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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.enter image description here

3 Answers 3

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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.

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    I will also write, for **not** uniformly continuous, $|g(x)-g(x')| \ge \epsilon$. But I found people simplify it to $|g(x)-g(x')| \gt \epsilon$. – 2017-01-31
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    @N1ng It's a bit stronger but not needed for the proof. – 2017-01-31
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Well x' is not equals to 1/delta it is assumed to be greater than there so as to get the contradiction

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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.