Visual interpretation of $\sup(A) \leq \sup(B) $

Question

By definition for Sup(S):

$\forall x \in S, x\leq \sup(S)$ and $\forall s_1\in S<\sup(S), \exists x > s_1 $

Is the below visualization of $\sup(A)\leq \sup(B)$ correct? With $X_A \in A, X_B \in B$.

If so would the following statements about it be equivalent to $Sup(A)\leq Sup(B)$ I think they are from the diagram.

1. $\forall X_B \exists X_A : X_B \geq X_A $
2. $\forall X_B $ and $ \epsilon >0, \exists X_A: X_A\leq X_B + \epsilon $

Answer 1

Here's a very interesting mistake you are making :

Even $\sup A \leq \sup B$ does not guarantee that every element of $A$ lies to the left of some element of $B$.

For example, simply take $A = [0,1]$ and $B = [0,1)$. Then, $\sup A \leq \sup B$ , since both are equal to $1$, but $A$ still contains an element which is bigger than every element of $B$, namely $1 \in A$.

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Lemma : $\sup A \leq \sup B$ implies that for all $a \in A, \color{blue}{a \neq \sup A}$, there exists $b \in B$ such that $a < b$.

Proof : If $\sup A \leq \sup B$, then fix $a \in A$. We know that $a \color{red}{<} \sup A $, so $a < \sup B$ and hence there is an element of $B$ which is greater than $a$.

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Lemma : However, $\sup A \leq \sup B$ is implied by : for all $a \in A$ there exists $b \in B$ such that $a < b$.

Proof : For all $\epsilon > 0$, there exists $a > \sup A - \epsilon$, now find $b > a > \sup A - \epsilon$, and it follows that $\sup B \geq b > a > \sup A - \epsilon$, for all $\epsilon$, so the lemma follows.

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In conclusion, $\sup A \leq \sup B$ is sandwiched between two very close looking but certainly different conditions :

• It is stronger than the condition that every element of $A$ bar the supremum(if it is in $A$) is strictly to the left of some element of $B$.

• It is weaker than the condition that every element of $A$ is to the left of some element of $B$.

Note that if $A$ does not contain its supremum, then the condition $\sup A \leq \sup B$ is equivalent to any of the above conditions.

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Therefore, a diagram wishing to illustrate $\sup A \leq \sup B$ must illustrate that it is sandwiched between the two cases.

Therefore, one diagram does not do the job : many are required. While you have demonstrated a single and correct example, it is still worth noting from above that the condition given to you is a very subtle one. I leave you to decide how to draw a diagram after this discussion.