=== 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: :$S' = \set {s \in S : \exists t \in T : s \le t}$ 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: {{begin-eqn}} {{eqn | l = s       | o = \le       | r = t }} {{eqn | ll= \leadsto       | l = s       | o = \le       | r = t < t + \epsilon       | c = as $t < t + \epsilon$ is true }} {{eqn | ll= \leadsto       | l = s       | o = <       | r = t + \epsilon       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \nexists t \in T : s       | o = \le       | r = t }} {{eqn | ll= \leadsto       | l = \forall t \in T : s       | o = >       | r = t       | c = so, $s$ is an upper bound for $T$ }} {{eqn | ll= \leadsto       | l = s       | o = \ge       | r = \sup T       | c = as (a) $s$ is an upper bound for $T$ and (b) $\sup T$ is the least upper bound of $T$ }} {{eqn | ll= \leadsto       | l = \sup S       | o = \ge       | r = s \ge \sup T       | c = as $\sup S \ge s$ is true since (a) $\sup S$ is an upper bound for $S$ and (b) $s \in S$ }} {{eqn | ll= \leadsto       | l = \sup T       | o = \ge       | r = \sup S \ge s \ge \sup T       | c = as $\sup S \le \sup T$ is true }} {{eqn | ll= \leadsto       | l = \sup T       | o = \ge       | r = s \ge \sup T }} {{eqn | ll= \leadsto       | l = s       | r = \sup T }} {{end-eqn}} So, $s = \sup T$ is the only element  of $S \setminus S'$. A $t \in T$ exists such that: {{begin-eqn}} {{eqn | l = \sup T - t       | o = <       | r = \epsilon       | c = by Supremum of Subset of Real Numbers is Arbitrarily Close }} {{eqn | ll= \leadsto       | l = \sup T       | o = <       | r = t + \epsilon }} {{eqn | ll= \leadsto       | l = s       | o = <       | r = t + \epsilon       | c = as $s = \sup T$ }} {{end-eqn}} 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: : $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$ {{qed|lemma}} === 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 $\epsilon \in \R_{>0}$. An $s \in S$ exists such that: {{begin-eqn}} {{eqn | l = \sup S - s       | o = <       | r = \epsilon       | c = by Supremum of Subset of Real Numbers is Arbitrarily Close }} {{eqn | ll = \iff       | l = s       | o = >       | r = \sup S - \epsilon }} {{end-eqn}} Let $s$ be such an element of $S$. A $t \in T$ exists such that: {{begin-eqn}} {{eqn | l = s       | o = <       | r = t + \epsilon       | c = as $\forall \epsilon' \in \R_{>0}: \forall s' \in S: \exists t \in T: s' < t + \epsilon'$ }} {{eqn | ll = \iff       | l = t       | o = >       | r = s - \epsilon }} {{end-eqn}} Let $t$ be such an element of $T$. Then: {{begin-eqn}} {{eqn | l = t       | o = >       | r = s - \epsilon }} {{eqn | ll= \leadsto       | l = t       | o = >       | r = s - \epsilon \text { and } s > \sup S - \epsilon        | c = as $s > \sup S - \epsilon$ is true }} {{eqn | ll= \leadsto       | l = t       | o = >       | r = s - \epsilon \text{ and } s - \epsilon > \sup S - 2 \epsilon }} {{eqn | ll= \leadsto       | l = t       | o = >       | r = s - \epsilon > \sup S - 2 \epsilon }} {{eqn | ll= \leadsto       | l = t       | o = >       | r = \sup S - 2 \epsilon }} {{eqn | ll= \leadsto       | l = \sup T       | o = \ge       | r = t > \sup S - 2 \epsilon       | c = as $\sup T \ge t$ is true since (a) $\sup T$ is an upper bound for $T$ and (b) $t \in T$ }} {{eqn | ll= \leadsto       | l = \sup T       | o = >       | r = \sup S - 2 \epsilon }} {{eqn | ll= \leadsto       | l = \sup T       | o = \ge       | r = \sup S       | c = as $\epsilon$ is an arbitrary element of $\R_{>0}$ }} {{eqn | ll= \leadsto       | l = \sup S       | o = \le       | r = \sup T }} {{end-eqn}} {{qed}}