1
$\begingroup$

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 $\sup S \leq \sup T. \blacksquare$

$\endgroup$
2
  • $\begingroup$ Your proof is wrong. $\sup T \in T$ is not always true. $\endgroup$
    – Mason
    Commented Oct 3, 2022 at 5:32
  • $\begingroup$ 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 $\sup T \in T$. $\endgroup$ Commented Oct 3, 2022 at 5:44

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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! $\endgroup$ Commented Oct 3, 2022 at 6:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .