What is the Property of Infimum for Sets?

[evinda]

Hello! (Wave)

I want to show that $\ell$ is the infimum of a set $A$ iff $\ell$ is a lower bound of $A$ and for each $\epsilon>0$ there exists an $a \in A$ such that $\ell+\epsilon>a$.

I have thought the following so far for the direction "$\Leftarrow$".
Let $\ell$ be a lower bound of $A$ such that for each $\epsilon>0$ there exists an $a \in A$ such that $\ell+\epsilon>a$.
Let $\ell<t$. We pick $\epsilon=t-\ell>0$. Then there is some $b \in A$ such that $\ell+\epsilon>b$.
But does this help somehow? I don't know how, right now... (Thinking)

 

[S.G. Janssens]

Hello! (Wave)

I want to show that $\ell$ is the infimum of a set $A$ iff $\ell$ is a lower bound of $A$ and for each $\epsilon>0$ there exists an $a \in A$ such that $\ell+\epsilon>a$.

I have thought the following so far for the direction "$\Leftarrow$".
Let $\ell$ be a lower bound of $A$ such that for each $\epsilon>0$ there exists an $a \in A$ such that $\ell+\epsilon>a$.
Let $\ell<t$. We pick $\epsilon=t-\ell>0$. Then there is some $b \in A$ such that $\ell+\epsilon>b$.
But does this help somehow? I don't know how, right now... (Thinking)

It is not clear what $t$ is, so that makes it a bit confusing, but I think you are almost there.

Namely, make explicit that $t$ is another lower bound of $A$ that is strictly greater than $\ell$. Define $\epsilon > 0$ as you do. Since plainly $\ell + \epsilon = t$ you have shown that there exists $b \in A$ that is less than $t$. Hence $t$ is not a lower bound of $A$ at all.

Conclusion: There cannot be another lower bound of $A$ that is strictly greater than $\ell$. This just means that $\ell$ is the infimum, indeed.

 

[evinda]

It is not clear what $t$ is, so that makes it a bit confusing, but I think you are almost there.

Namely, make explicit that $t$ is another lower bound of $A$ that is strictly greater than $\ell$. Define $\epsilon > 0$ as you do. Since plainly $\ell + \epsilon = t$ you have shown that there exists $b \in A$ that is less than $t$. Hence $t$ is not a lower bound of $A$ at all.

Conclusion: There cannot be another lower bound of $A$ that is strictly greater than $\ell$. This just means that $\ell$ is the infimum, indeed.

I see. (Smile)
At the other direction, we want to show that if $\ell$ is the infimum of $A$ then $\ell$ is a lower bound of $A$ and for each $\epsilon>0$ there is an $a \in A$ such that $\ell+\epsilon>a$.

So let $\ell$ be the infimum of $A$. By definiton, the infimum is the greatest lower bound of the set. Thus $\ell$ is a lower bound of $A$. Let $\epsilon>0$. We note that $\ell<\ell+\epsilon$. Can we just set $a=\ell$ ? (Thinking)

 

[S.G. Janssens]

I see. (Smile)
At the other direction, we want to show that if $\ell$ is the infimum of $A$ then $\ell$ is a lower bound of $A$ and for each $\epsilon>0$ there is an $a \in A$ such that $\ell+\epsilon>a$.

So let $\ell$ be the infimum of $A$. By definiton, the infimum is the greatest lower bound of the set. Thus $\ell$ is a lower bound of $A$. Let $\epsilon>0$. We note that $\ell<\ell+\epsilon$. Can we just set $a=\ell$ ? (Thinking)

No, because $\ell$ need not be in $A$. (Think of $A = (1,2)$ with infimum $1$.).

Surely, $\ell$ is a lower bound, so that is done.
For the rest of the statement, consider the negation, i.e. consider what happens when there would exist $\epsilon > 0$ such that there is no $a \in A$ with $\ell + \epsilon > a$. Then $\ell + \epsilon \le a$ for all $a \in A$.

What does this tell you about the quantity $\ell + \epsilon$?
And what does that, in turn, say about $\ell$ itself?

 

[evinda]

No, because $\ell$ need not be in $A$. (Think of $A = (1,2)$ with infimum $1$.).

Surely, $\ell$ is a lower bound, so that is done.
For the rest of the statement, consider the negation, i.e. consider what happens when there would exist $\epsilon > 0$ such that there is no $a \in A$ with $\ell + \epsilon > a$. Then $\ell + \epsilon \le a$ for all $a \in A$.

What does this tell you about the quantity $\ell + \epsilon$?

This tells us that $\ell+\epsilon$ is a lower bound of $A$, right?

 

[evinda]

And since $\ell<\ell+\epsilon$, $\ell$ cannot be the infimum. So we get a contradiction, right? (Thinking)

 

[Olinguito]

So let $\ell$ be the infimum of $A$. By definiton, the infimum is the greatest lower bound of the set. Thus $\ell$ is a lower bound of $A$. Let $\epsilon>0$. We note that $\ell<\ell+\epsilon$. Can we just set $a=\ell$ ? (Thinking)

You’re almost there! Since $\ell$ is the greatest lower bound of $A$ and $\ell+\epsilon$ is greater than $\ell$, $\ell+\epsilon$ can’t be a lower bound of $A$. Therefore …?

 

[S.G. Janssens]

And since $\ell<\ell+\epsilon$, $\ell$ cannot be the infimum. So we get a contradiction, right? (Thinking)

Exactly.

 

[evinda]

You’re almost there! Since $\ell$ is the greatest lower bound of $A$ and $\ell+\epsilon$ is greater than $\ell$, $\ell+\epsilon$ can’t be a lower bound of $A$. Therefore …?

Therefore, $\forall \epsilon >0 \ \exists a \in A$ such that $\ell+\epsilon>a$... (Smile) Thanks for answering!

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Exactly.

Thanks a lot... (Smirk)