# Comparison Property of Supremum Proof

In Apostol's Mathematical Analysis the following theorem's proof was left as an exercise:

Theorem 1.16 (Comparison Property). $$\text{Given nonempty subsets S and T of } \mathbb{R}$$ $$\text{ such that s ≤ t for every s in S and t in T, }$$ $$\text{ if T has a supremum then S has a supremum and }$$ $$\sup S \leq \sup T.$$

I would like to know if my following proof is valid,

Proof. For every s in S and t in T we have, $$s \leq t$$ so S is bounded above by T. Also S is a nonempty subset of $$\mathbb{R}$$ so by the Completeness Axiom $$\exists \sup S$$ and $$\sup T$$ is already given. Now, $$s \leq t \leq \sup T \quad \forall s \in S \ \forall t \in T$$ $$\Rightarrow \sup T \in T$$ and $$s \leq \sup S \quad \forall s \in S$$ $$\Rightarrow \sup S \in T.$$ Assume for contradiction that $$\sup S > \sup T.$$ Then $$\sup S$$ is an upper bound of T but not the least one, so $$t \leq \sup T < \sup S \quad \forall t \in T$$ $$\Rightarrow t < \sup S \quad \forall t \in T$$ $$\Rightarrow \sup S \notin T$$ which is a contradiction as by definition of the supremum $$s \leq \sup S \quad \forall s \in S$$ $$\Rightarrow \sup S \in T.$$ Thus on negating our assumption we get $$\sup S \leq \sup T. \blacksquare$$

• Your proof is wrong. $\sup T \in T$ is not always true. Oct 3, 2022 at 5:32
• Is the part about $\sup S \in T$ true at least? Also I am aware that the supremum of T doesn't necessarily have to be in T in general, but in this specific case $s \leq \sup T$ for all s in S so T's supremum is also an upper bound of S and as T is the set of upper bounds of S we have $\sup T \in T$. Oct 3, 2022 at 5:44

Your proof is incorrect. The line $$s\leq t \leq \sup{T}$$ implies that $$\sup{T} \in T$$ does not hold true in general. As a counterexample, consider $$S = (0,1)$$ and $$T = (1,2).$$ You can show that $$\sup{S} = 1 \leq 2 = \sup{T}$$ but neither $$1$$ nor $$2$$ belong in $$T.$$

Here is my proof of this assertion.

Let $$S, T$$ be two nonempty subsets of the set of real numbers such that, for all $$s\in S, t \in T,$$ $$s \leq t.$$ Further, suppose $$T$$ has a supremum. We claim that $$\sup{S}$$ exists and $$\sup{S}\leq\sup{T}.$$

As $$S,T$$ are nonempty, let $$s\in S, t\in T.$$ By definition of the supremum, $$t \leq \sup{T}.$$ By definition of the sets $$S$$ and $$T,$$ $$s \leq t \leq \sup{T}.$$ Consequently, $$S$$ is bounded above by $$\sup{T}$$ and the supremum of $$S$$ exists by the Completeness Axiom. Since $$\sup{T}$$ is an upper bound of $$S$$, it follows that $$\sup{S}\leq \sup{T}$$ by definition of the least upper bound.

Please see if this proof makes sense to you.

• Ahh, dang it, it was so simple I get it now. The upper bound is ofc greater than or equal to the least upper bound... thanks a lot! Oct 3, 2022 at 6:51