Another supremum and infimum problem

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Discussion Overview

The discussion centers around a problem involving the supremum and infimum of non-empty subsets of the real numbers, specifically proving that if every element of one set is less than or equal to every element of another set, then the supremum of the first set is less than or equal to the infimum of the second set. The scope includes mathematical reasoning and proofs related to real analysis.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants assert that since \( S \) and \( T \) are non-empty subsets of \( \mathbb{R} \), the infimum of \( T \) and the supremum of \( S \) exist.
  • There is a claim that for all \( s \in S \), it holds that \( s \le \inf(T) \), but participants question the reasoning behind this assertion.
  • One participant suggests that every \( s \in S \) acts as a lower bound for \( T \), leading to the conclusion that \( s \le \inf(T) \).
  • Another participant proposes that if \( \sup S > \inf T \), then there must exist an \( s \in S \) such that \( s > \inf T \), which would create a contradiction, prompting further discussion on the correctness of this reasoning.
  • A later reply confirms the correctness of the contradiction argument presented regarding the relationship between \( \sup S \) and \( \inf T \).

Areas of Agreement / Disagreement

Participants express uncertainty about the reasoning behind certain claims, particularly regarding the relationship between elements of \( S \) and \( T \). While some points are affirmed, the discussion remains unresolved on the foundational reasoning for the inequalities presented.

Contextual Notes

Participants note the need for clarity on the definitions of supremum and infimum, as well as the implications of the properties of lower and upper bounds in the context of the sets involved.

alexmahone
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Let $S$ and $T$ be non-empty subsets of $\mathbb{R}$, and suppose that for all $s\in S$ and $t\in T$, we have $s\le t$. Prove that $\sup S\le\inf T$.
 
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Alexmahone said:
Let $S$ and $T$ be non-empty subsets of $\mathbb{R}$, and suppose that for all $s\in S$ and $t\in T$, we have $s\le t$. Prove that $\sup S\le\inf T$.
Clearly $ \inf(T)$ & $\sup(S) $ exist. WHY?

$\forall s\in S$ we know $s\le\inf(T)~.$ WHY?$
 
Last edited:
Plato said:
Clearly $ \inf(T)$ & $\sup(S) $ exist. WHY?

Because $S$ and $T$ are non-empty subsets of $\mathbb{R}$.

$\forall s\in S$ we know $s\le\inf(T)~.$ WHY?

I don't know. Why?
 
Alexmahone said:
I don't know. Why?
Well every $s\in S$ is a lower bound for $T$.
Therefore $s\le\inf(T)$.
 
Plato said:
Well every $s\in S$ is a lower bound for $T$.
Therefore $s\le\inf(T)$.

So if $\sup S>\inf T$, there must be an $s\in S$ such that $s>\inf T$. (Otherwise, $\inf T$ is a smaller upper bound for $S$ than $\sup S$.) So we get a contradiction. (Is that correct?)
 
Alexmahone said:
So if $\sup S>\inf T$, there must be an $s\in S$ such that $s>\inf T$. (Otherwise, $\inf T$ is a smaller upper bound for $S$ than $\sup S$.) So we get a contradiction. (Is that correct?)
Yes. It is correct.
 

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