(Riemann Integrals) Three proofs from elementary analysis Lemma 1: Suppose $f,g:[a,b] \to \mathbb{R}$ are bounded functions.
Prove:
(a) $\sup\{f(x)+g(x)| x \in [a,b]\} \le \sup\{f(x)|x\in[a,b]\} + \sup\{g(x)|x\in[a,b]\}$  
Lemma 2: Suppose that $A$ and $B$ are non-empty subsets of $\mathbb{R}$. Suppose
that whenever $x\in A$ and $y \in B$, $x\le y$
(a) Prove that $\sup A \le \inf B$  
Lemma 3: Suppose $A$ and $B$ are subsets of $\mathbb{R}$ and $A$ is a subset of $B$, $A$ is non-empty.  
Prove $\sup A \le \sup B$.  
I understand all of these intuitively, but I cannot come up with a proof. (That would
be deemed acceptable.)
 A: The sup (resp. inf) of a nonempty set is the least upper bound (resp. greatest lower bound), when an upper bound (resp. lower bound) exists. Otherwise, it is $+\infty$ (resp. $-\infty$). It is in the definition, but it's good to stress ou that the sup (resp. the inf) is itself an upper bound (resp. a lower bound). 
Standard strategy to prove that $\sup S\leq M$ for a set $S$ and a number $M$: prove that $M$ is an upper bound for $S$. That's as simple as that. When the inequality involves several sup/inf, one usually tries to fix some elements on one side and to get a uniform estimate on the other side. 
Lemma 1: Following a common usage, I'll denote $\|h\|_\infty:=\sup\{h(x)\;;\;x\in[a,b]\}$ for a real valued function $h$ defined on $[a,b]$. That is the sup of the range of $h$. Of course, for every $x\in[a,b]$, $h(x)\leq \|h\|_\infty$. Now observe that 
$$
(f+g)(x)=f(x)+g(x)\leq\|f\|_\infty+\|g\|_\infty\qquad\forall x\in[a,b].
$$
In words, the number $\|f\|_\infty+\|g\|_\infty$ is either infinite or an upper bound of the range of the function $f+g$. So its least upper bound is not greater than that, i.e.
$$
\|f+g\|_\infty\leq \|f\|_\infty+\|g\|_\infty.
$$
Lemma 2:  Fix $x\in A$ first. Then 
$$
x\leq y\qquad\forall y\in B.
$$
This means that $x$ is a lower bound for $B$. So it does not exceed the greatest lower bound $\inf B$:
$$
x\leq \inf B.
$$
Now set $x$ free and observe the above holds for every $x\in A$. So $\inf B$ is an upper bound for $A$. Hence
$$
\sup A\leq \inf B.
$$
Lemma 3: This one is slightly different. For every $b\in B$, we have $b\leq \sup B$, as it is an upper bound. Since $A\subseteq B$, this is true a fortiori for all $a\in A$:
$$
a\leq \sup B\qquad\forall a\in A.
$$
So $\sup B$ is either infinite or an upper bound for $A$. Hence
$$
\sup A\leq \sup B.
$$
Note: it makes things slightly easier to write to allow $+\infty$ to be an upper bound, and $-\infty$ a lower bound.
