# Thread: show that A has a glb

1. ## show that A has a glb

If A is a nonempty subset of R that is bounded below, show that A has a greatest lower bound. That is, show that there is a number $m \in R$satisfying: 1) m is a lower bound for A; and 2) if x is a lower bound for A, then x<=m. [Hint: Consider the set -A={ $-a: a \in A$} and show that m = -sup(-A) works.]

The hint is confusing me, I'm really not good at proving, any help is appreaciated!

2. ## Re: show that A has a glb

so should I just show that -m = sup(-A), then we have m= - sup(-A) by the least upper bound axiom?

3. ## Re: show that A has a glb

Originally Posted by wopashui
If A is a nonempty subset of R that is bounded below, show that A has a greatest lower bound. That is, show that there is a number $m \in R$satisfying: 1) m is a lower bound for A; and 2) if x is a lower bound for A, then x<=m. [Hint: Consider the set -A={ $-a: a \in A$} and show that m = -sup(-A) works.]

The hint is confusing me, I'm really not good at proving, any help is appreaciated!
Hi wopashui,

This problem is incorrect. There are sets for which there exist lower bounds but not a greatest lower bound. For example let,

$\mathbb{Q}$ be the set of Rational numbers and $A=\left\{a\in\mathbb{Q}~|~a\in\left(\sqrt{2},5 \right) \right\}.$ Clearly there are lower bounds for the set A, such as, 1,0,-1 etc. But a greatest lower bound does not exist. Since if you take any lower bound (say $l\in\mathbb{Q}$) there exist infinity many rational elements in between $l$ and $\sqrt{2}$.

4. ## Re: show that A has a glb

Originally Posted by Sudharaka
Hi wopashui,

This problem is incorrect. There are sets for which there exist lower bounds but not a greatest lower bound. For example let,

$\mathbb{Q}$ be the set of Rational numbers and $A=\left\{a\in\mathbb{Q}~|~a\in\left(\sqrt{2},5 \right) \right\}.$ Clearly there are lower bounds for the set A, such as, 1,0,-1 etc. But a greatest lower bound does not exist. Since if you take any lower bound (say $l\in\mathbb{Q}$) there exist infinity many rational elements in between $l$ and $\sqrt{2}$.

The set has to be bounded below, the one you have given is not.

5. ## Re: show that A has a glb

actually, the glb for your example is root of 2, isn't it? Oh because the problem I have is in R, the example you gave is in Q, yes R and Q appear quite different when we examine the existence of lub or glb fir biunded sets.

6. ## Re: show that A has a glb

Originally Posted by wopashui
The set has to be bounded below, the one you have given is not.
Of course it is. As I have mentioned in my previous post examples of lower bounds for A include, 1,0,-1 etc.

Originally Posted by wopashui
actually, the glb for your example is root of 2, isn't it?
The greatest lower bound should be inside the superset. In my example, $\sqrt{2}\notin\mathbb{Q}$. I think you have to carefully go though the definitions Lower and Upper bound of a set, Supremum and Infimum.

Oh because the problem I have is in R, the example you gave is in Q, yes R and Q appear quite different when we examine the existence of lub or glb fir biunded sets.
Did you denote the the set of Real numbers by R? I thought it was a general set not specifically $\Re$.

7. ## Re: show that A has a glb

Originally Posted by wopashui
If A is a nonempty subset of R that is bounded below, show that A has a greatest lower bound. That is, show that there is a number $m \in R$satisfying: 1) m is a lower bound for A; and 2) if x is a lower bound for A, then x<=m.
I do not like that hint ether.
Let $B = \left\{ {t:\left( {\forall x \in A} \right)\left[ {t < x} \right]} \right\}$
We know that $B\ne\emptyset$ because $A$ has a lower bound.
Because $\left( {\exists s \in A} \right)$ the set $B$ is bounded above.
Now prove the $\sup(B)=\inf(A)$.