$\inf$ and $\sup$ of subsets of $\mathbb{R}$ 
Suppose $A, B$ are non-empty subsets of $\mathbb{R}$, with the following property: $$\forall x\in A, y\in B, x\le y,$$ and $B$ is bounded above. Show that $\sup A \le \inf B$.

We have that $A$ is bounded above and $B$ is bounded below, and any element $x$ of $A$ is a lower bound of $B$, and each $y$ in $B$ is a upper bound of $A$. But I cant go on from here. 
 A: Note that $B$ is contained in the set $U(A)$ of upper bounds of $A$. Therefore
$$
\inf B\ge\inf U(A)=\min U(A)=\sup A
$$
Here I use the property that if $X\subseteq Y$ and both sets are lower bounded, then $\inf X\ge\inf Y$.
Proof. $\inf X$ is the greatest lower bound of $X$; every lower bound of $Y$ is also a lower bound of $X$, so the greatest lower bound of $Y$ is a lower bound of $X$. This means that $\inf Y\le\inf X$.
A: I think this is a mildly interesting way to get the answer.
Choose any two points $x_0 \in A$ and $y_0 \in B$. To find the $\sup$ and $\inf$, we can assume that both $A$ and $B$ are contained in the closed interval $K =  [x_0, y_0]$ (a singleton interval is OK here).
The intersection of any finite number of closed intervals of the form $[x, y]$ with $x \in A$ and $y \in B$ is another non-empty closed interval with the left endpoint in $A$ and the right endpoint in $B$. Since $K$ is compact, the intersection of all these closed sets is a non empty closed set, and it would have to also be a closed interval, say, 
$[a, b]$
If $c \in [a, b]$ then $\sup A \le c \le \inf B$. 
Proof: Since $c$ belong to every interval $[x,y]$, it is both an upper bound of $A$ and a lower bound of $B$.
For extra credit show that 
$\sup A = a $
and 
$\inf B = b $
Note: The only requirements here are 
$A \ne \emptyset$
$B \ne \emptyset$
$\forall (x\in A, y\in B), \, x\le y$
A: Let $\beta = \inf B$ and $ \alpha=\sup A$. 
We have that $x \le \beta$ for all $x\in A$. In fact, if there's a $x_0\in A$ such that $\beta < x_0$, then $x_0$ would be a lower bound of $B$ greater than $\beta$, which is absurd.
Then if $\beta < \alpha$, $\beta$ would be a upper bound of $A$ lower than $\alpha$, which is absurd. Hence we must have $\alpha \le \beta.$
A: $A,B \subset \mathbb{R}$
$\forall x\in A, y \in B$,  $x \le y$.
$A$ is bounded  above : $\sup (A)$ exists.
$B$ is bounded below: $\inf (B)$ exists.
1)There exists a sequence $x_n \in A , n\in \mathbb{N},$ with
$\lim_{n \to \infty} x_n = \sup(A)$.
2)There exists a sequence $y_n \in B , n\in \mathbb{N},$ with
$\lim_{n \to \infty} y_n = \inf(B)$.
Since $x_n \le y_n$,  $n \in \mathbb{N},$ we have 
$\lim_{n \to \infty} x_n \le \lim_{n \to \infty} y_n$, 
$\sup(A) \le \inf(B)$.
A: By contradiction. Let $a=\sup(A)$ and $b=\inf(B)$. Suppose $a>b$. We know that $A\cap (a-\epsilon,a]\neq\emptyset~~~\forall\epsilon>0$ and $B\cap[b,b+\epsilon)\neq\emptyset~~~\forall\epsilon>0$. Since $a>b$ this means that we can choose $\epsilon$'s such that $B\cap[b,\beta)\neq\emptyset$ and $A\cap(\alpha,a]\neq\emptyset$ where $\beta<\alpha$. Thus, there exists $x\in B:x< \beta$ and $y\in A:y>\alpha$. Hence, $x<\beta<\alpha<y$, which is a contradiction. 
A: Let $a=\sup(A)$ and $b=\inf(B)$. Suppose $a \gt b$. 
Set $\gamma = .5 (a + b)$.
Select $x \in A \cap (\gamma,a]$ and $y \in B \cap [b, \gamma)$ so that  $x \gt \gamma \gt y$.
This is a contradiction.
