Supremum equation proof. I want to show that
$$\overbrace{\sup\{f(x)-f(y):x,y\in X\}}^{(1)}=\overbrace{\sup\{|f(x)-f(y)|:x,y\in X\}}^{(2)}$$
where $f:X\to\mathbb{R}$ is a bounded function. 

It seems obvious, but I don't quite know how to prove it.  

Here is what I have so far:
I need to show that $(1)\leq (2)$ and $(1)\geq (2)$, which implies that $(1)=(2)$.
$(1)\leq (2):$
Let's take some $x_0, y_0\in X$, then
$$ f(x_0)-f(y_0)\leq |f(x_0)-f(y_0)|\leq \sup\{|f(x)-f(y)|:x,y\in X\}$$
$$ f(x_0)-f(y_0)\leq \sup\{|f(x)-f(y)|:x,y\in X\}$$

Now I would think that

$$\sup\{f(x)-f(y):x,y\in X\}\leq \sup\{|f(x)-f(y)|:x,y\in X\}$$

Is that correct?

$(2)\leq (1):$

I don't have ideas here. 

How to prove it? Any ideas or tips?
 A: $\boxed{\leq \text{inequality}}$ Note that
$$
f(x)-f(y)\leq\left|f(x)-f(y)\right|
$$
holds for all pairs $(x,y)$. Therefore, taking the supremum of both
sides,
$$
\sup_{x,y} \left\{ f(x)-f(y) \right\}
\leq \sup_{x,y} \left| f(x)-f(y) \right| 
$$

$\boxed{\geq \text{inequality}}$ Since $f$ is bounded, we know that for each $\epsilon > 0$, we can find $(u,w)$ such that
$$
\sup_{x,y} \left| f(x)-f(y) \right| \leq  \left| f(u)-f(w) \right| + \epsilon.
$$
Now, define the pair $(u^\prime,w^\prime)$ by
$$
(u^{\prime},w^{\prime})=\begin{cases}
(u,w) & \text{if }f(u)\geq f(w)\\
(w,u) & \text{otherwise}
\end{cases}
$$
so that
$$
\left| f(u)-f(w) \right| = f(u^\prime) - f(w^\prime).
$$
What can you conclude? (remember, $\epsilon$ is arbitrary)
A: Note that for every $x,y \in X$, we have (1) $\geq f(x)-f(y)$. Therefore, given $a,b\in X$, (1) $\geq f(a)-f(b)$ and (1) $\geq f(b)-f(a) =-(f(a)-f(b))$. This implies (1) $\geq |f(a)-f(b)|$. Now, we conclude (1) is an upper bound of $\{|f(x)-f(y)|:x,y\in X\}$.
Since (2) is the supremum, we have the inequality desired. 
A: There's a mildly distracting complication: Let $Y = \{f(x) : x\in X\}$ be the image of the function $X.$
Thus instead of calling our set $\{f(x)-f(y) : x,y\in X\},$ we can call it $\{a-b : a,b\in Y\}.$ In other words, get rid of the function $f$ and state the proposition like this:
$$
\sup\{|a-b| : a,b\in Y\} = \sup\{a-b: a,b\in Y\}.
$$
We have
$$
\{|a-b|:a,b\in Y\} = \{a-b: a,b\in Y\ \&\ a\ge b\} \subseteq \{a-b: a,b\in Y\}. \tag {inclusion}
$$
Do you know how to prove that this inclusion implies that
$$
\sup\{|a-b|:a,b\in Y\} \le \sup\{a-b:a,b\in Y\} \text{ ?}
$$
The other inequality follows from $a-b\le |a-b|.$
