Supremum property of functions 
Let $f$ and $g$ be functions from $\mathbb{R}$ to $\mathbb{R}$. Prove that:
a.) $\sup\{f(x)+g(x):x\in\mathbb{R}\}\leq \sup\{f(x):x\in\mathbb{R}\}+\sup\{g(x):x\in\mathbb{R}\}$
b.) $\sup\{f(x)+g(y):x,y\in\mathbb{R}\} =  \sup\{f(x):x\in\mathbb{R}\}+\sup\{g(y):y\in\mathbb{R}\}$

Attempted proof a.) - Since $f(x)\leq \sup\{f(x):x\in\mathbb{R}$ and $g(x)\leq\sup\{g(x):x\in\mathbb{R}\}$ then we have $$f(x) + g(x) \leq \sup\{f(x):x\in\mathbb{R}\} + \sup\{g(x):x\in\mathbb{R}\}$$ Thus, $f(x) + g(x)$ are bounded from above by $\sup\{f(x):x\in\mathbb{R}\} + \sup\{g(x):x\in\mathbb{R}\}$. So, $$\sup\{f(x)+g(x):x\in\mathbb{R}\}\leq \sup\{f(x):x\in\mathbb{R}\}+\sup\{g(x):x\in\mathbb{R}\}$$
I am not exactly sure how to show b.) any suggestions is greatly appreciated.
 A: b) Let $x, y \in \mathbb{R}$ 
Then $f(x)+g(y) \leq \sup f +\sup g$. Taking the supremum for all $x,y$ we get that:
$\sup \{f(x)+g(y)| x,y \in \mathbb{R}\} \leq \sup f +\sup g$.
Besides, is obvious that for $x,y \in \mathbb{R}, f(x)+g(y) \leq \sup \{f(x)+g(y)| x,y \in \mathbb{R}\} $
Taking the supremum in $x$, we get $\sup f +g(y) \leq \sup \{f(x)+g(y)| x,y \in \mathbb{R}\} $. 
Then taking the supremum in $y$, we get $\sup f + \sup g \leq \sup \{f(x)+g(y)| x,y \in \mathbb{R}\} $
Finally $\sup \{f(x)+g(y)| x,y \in \mathbb{R}\} = \sup f + \sup g $
A: Note that in order to prove part b) it is suffices to prove that 
$$
\sup(A+B)=\sup A+\sup B
$$
where $A+B=\{x+y:x\in A,y\in B\}$. Set $\sup A=a$ and $\sup B=b$. We shall prove that $\sup(A+B)=a+b$. We need to prove that 
$$
\begin{array}{ll}
i)  & \hbox{$\sup(A+B)\leqslant a+b$} \\
ii) & \hbox{for all $\epsilon>0$, there exist $z_0\in A+B$ such that $a+b-\epsilon<z_0$}
\end{array}
$$
Proof: i) Since $\sup A=a$ and $\sup B=b$, then $x\leqslant a$ and $y\leqslant b$ for every $x\in A$ and $y\in B$. Therefore $x+y\leqslant a+b$ for every $x+y
\in A+B$, as required.
ii) Let $\epsilon>0$. Since $\sup A=a$ and $\sup B=b$, there exist $x_0\in A$ and 
$y_0\in B$ such that $a-\epsilon/2<x_0$ and $b-\epsilon/2<y_0$. Therefore, $a+b-\epsilon<x_0+y_0$. The assertion then follows. 
To prove part a) note that 
$$
\{f(x)+g(x):x\in R\}\subseteq\{f(x):x\in R\}+\{g(x):x\in R\}\}
$$
Therefore 
$$
\sup\{f(x)+g(x):x\in R\}\leqslant\sup(\{f(x):x\in R\}+\{g(x):x\in R\}\})
$$
The assertion then follows using the identity $\sup(A+B)=\sup A+\sup B$
