Proof on greatest lower bound and least upper bound.

If $A$ is a non-empty bounded subset of $mathbb{R}$, and $B$ is the set of all upper bounds of $A$, prove that
$$glb(B)= lub(A)$$

My reasoning and thinking:

The set of upper bounds may be infinite countable set, right?
Also all the upper bounds (elements of set $B$) are greater than or equal to $x$ for all $x$ in $A$, using definition of upper bound.
Then it is $glb(B)$, I am not getting what does it mean? As $y$ is an element of $B$, then is $glb$ of $y$ is $y$ itself?

Any help/hint please.

Answer

First, observe that for any $b in B$, $lub(A) le b$,
because $b in B$ is an upper bound of $A$ and $lub(A)$ is the least of them. So $lub(A)$ is a lower bound of B, hence:
$$lub(A) le glb(B) text{.}
$$

Because $lub(A)$ is an upper bound of $A$, $lub(A) in B$ by definition of $B$. Thus,
$$glb(B) le lub(A)text{.}
$$
Hence $lub(A) = glb(B)$.

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