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The definition my textbook gives for continuity is of a function $f: X \to Y$ is: $f$ is continuous $\iff \forall x,x' \in X, \forall \epsilon > 0, \exists \delta > 0: d_X(x,x')<\delta \implies d_Y(f(x),f(x')) < \epsilon$.

To show that a function is not continuous can I show that the negation is true?

i.e. $f$ is not continuous $\iff \exists \epsilon > 0: \forall \delta > 0, d_X(x,x') < \delta$ does not imply $d_Y(f(x),f(x')) < \epsilon$

How do I actually do it?

Thanks for you help.

Matthew Riley
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    Yes, if you can prove its negation, you've proven the function not to be continous. However, continuity is a local behaviour thus you need to show that

    $ \exists\varepsilon > 0\ \forall\delta > 0\ \exists x\in X : d(x,x') < \delta \wedge d(f(x),f(x')) < \varepsilon$

     – Zest Oct 04 '18 at 15:53
  • Your definition of continuity is wrong. The quantifier over $x'$ (or over $x$, but only one of the two) should be on the inside. Otherwise you could trivially always take $\delta = d_X(x, x')/2$ and every function would be continuous. – Mees de Vries Oct 04 '18 at 15:56
  • @Zest I believe you need $\exists x'$ before the $\exists \varepsilon$. – Yanko Oct 04 '18 at 15:57
  • Considering $ \neg(\forall\varepsilon > \exists\delta > 0 \forall x \in X : d(x,x') < \delta \wedge d(f(x),f(x')) < \varepsilon)$ you get $\exists\varepsilon > 0 \forall\delta >0 \exists x \in X: d(x,x') < \delta \wedge d(f(x),f(x')) > \varepsilon$ – Zest Oct 04 '18 at 16:09

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Keep in mind that continuity is a local behaviour. A function $f: X \to Y $ between metric spaces is called continuous if it is continuous for every $x \in X$. Thus we call a function not continuous if there is an $x \in X$ at which $f$ is not continuous.

In order to negate the $\varepsilon, \delta$ criteria for continuity you'd like to show that

$$ \neg(\forall\varepsilon > \exists\delta > 0 \forall x \in X : d(x,x') < \delta \wedge d(f(x),f(x')) < \varepsilon)$$ which is $$\exists\varepsilon > 0 \forall\delta >0 \exists x \in X: d(x,x') < \delta \wedge d(f(x),f(x')) > \varepsilon$$

Which basically means that there is an $\varepsilon$ such that for all $\delta > 0$ there is an $x \in X$ for which it holds that $$d(x,x') < \delta$$ but $$d(f(x),f(x')) > \varepsilon$$

thus you need to find the exception $x \in X$ (and additionally an $\varepsilon$) at which the $\varepsilon,\delta$ criteria fails.

Zest
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As an example, consider the function

$f(x) = 0 \quad x \le 0 \\ f(x) = 1 \quad x > 0 $

and test for continuity at $x=0$. No matter how small we make $\delta$ we can always find an $x'$ closer than $\delta$ to $0$ where $f(x')=1$ and so $|f(x')-f(x)|=1$. For instance, we could use $x'=\frac{\delta}{2}$.

So if we take a value of $\epsilon$ that is less than 1 - say $\epsilon = \frac{1}{2}$ - then there is no value of $\delta$ that ensures that $|f(x')-f(0)| < \epsilon$ for all $x'$ such that $|x'-0| < \delta$. Therefore $f(x)$ is not continuous at $x=0$.

gandalf61
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