Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have to describe what it means, when a function is discontinuous in a. I also have to use quantification notation.

I have tried following:

$$\exists \varepsilon > 0, \exists \delta > 0 : |x-a| < \delta \Rightarrow |f(x)-f(a)| \geq \varepsilon$$

But when I went to the classes they used a different notation. But is this wrong?

share|cite|improve this question
Unfortunately it's wrong, irrespective of the notation. The function $$f(x) = \begin{cases} 0 & x < 0 \\ 1 & x \geq 0 \end{cases}$$ is certainly discontinuous at $0$. – Dylan Moreland Sep 7 '11 at 16:17
In essence, this is the same question as this one and my answer fits here as well. In this question $P(x,y)=|x-y|<\delta$ and $Q(x,y)=|f(x)-f(y)|<\varepsilon$ (these are the notations from the linked question, of course). – Asaf Karagila Sep 7 '11 at 16:22
up vote 3 down vote accepted

I'm afraid that your negation of continuity is incorrect. First, let us remember the definition of "continuous at $a$":

$\forall \epsilon\gt 0\Biggl( \exists \delta\gt 0\biggl(\forall x\bigl( |x-a|\lt \delta \Rightarrow |f(x)-f(a)|\lt \epsilon\bigr)\biggr)\Biggr)$.

Now, how do we write down the negation? (I'm assuming that "discontinuous at $a$" means "defined at $a$ but not continuous, by the way)?

Let's break it down: what is the negation of a "For all" statement?

The negation of $\forall x (P(x))$ is $\exists x(\neg P(x))$.

That is, the negation of "For every $x$, $P(x)$ is true" is "There is at least one $x$ for which $P(x)$ is false". So, the negation of the definition of continuity will be:

$\exists \epsilon\gt 0\Biggl(\neg\Biggl(\forall \delta\gt 0 \biggl(\forall x\bigl( |x-a|\lt \delta \Rightarrow |f(x)-f(a)|\lt \epsilon\bigr)\biggr)\Biggr)\Biggr)$.

Now we need to find the negation of the statement inside the "For all", which is a statement of the form "There exists..." What is the negation of a "There exists" statement?

The negation of $\exists y(Q(y))$ is $\forall y(\neg Q(y))$.

That is: the negation of "There exists a $y$ such that $Q(y)$ is true" is "For every $y$, $Q(y)$ is false". So, to negate that "There exists" statement, we have:

$\exists \epsilon\gt 0\forall\delta\gt 0\Biggl(\neg\Biggl(\forall x\bigl( |x-a|\lt \delta\Rightarrow |f(x)-f(a)|\lt \epsilon\bigr)\Biggr)\Biggr)$.

Now we need to negate a "for all" statement. We already know how to do that:

$\exists \epsilon\gt 0 \forall\delta\gt 0 \Biggl( \exists x\Bigl( \neg\bigl( |x-a|\lt\delta\Rightarrow |f(x)-f(a)\lt\epsilon\bigr)\Bigr)\Biggr)$.

Now we need to find the negation of the implication. What is the negation of an implication?

The negation of $A\Rightarrow B$ is $A\text{ and }\neg B$.

That is, the negation of "If $A$, then $B$" is "$A$, and not $B$." So:

$\exists \epsilon\gt 0 \forall \delta\gt 0\Biggl( \exists x\bigl( |x-a|\lt\delta \text{ and }\neg(|f(x)-f(a)|\lt\epsilon)\bigr)\Biggr)$.

Finally, we need to find the negation of "$|f(x)-f(a)|\lt\epsilon$". That's $|f(x)-f(a)|\geq \epsilon$. So we finally get:

$\exists \epsilon\gt 0\forall\delta \gt 0 \Biggl( \exists x\bigl( |x-a|\lt\delta\text{ and }|f(x)-f(a)|\geq \epsilon\bigr)\Biggr)$.

In words: "there is an $\epsilon\gt 0$ such that, no matter what $\delta\gt 0$ you pick, there is an $x$ which is $\delta$-close to $a$, but with $f(x)$ not $\epsilon$-close to $f(a)$."

Note. Some people say that $f$ is discontinuous at $a$ if and only if it is defined at $a$ but not continuous at $a$. Less common is to say that $f$ is discontinuous at $a$ if and only if it is not continuous at $a$. Under the latter (in my experience uncommon) definition, "$f$ is discontinuous at $a$" would be a disjunction between the formula above, and "$f$ is not defined at $a$".

share|cite|improve this answer
Thank you so much. Your explanation is really good, better than my teacher assistant's explanation :) – Brugerfugl Sep 7 '11 at 16:46

Your attempt is lacking a quantifier for $x$. It will be wrong no matter where you place it, but in different ways. What you should have done is take the definition for being continuous at a: $$\forall \epsilon > 0\; \exists \delta > 0\; \forall x\;\big( |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon \big)$$ and then negated that, moving the negation inwards $$\neg \forall \epsilon > 0\; \exists \delta > 0\; \forall x\;\big( |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon \big)$$ $$\exists \epsilon > 0\; \neg\exists \delta > 0\; \forall x\;\big( |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon \big)$$ $$\exists \epsilon > 0\; \forall \delta > 0\; \neg\forall x\;\big( |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon \big)$$ $$\exists \epsilon > 0\; \forall \delta > 0\; \exists x\;\neg\big( |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon \big)$$ $$\exists \epsilon > 0\; \forall \delta > 0\; \exists x\;\big( |x-a|<\delta \land |f(x)-f(a)|\ge\epsilon \big)$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.