$ is non-empty and bounded below by $0$, since $1/n>0forall ninBbb N$. So $inf A$ exists, let it be $s$.
Since $0$ is a lower bound, $sge0$, by the basic definition of infimum. But if $s>0,exists ninBbb N|1/n<s$ by the Archimedian property. Thus $snot>0$ because you can always find elements of $A$ smaller than $s$ in that case. Thus, $s=0$.
There is no need to allude to the supremum of $A$.
answered Feb 2 at 7:13
Shubham JohriShubham Johri
5,613818
$endgroup$
add a comment |
$begingroup$
You get the logic, but you fall short in formalizing it rigorously.
$A$ is non-empty and bounded below by $0$, since $1/n>0forall ninBbb N$. So $inf A$ exists, let it be $s$.
Since $0$ is a lower bound, $sge0$, by the basic definition of infimum. But if $s>0,exists ninBbb N|1/n<s$ by the Archimedian property. Thus $snot>0$ because you can always find elements of $A$ smaller than $s$ in that case. Thus, $s=0$.
There is no need to allude to the supremum of $A$.
answered Feb 2 at 7:13
Shubham JohriShubham Johri
5,613818
$endgroup$
You get the logic, but you fall short in formalizing it rigorously.
$A$ is non-empty and bounded below by $0$, since $1/n>0forall ninBbb N$. So $inf A$ exists, let it be $s$.
Since $0$ is a lower bound, $sge0$, by the basic definition of infimum. But if $s>0,exists ninBbb N|1/n<s$ by the Archimedian property. Thus $snot>0$ because you can always find elements of $A$ smaller than $s$ in that case. Thus, $s=0$.
There is no need to allude to the supremum of $A$.
answered Feb 2 at 7:13
Shubham JohriShubham Johri
5,613818
answered Feb 2 at 7:13
Shubham JohriShubham Johri
5,613818
answered Feb 2 at 7:13
Shubham JohriShubham Johri
5,613818
answered Feb 2 at 7:13
Shubham JohriShubham Johri
5,613818
5,613818
add a comment |
add a comment |
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$begingroup$
When you suppose $inf Ane 0$ how do you know inf A exists at all?
$endgroup$
– fleablood
Feb 2 at 6:40
1
$begingroup$
Yous say let $s $ be *some* lower bound then for any other lower bound you say $sge b $. Why? You never said $s $ was the greatest lower bound. Then you say $s>0$. Why? You've never said $0$ is a lower bound.
$endgroup$
– fleablood
Feb 2 at 6:49