What happens to inequalities when $epsilon$ $to$ $0$?
$begingroup$
I recently came across a proof in topology. There was an inequality in which
$A$$<$$B$ + $epsilon$ (strictly less than),But when $epsilon$ $to$ $0$ , then it was inferred that $A$<=$B$.
Here $B$ is the infimum of a sequence and we are using the basic definition of infimum.(There is always a number between infimum and infimum + $epsilon$)
How is this possible?
real-analysis general-topology measure-theory
asked Jan 15 at 10:15
CosmicCosmic
8810
$endgroup$
add a comment |
$begingroup$
I recently came across a proof in topology. There was an inequality in which
$A$$<$$B$ + $epsilon$ (strictly less than),But when $epsilon$ $to$ $0$ , then it was inferred that $A$<=$B$.
Here $B$ is the infimum of a sequence and we are using the basic definition of infimum.(There is always a number between infimum and infimum + $epsilon$)
How is this possible?
real-analysis general-topology measure-theory
asked Jan 15 at 10:15
CosmicCosmic
8810
$endgroup$
$begingroup$
I think we will need a bit more context. What are $A$ and $B$, exactly? The infimum of what sequence?
$endgroup$
– Arthur
Jan 15 at 10:17
1
$begingroup$
For future reference, please refrain from using ALL CAPS in a single sentence. There's a high chance your question would be viewed as low quality by some.
$endgroup$
– BigbearZzz
Jan 15 at 10:28
$begingroup$
@BigbearZzz thanks, will make sure.
$endgroup$
– Cosmic
Jan 15 at 10:39
add a comment |
$begingroup$
I recently came across a proof in topology. There was an inequality in which
$A$$<$$B$ + $epsilon$ (strictly less than),But when $epsilon$ $to$ $0$ , then it was inferred that $A$<=$B$.
Here $B$ is the infimum of a sequence and we are using the basic definition of infimum.(There is always a number between infimum and infimum + $epsilon$)
How is this possible?
real-analysis general-topology measure-theory
asked Jan 15 at 10:15
CosmicCosmic
8810
$endgroup$
I recently came across a proof in topology. There was an inequality in which
$A$$<$$B$ + $epsilon$ (strictly less than),But when $epsilon$ $to$ $0$ , then it was inferred that $A$<=$B$.
Here $B$ is the infimum of a sequence and we are using the basic definition of infimum.(There is always a number between infimum and infimum + $epsilon$)
How is this possible?
real-analysis general-topology measure-theory
real-analysis general-topology measure-theory
asked Jan 15 at 10:15
CosmicCosmic
8810
asked Jan 15 at 10:15
CosmicCosmic
8810
edited Jan 15 at 11:21
Cosmic
asked Jan 15 at 10:15
CosmicCosmic
8810
asked Jan 15 at 10:15
CosmicCosmic
8810
asked Jan 15 at 10:15
CosmicCosmic
8810
8810
$begingroup$
I think we will need a bit more context. What are $A$ and $B$, exactly? The infimum of what sequence?
$endgroup$
– Arthur
Jan 15 at 10:17
1
$begingroup$
For future reference, please refrain from using ALL CAPS in a single sentence. There's a high chance your question would be viewed as low quality by some.
$endgroup$
– BigbearZzz
Jan 15 at 10:28
$begingroup$
@BigbearZzz thanks, will make sure.
$endgroup$
– Cosmic
Jan 15 at 10:39
add a comment |
$begingroup$
I think we will need a bit more context. What are $A$ and $B$, exactly? The infimum of what sequence?
$endgroup$
– Arthur
Jan 15 at 10:17
1
$begingroup$
For future reference, please refrain from using ALL CAPS in a single sentence. There's a high chance your question would be viewed as low quality by some.
$endgroup$
– BigbearZzz
Jan 15 at 10:28
$begingroup$
@BigbearZzz thanks, will make sure.
$endgroup$
– Cosmic
Jan 15 at 10:39
$begingroup$
I think we will need a bit more context. What are $A$ and $B$, exactly? The infimum of what sequence?
$endgroup$
– Arthur
Jan 15 at 10:17
$begingroup$
I think we will need a bit more context. What are $A$ and $B$, exactly? The infimum of what sequence?
$endgroup$
– Arthur
Jan 15 at 10:17
1
1
$begingroup$
For future reference, please refrain from using ALL CAPS in a single sentence. There's a high chance your question would be viewed as low quality by some.
$endgroup$
– BigbearZzz
Jan 15 at 10:28
$begingroup$
For future reference, please refrain from using ALL CAPS in a single sentence. There's a high chance your question would be viewed as low quality by some.
$endgroup$
– BigbearZzz
Jan 15 at 10:28
$begingroup$
@BigbearZzz thanks, will make sure.
$endgroup$
– Cosmic
Jan 15 at 10:39
$begingroup$
@BigbearZzz thanks, will make sure.
$endgroup$
– Cosmic
Jan 15 at 10:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $A<B+epsilon$ for all $epsilon > 0$, then we can conclude that $Aleq B$. In fact, this is the most you can conclude. It is in fact rather easy to prove the following statement:
$A$ is less than or equal to $B$ if and only if, for all $epsilon>0$, $A$ is less than $B+epsilon$. In symbols, this is:
$$(Aleq B) iff (forall epsilon > 0: A<B+epsilon)$$
We cannot conclude that $A<B$, and we cannot conclude $A=B$.
answered Jan 15 at 10:18
5xum5xum
91.8k394161
$endgroup$
$begingroup$
Thanks, will make the edit, I found out that indeed the statement you provided was the actual doubt.
$endgroup$
– Cosmic
Jan 15 at 10:46
$begingroup$
Can you give the proof for the statement you provided.
$endgroup$
– Cosmic
Jan 15 at 10:46
$begingroup$
@Cosmic I think it's good practice if you try to prove it yourself!
$endgroup$
– 5xum
Jan 15 at 10:49
$begingroup$
Will try doing that
$endgroup$
– Cosmic
Jan 15 at 10:50
1
$begingroup$
@Cosmic The proof to the right is fairly straightforward. Proving to the left is best done through a proof by contradiction (or more simply, contrapositive)
$endgroup$
– 5xum
Jan 15 at 10:54
add a comment |
$begingroup$
I think you're misguided.
In general, for any $a,binBbb R$ such that $a<b+varepsilon$ for all $varepsilon>0$, we can deduce only that $ale b$. Indeed, to prove the claim we can assume that $a>b$ and take $varepsilon=frac {a-b}2$ to get a contradiction.
Since I've seen you used the tag measure theory I'll assume further about how the proof you've seen conclude $a=b$ . Normally, when you want to prove that $a=b$ you'd need to show that $age b$ and $ale b$. In measure theory, it is often the case that the $age b$ direction is obvious or easy to show, hence you only need to do the other half. Proving $ale b$ directly could still be difficult, hence we instead prove that $a<b+varepsilon$ for all $varepsilon>0$ and then deduce that $ale b$.
answered Jan 15 at 10:23
BigbearZzzBigbearZzz
8,94521652
$endgroup$
$begingroup$
Thanks, I found out that it was the case in the question.
$endgroup$
– Cosmic
Jan 15 at 10:42
$begingroup$
@GitGud Right, thank you I'll make the edit.
$endgroup$
– BigbearZzz
Jan 15 at 10:46
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $A<B+epsilon$ for all $epsilon > 0$, then we can conclude that $Aleq B$. In fact, this is the most you can conclude. It is in fact rather easy to prove the following statement:
$A$ is less than or equal to $B$ if and only if, for all $epsilon>0$, $A$ is less than $B+epsilon$. In symbols, this is:
$$(Aleq B) iff (forall epsilon > 0: A<B+epsilon)$$
We cannot conclude that $A<B$, and we cannot conclude $A=B$.
answered Jan 15 at 10:18
5xum5xum
91.8k394161
$endgroup$
$begingroup$
Thanks, will make the edit, I found out that indeed the statement you provided was the actual doubt.
$endgroup$
– Cosmic
Jan 15 at 10:46
$begingroup$
Can you give the proof for the statement you provided.
$endgroup$
– Cosmic
Jan 15 at 10:46
$begingroup$
@Cosmic I think it's good practice if you try to prove it yourself!
$endgroup$
– 5xum
Jan 15 at 10:49
$begingroup$
Will try doing that
$endgroup$
– Cosmic
Jan 15 at 10:50
1
$begingroup$
@Cosmic The proof to the right is fairly straightforward. Proving to the left is best done through a proof by contradiction (or more simply, contrapositive)
$endgroup$
– 5xum
Jan 15 at 10:54
add a comment |
$begingroup$
If $A<B+epsilon$ for all $epsilon > 0$, then we can conclude that $Aleq B$. In fact, this is the most you can conclude. It is in fact rather easy to prove the following statement:
$A$ is less than or equal to $B$ if and only if, for all $epsilon>0$, $A$ is less than $B+epsilon$. In symbols, this is:
$$(Aleq B) iff (forall epsilon > 0: A<B+epsilon)$$
We cannot conclude that $A<B$, and we cannot conclude $A=B$.
answered Jan 15 at 10:18
5xum5xum
91.8k394161
$endgroup$
$begingroup$
Thanks, will make the edit, I found out that indeed the statement you provided was the actual doubt.
$endgroup$
– Cosmic
Jan 15 at 10:46
$begingroup$
Can you give the proof for the statement you provided.
$endgroup$
– Cosmic
Jan 15 at 10:46
$begingroup$
@Cosmic I think it's good practice if you try to prove it yourself!
$endgroup$
– 5xum
Jan 15 at 10:49
$begingroup$
Will try doing that
$endgroup$
– Cosmic
Jan 15 at 10:50
1
$begingroup$
@Cosmic The proof to the right is fairly straightforward. Proving to the left is best done through a proof by contradiction (or more simply, contrapositive)
$endgroup$
– 5xum
Jan 15 at 10:54
add a comment |
$begingroup$
If $A<B+epsilon$ for all $epsilon > 0$, then we can conclude that $Aleq B$. In fact, this is the most you can conclude. It is in fact rather easy to prove the following statement:
$A$ is less than or equal to $B$ if and only if, for all $epsilon>0$, $A$ is less than $B+epsilon$. In symbols, this is:
$$(Aleq B) iff (forall epsilon > 0: A<B+epsilon)$$
We cannot conclude that $A<B$, and we cannot conclude $A=B$.
answered Jan 15 at 10:18
5xum5xum
91.8k394161
$endgroup$
If $A<B+epsilon$ for all $epsilon > 0$, then we can conclude that $Aleq B$. In fact, this is the most you can conclude. It is in fact rather easy to prove the following statement:
$A$ is less than or equal to $B$ if and only if, for all $epsilon>0$, $A$ is less than $B+epsilon$. In symbols, this is:
$$(Aleq B) iff (forall epsilon > 0: A<B+epsilon)$$
We cannot conclude that $A<B$, and we cannot conclude $A=B$.
answered Jan 15 at 10:18
5xum5xum
91.8k394161
answered Jan 15 at 10:18
5xum5xum
91.8k394161
answered Jan 15 at 10:18
5xum5xum
91.8k394161
answered Jan 15 at 10:18
5xum5xum
91.8k394161
91.8k394161
$begingroup$
Thanks, will make the edit, I found out that indeed the statement you provided was the actual doubt.
$endgroup$
– Cosmic
Jan 15 at 10:46
$begingroup$
Can you give the proof for the statement you provided.
$endgroup$
– Cosmic
Jan 15 at 10:46
$begingroup$
@Cosmic I think it's good practice if you try to prove it yourself!
$endgroup$
– 5xum
Jan 15 at 10:49
$begingroup$
Will try doing that
$endgroup$
– Cosmic
Jan 15 at 10:50
1
$begingroup$
@Cosmic The proof to the right is fairly straightforward. Proving to the left is best done through a proof by contradiction (or more simply, contrapositive)
$endgroup$
– 5xum
Jan 15 at 10:54
add a comment |
$begingroup$
Thanks, will make the edit, I found out that indeed the statement you provided was the actual doubt.
$endgroup$
– Cosmic
Jan 15 at 10:46
$begingroup$
Can you give the proof for the statement you provided.
$endgroup$
– Cosmic
Jan 15 at 10:46
$begingroup$
@Cosmic I think it's good practice if you try to prove it yourself!
$endgroup$
– 5xum
Jan 15 at 10:49
$begingroup$
Will try doing that
$endgroup$
– Cosmic
Jan 15 at 10:50
1
$begingroup$
@Cosmic The proof to the right is fairly straightforward. Proving to the left is best done through a proof by contradiction (or more simply, contrapositive)
$endgroup$
– 5xum
Jan 15 at 10:54
$begingroup$
Thanks, will make the edit, I found out that indeed the statement you provided was the actual doubt.
$endgroup$
– Cosmic
Jan 15 at 10:46
$begingroup$
Thanks, will make the edit, I found out that indeed the statement you provided was the actual doubt.
$endgroup$
– Cosmic
Jan 15 at 10:46
$begingroup$
Can you give the proof for the statement you provided.
$endgroup$
– Cosmic
Jan 15 at 10:46
$begingroup$
Can you give the proof for the statement you provided.
$endgroup$
– Cosmic
Jan 15 at 10:46
$begingroup$
@Cosmic I think it's good practice if you try to prove it yourself!
$endgroup$
– 5xum
Jan 15 at 10:49
$begingroup$
@Cosmic I think it's good practice if you try to prove it yourself!
$endgroup$
– 5xum
Jan 15 at 10:49
$begingroup$
Will try doing that
$endgroup$
– Cosmic
Jan 15 at 10:50
$begingroup$
Will try doing that
$endgroup$
– Cosmic
Jan 15 at 10:50
1
1
$begingroup$
@Cosmic The proof to the right is fairly straightforward. Proving to the left is best done through a proof by contradiction (or more simply, contrapositive)
$endgroup$
– 5xum
Jan 15 at 10:54
$begingroup$
@Cosmic The proof to the right is fairly straightforward. Proving to the left is best done through a proof by contradiction (or more simply, contrapositive)
$endgroup$
– 5xum
Jan 15 at 10:54
add a comment |
$begingroup$
I think you're misguided.
In general, for any $a,binBbb R$ such that $a<b+varepsilon$ for all $varepsilon>0$, we can deduce only that $ale b$. Indeed, to prove the claim we can assume that $a>b$ and take $varepsilon=frac {a-b}2$ to get a contradiction.
Since I've seen you used the tag measure theory I'll assume further about how the proof you've seen conclude $a=b$ . Normally, when you want to prove that $a=b$ you'd need to show that $age b$ and $ale b$. In measure theory, it is often the case that the $age b$ direction is obvious or easy to show, hence you only need to do the other half. Proving $ale b$ directly could still be difficult, hence we instead prove that $a<b+varepsilon$ for all $varepsilon>0$ and then deduce that $ale b$.
answered Jan 15 at 10:23
BigbearZzzBigbearZzz
8,94521652
$endgroup$
$begingroup$
Thanks, I found out that it was the case in the question.
$endgroup$
– Cosmic
Jan 15 at 10:42
$begingroup$
@GitGud Right, thank you I'll make the edit.
$endgroup$
– BigbearZzz
Jan 15 at 10:46
add a comment |
$begingroup$
I think you're misguided.
In general, for any $a,binBbb R$ such that $a<b+varepsilon$ for all $varepsilon>0$, we can deduce only that $ale b$. Indeed, to prove the claim we can assume that $a>b$ and take $varepsilon=frac {a-b}2$ to get a contradiction.
Since I've seen you used the tag measure theory I'll assume further about how the proof you've seen conclude $a=b$ . Normally, when you want to prove that $a=b$ you'd need to show that $age b$ and $ale b$. In measure theory, it is often the case that the $age b$ direction is obvious or easy to show, hence you only need to do the other half. Proving $ale b$ directly could still be difficult, hence we instead prove that $a<b+varepsilon$ for all $varepsilon>0$ and then deduce that $ale b$.
answered Jan 15 at 10:23
BigbearZzzBigbearZzz
8,94521652
$endgroup$
$begingroup$
Thanks, I found out that it was the case in the question.
$endgroup$
– Cosmic
Jan 15 at 10:42
$begingroup$
@GitGud Right, thank you I'll make the edit.
$endgroup$
– BigbearZzz
Jan 15 at 10:46
add a comment |
$begingroup$
I think you're misguided.
In general, for any $a,binBbb R$ such that $a<b+varepsilon$ for all $varepsilon>0$, we can deduce only that $ale b$. Indeed, to prove the claim we can assume that $a>b$ and take $varepsilon=frac {a-b}2$ to get a contradiction.
Since I've seen you used the tag measure theory I'll assume further about how the proof you've seen conclude $a=b$ . Normally, when you want to prove that $a=b$ you'd need to show that $age b$ and $ale b$. In measure theory, it is often the case that the $age b$ direction is obvious or easy to show, hence you only need to do the other half. Proving $ale b$ directly could still be difficult, hence we instead prove that $a<b+varepsilon$ for all $varepsilon>0$ and then deduce that $ale b$.
answered Jan 15 at 10:23
BigbearZzzBigbearZzz
8,94521652
$endgroup$
I think you're misguided.
In general, for any $a,binBbb R$ such that $a<b+varepsilon$ for all $varepsilon>0$, we can deduce only that $ale b$. Indeed, to prove the claim we can assume that $a>b$ and take $varepsilon=frac {a-b}2$ to get a contradiction.
Since I've seen you used the tag measure theory I'll assume further about how the proof you've seen conclude $a=b$ . Normally, when you want to prove that $a=b$ you'd need to show that $age b$ and $ale b$. In measure theory, it is often the case that the $age b$ direction is obvious or easy to show, hence you only need to do the other half. Proving $ale b$ directly could still be difficult, hence we instead prove that $a<b+varepsilon$ for all $varepsilon>0$ and then deduce that $ale b$.
answered Jan 15 at 10:23
BigbearZzzBigbearZzz
8,94521652
edited Jan 15 at 10:47
answered Jan 15 at 10:23
BigbearZzzBigbearZzz
8,94521652
answered Jan 15 at 10:23
BigbearZzzBigbearZzz
8,94521652
answered Jan 15 at 10:23
BigbearZzzBigbearZzz
8,94521652
8,94521652
$begingroup$
Thanks, I found out that it was the case in the question.
$endgroup$
– Cosmic
Jan 15 at 10:42
$begingroup$
@GitGud Right, thank you I'll make the edit.
$endgroup$
– BigbearZzz
Jan 15 at 10:46
add a comment |
$begingroup$
Thanks, I found out that it was the case in the question.
$endgroup$
– Cosmic
Jan 15 at 10:42
$begingroup$
@GitGud Right, thank you I'll make the edit.
$endgroup$
– BigbearZzz
Jan 15 at 10:46
$begingroup$
Thanks, I found out that it was the case in the question.
$endgroup$
– Cosmic
Jan 15 at 10:42
$begingroup$
Thanks, I found out that it was the case in the question.
$endgroup$
– Cosmic
Jan 15 at 10:42
$begingroup$
@GitGud Right, thank you I'll make the edit.
$endgroup$
– BigbearZzz
Jan 15 at 10:46
$begingroup$
@GitGud Right, thank you I'll make the edit.
$endgroup$
– BigbearZzz
Jan 15 at 10:46
add a comment |
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$begingroup$
I think we will need a bit more context. What are $A$ and $B$, exactly? The infimum of what sequence?
$endgroup$
– Arthur
Jan 15 at 10:17
1
$begingroup$
For future reference, please refrain from using ALL CAPS in a single sentence. There's a high chance your question would be viewed as low quality by some.
$endgroup$
– BigbearZzz
Jan 15 at 10:28
$begingroup$
@BigbearZzz thanks, will make sure.
$endgroup$
– Cosmic
Jan 15 at 10:39