Uniform Continuity exercise I am stuck at this example because I cannot figure out from where comes delta over 2. How can x´be 1/delta. Sorry for the terminologies. Thanks for reading.
 A: The definition of uniform continuity of $g$:
$$\forall \varepsilon > 0 \exists \delta > 0: \forall x,x' \in (-\infty ,+\infty) : ( |x -x'| < \delta ) \rightarrow |g(x) - g(x')| < \varepsilon$$
So the negation (by pure logic, and knowing that an implication is false iff the antecedent is true and the conclusion false:
$g$ is not uniformly  continuous iff
$$\exists \varepsilon > 0 \forall \delta > 0: \exists x,x' \in (-\infty ,+\infty) : ( |x -x'| < \delta ) \land (|g(x) - g(x')| \ge \varepsilon)$$
This is what the quoted proof does: it chooses $\varepsilon= 1$ and for any $\delta> 0$, the constructed points $x, x'$ indeed satisfy $|x -x'| = \frac{\delta}{2} < \delta$ and $|g(x) - g(x') | \ge 1$, as required. 
A: Well  x' is not equals to 1/delta it is assumed to be greater than there so as to  get the contradiction 
A: The definition of uniform continuity says that for any $\epsilon>0$  there is some $\delta$ such that for every pair of points $x,x'$   within $\delta$  of each other, $|g(x)-g(x')|<\epsilon$.
SInce that last $<$ is awkward, they pick a number $\delta/2$ which is assuredley less than $\delta$, and work with that.
