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

[evinda]

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)

 

[I like Serena]

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)

 

[evinda]

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)

 

[I like Serena]

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)

 

[evinda]

Yep. (Nod)

Great... Thank you (Happy)