What happens to inequalities when $epsilon$ $to$ $0$?












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 11:21







Cosmic

















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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















4












$begingroup$

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 $A=B$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















2












$begingroup$

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 $ale b$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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 $A=B$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















4












$begingroup$

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 $A=B$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














4












4








4





$begingroup$

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 $A=B$.






share|cite|improve this answer









$endgroup$



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 $A=B$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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











2












$begingroup$

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 $ale b$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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 $ale b$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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 $ale b$.






share|cite|improve this answer











$endgroup$



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 $ale b$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 15 at 10:47

























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


















  • $begingroup$
    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


















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