Disproving continuity using epsilon delta

Suppose I wanted to show that $\ln(x)$ is not uniformly continuous on $\mathbb{R}_{>0}$.

I know if I negate the definition of uniform continuity, that I have to find an $\varepsilon$ such that

$\forall \delta > 0 \quad \exists x$ :

$$|x-y|<\delta \implies |f(x)-f(y)|\geq \varepsilon$$

Am I correct in saying I can choose epsilon without any restrictions? And for $x,y$ I have to make sure, that their absolute difference remains smaller than delta, for any $\delta$ chosen, now matter how small?

Would it then be correct to state that $\varepsilon:=\frac{\ln(2)}{2}$ and $x:= \delta$ and $y:= \frac{\delta}{2}$ work, but I couldn't for example choose $x:= e$ and $y:= 1$ ?

• $\forall\,\delta>0, x_\delta := \delta/2, y_\delta := \delta$. $$|f(x)-f(y)| = |\ln(x) - \ln(y)| = |\ln(x/y)| = \ln2 = 2\varepsilon > \varepsilon$$ – GNUSupporter 8964民主女神 地下教會 Feb 17 '17 at 15:37

1 Answer

The definition of being uniformly continuous on $I\subseteq \mathbb{R}$ is $$\forall \varepsilon > 0 \; \exists \delta > 0 \quad \forall x,y \in {I}, \; |x-y|<\delta \Rightarrow |f(x)-f(y)|<\varepsilon$$

So its negation is (remembering that "not $\forall$ = $\exists$") is $$\exists \varepsilon > 0 \; \forall \delta > 0 \quad \exists x,y \in {I}, \; |x-y|<\delta \text { and } |f(x)-f(y)|>\varepsilon$$

So yes, you would be correct. Of course there are some $x$ and $y$ such that $|f(x)-f(y)|$ is still going to be less than $\varepsilon$, but you just need a pair for which it's not, and whose distance is less than $\delta$, to provide the counterexample.

• So as a kind of "strategy", you'd often try to construct some x and y so that they cancel out? As in, I try to construct something that makes $|f(x)-f(y)|$ greater than a certain value but still has the property of $|x-y|< \delta$ for all deltas and then choose epsilon accordingly? – Jonathan Feb 17 '17 at 15:41
• Well, you can usually see where things go wrong by trying to prove that $f$ $\textit{is}$ uniformly continuous, like, in your example you would try to derive some inequality for $h$ from $$ln(x+h)-ln(x) < \varepsilon$$ – AnalysisStudent0414 Feb 17 '17 at 15:44