Suprema of two Real Sets

Theorem
Let $S$ and $T$ be real sets.

Let $S$ and $T$ admit suprema.

Then:


 * $\sup S \le \sup T \iff \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$

Proof
Note that because $S$ and $T$ both admit suprema, it follows that they are not empty.

Necessary Condition
Let $\sup S \le \sup T$.

The aim is to establish that:
 * $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$

Let $\epsilon \in \R_{>0}$.

Let


 * $S' = \left\{{s \in S : \exists t \in T : s \le t}\right\}$

There are two cases for an $s \in S$:


 * $s \in S'$, in other words: $\exists t \in T : s \le t$

and:


 * $s \in S \setminus S'$, in other words: $\nexists t \in T : s \le t$

First, consider the case that $s \in S'$.

Let $s \in S'$.

Then $t \in T$ exists such that:

We have found:


 * $\forall \epsilon \in \R_{>0}: \forall s \in S': \exists t \in T: s < t + \epsilon$

Next, consider the second case, that $s \in S \setminus S'$.

Let $s \in S \setminus S'$.

Then:

So, $s = \sup T$ is the only element of $S \setminus S'$.

Observe that $T$ is non-empty as the empty set does not admit a supremum (in $\R$).

A $t \in T$ exists such that:

We have found:


 * $\forall \epsilon \in \R_{>0}: \forall s \in S \setminus S': \exists t \in T: s < t + \epsilon$

We conclude by combining the results for the two cases:


 * $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$

Sufficient Condition
Let $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$.

The aim is to establish that $\sup S \le \sup T$.

Let $\epsilon \in \R_{>0}$.

Observe that $S$ is non-empty as the empty set does not admit a supremum (in $\R$).

An $s \in S$ exists such that:

Let $s$ be such an element of $S$.

A $t \in T$ exists such that:

Let $t$ be such an element of $T$.

Then:

Also see

 * Infima of two Real Sets