Are these two propositions equivalent when dealing with subsets of real numbers?

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Discussion Overview

The discussion revolves around the equivalence of two propositions concerning subsets of real numbers: the existence of least upper bounds for non-empty upper bounded subsets and the existence of greatest lower bounds for non-empty lower bounded subsets. The scope includes mathematical reasoning and proof verification.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant proposes a proof that if each non-empty upper bounded subset of real numbers has a least upper bound, then each non-empty lower bounded subset has a greatest lower bound, using the transformation of sets.
  • Another participant challenges the proof by stating it only applies to sets of the form $-A$ and suggests that the proof should start with any non-empty lower bounded subset.
  • A later reply agrees with the challenge and reformulates the proof, asserting that if (1) holds, then for any non-empty lower bounded subset $B$, the corresponding set $A = -B$ is upper bounded and thus has a least upper bound.
  • Participants express uncertainty about the completeness and correctness of the proof, seeking confirmation from others.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the completeness of the proof. There is acknowledgment of the need for a more general approach, but no final agreement on the validity of the arguments presented.

Contextual Notes

The discussion highlights the need for careful consideration of definitions and the conditions under which the propositions hold. There are unresolved aspects regarding the generality of the proof and the assumptions made about the subsets.

evinda
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Hello! (Wave)

I want to show that the following is equivalent:

  1. each non-empty and upper bounded subset of real numbers has a least upper bound.
  2. each non-empty and lower bounded subset of real numbers has a greatest lower bound.

I have thought the following so far:

$1. \Rightarrow 2.$:

Let $A$ be an arbitrary non-empty and upper bounded subset of real numbers with least upper bound $a$.
Let $B=-A=\{ -x \mid x \in A\}$.
We have that $x \leq a, \forall x \in A$. Then $-x \geq -a$, so $y \geq -a, \forall y \in B$ and thus $B$ has $-a$ as its greatest lower bound.

Is the proof that 1. implies 2. as above complete? (Thinking)
 
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evinda said:
Hello! (Wave)

I want to show that the following is equivalent:

  1. each non-empty and upper bounded subset of real numbers has a least upper bound.
  2. each non-empty and lower bounded subset of real numbers has a greatest lower bound.

I have thought the following so far:

$1. \Rightarrow 2.$:

Let $A$ be an arbitrary non-empty and upper bounded subset of real numbers with least upper bound $a$.
Let $B=-A=\{ -x \mid x \in A\}$.
We have that $x \leq a, \forall x \in A$. Then $-x \geq -a$, so $y \geq -a, \forall y \in B$ and thus $B$ has $-a$ as its greatest lower bound.

Is the proof that 1. implies 2. as above complete? (Thinking)

Hey evinda!

You have proven it only for a $B$ that is of the form $-A$, but formally it should be for each $B$ that is non-empty and has a lower bounded subset.
That is, we should start with a $B$ that is a non-empty and lower bounded subset of real numbers.
And then, using (1), show that $B$ has a greatest lower bound, shouldn't we? (Thinking)
 
I like Serena said:
Hey evinda!

You have proven it only for a $B$ that is of the form $-A$, but formally it should be for each $B$ that is non-empty and has a lower bounded subset.
That is, we should start with a $B$ that is a non-empty and lower bounded subset of real numbers.
And then, using (1), show that $B$ has a greatest lower bound, shouldn't we? (Thinking)

Oh yes, right... (Nod)

So, we suppose that (1) holds.
Let $B$ be an arbitrary non-empty and lower bounded subset of real numbers. Then the set $A=-B=\{ -y \mid y \in B\}$ is non-empty and upper bounded. Since (1) holds, $A$ has a least upper bound, i.e. $-x \leq a, \forall x \in B$, impying that $B$ has a greatest lower bound, since then $x \geq -a, \forall x \in B$.

Is it right like that? (Thinking)
 
evinda said:
Oh yes, right...

So, we suppose that (1) holds.
Let $B$ be an arbitrary non-empty and lower bounded subset of real numbers. Then the set $A=-B=\{ -y \mid y \in B\}$ is non-empty and upper bounded. Since (1) holds, $A$ has a least upper bound, i.e. $-x \leq a, \forall x \in B$, impying that $B$ has a greatest lower bound, since then $x \geq -a, \forall x \in B$.

Is it right like that?

Yep. (Nod)
 
Klaas van Aarsen said:
Yep. (Nod)

Great... Thank you (Happy)
 

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