Proving that the least upper bound of is less than or equal to the greatest lower bound of B

Question

I'm failing to understand the basics concepts of upper/lower bounds and I was hoping I could get some guidance through a simple question.

Suppose that $A$ and $B$ are two non-empty bounded sets of real numbers, such that $x \leq y$ for every $x \in A$ and $y \in B$. Show $lubA \leq glbB$

I have no idea how to even start this proof. I obviously need to somehow relate $x \leq y$ to the boundaries of their sets, that is $lubA \leq glbB$. Although I'm not even sure if my idea is correct. Any help?

Answer 1

I'm going to use $\sup$ and $\inf$ for lub and glb, respectively, as these are more natural for me. These are exactly equivalent terms.

Pick $x \in A$. (We can do this because $A$ is nonempty.) Then $x \leq \sup A$ (by definition). But $x \leq y$ for all $y \in B$, so this $x$ is a lower bound of $B$. But any lower bound of $B$, by definition, must be less than or equal to the greatest lower bound of $B$. So $x \leq \inf B$. Since this $x$ was arbitrary, we have $x \leq \inf B$ for all $x \in A$. But then $\inf B$ is an upper bound of $A$, so that means that it cannot possibly be less than the least upper bound of $A$, so $\sup A \leq \inf B$.

Answer 2

Hints: (assuming $A,B\neq \varnothing$)since $x\leq y$ for every $x\in A$ Therefore $y$ is an upper bound of $A$. Thus $\text{lub}A\leq y$. Similarly, $x\leq \text{glb}B$.

Answer 3

The least upper bound of a set is called its supremum ($\text{sup}$) and the greatest lower bound of a set is called its infimum ($\text{inf}$).

Firstly, we have,

$\forall x\in A\text{ and } y\in B, x\leq y \implies y\text{ is an upper bound for every }x \implies\text{sup}A\leq y,\text{ }\forall y\in B$ as $\text{sup}A$ is the least upper bound of the set $A$.

Now we have,

$\forall y\in B, \text{ sup}A\leq y\implies\text{sup}A\leq\text{inf}B$ as $\text{inf} B$ is the greatest lower bound of the set $B$.

Thus, proved.