Since I wrote the original post, I have received a lot of answers. However, all of them say why my alternatives do not work. None of them directly answer the question I asked. What makes the definition of continuity superior to the others? I am more interested in knowing why the epsilon-delta definition of continuity works rather than knowing why the variants do not work.


I want to expand on a question that was asked a few years ago.

The $\epsilon, \delta$-definition of continuity is:

$$\forall \varepsilon > 0\ \exists \delta > 0\ \text{s.t. } 0 < |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon $$

What makes the definiton above superior to all of the ones below?

$$\forall \varepsilon > 0\ \exists \delta > 0\ \text{s.t. } |f(x) - f(x_0)| < \varepsilon \implies 0 < |x - x_0| < \delta $$

$$\forall \delta > 0\ \exists \varepsilon > 0\ \text{s.t. } 0 < |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon $$

$$\forall \delta > 0\ \exists \varepsilon > 0\ \text{s.t. } |f(x) - f(x_0)| < \varepsilon \implies 0 < |x - x_0| < \delta $$

And, finally, the one that nags me the most (considering most "intuitive" explanations I have heard, like the one at Khan Academy, this should be almost equivalent to the one above): $$\forall \delta > 0\ \exists \varepsilon > 0\ \text{s.t. } 0< |f(x) - f(x_0)| < \varepsilon \implies |x - x_0| < \delta $$

I can go on making other variations, but I reckon you get the point...

  • 5
    $\begingroup$ All of your proposed definitions seem to lack a quantifier on $x$. $\endgroup$ Commented Sep 16, 2015 at 23:28
  • $\begingroup$ (Deleted my last comment because it was incorrect, but I suggested calling something described by the last definition "4-continuous" since it's the 4th alternate definition you suggested, and similarly for the others.) $\endgroup$ Commented Sep 16, 2015 at 23:33
  • 1
    $\begingroup$ Actually, no, I don't get the point. Is the point that you don't understand the idea that the definition of continuity is supposed to capture? Or that you do understand it but don't understand why the definition does it and the alternatives don't? Or is the point something else entirely? $\endgroup$
    – David
    Commented Sep 16, 2015 at 23:34
  • $\begingroup$ $f(x)=x^2$ is not 4-continuous at any point other than the origin. Consider $x=1$. If we choose $1$ to be our delta, what will be our epsilon? No matter what, it won't work for $x_0=-1$. $\endgroup$ Commented Sep 16, 2015 at 23:37
  • $\begingroup$ The last one would allow $x$ values that are infinitesimally close to $x_0$ to produce values of $f(x)$ which are not infinitesimally close to $f(x_0)$ $\endgroup$
    – WW1
    Commented Sep 16, 2015 at 23:41

3 Answers 3


The definition of continuity of $f$ at $x_0$ is actually not what you write, but $$ \forall \varepsilon > 0\ \exists \delta > 0\ \forall x \big[ 0 < |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon \big] $$ Note that there's a quantifier on $x$ which is missing in your rendering -- if you omit that, "continuity" would be a relation between an $f$ and an $x_0$ and an $x$.

Assuming you meant to universally quantify all of your $x$s in the usual place, let's take your alternatives one by one:

  1. $ \forall \varepsilon > 0\ \exists \delta > 0\ \forall x \big[ |f(x) - f(x_0)| < \varepsilon \implies 0 < |x - x_0| < \delta \big] $

This definition is not satisfied for anything, because no matter what $f$, $x_0$, $\varepsilon$ and $\delta$ are, we can choose $x=x_0$, which makes $|f(x) - f(x_0)| < \varepsilon$ true and $0 < |x - x_0|$ false.

  1. $ \exists \delta > 0\ \forall \varepsilon > 0\ \forall x \big[ 0 < |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon \big] $

This will make something like Dirichlet's function $f(x)=\begin{cases} 1 & x\in\mathbb Q \\ 0 & x\in\mathbb R\setminus\mathbb Q \end{cases}$ -- or indeed any bounded function, however wild it is otherwise -- "continuous", because for every $\delta$ we can take $\varepsilon=2$ and make the inner bracket true. This obviously doesn't capture any intuitive notion of continuity.

  1. $ \exists \delta > 0\ \forall \varepsilon > 0 \ \forall x \big[ |f(x) - f(x_0)| < \varepsilon \implies 0 < |x - x_0| < \delta \big] $

This has exactly the same problem as 1: It is never satisfied.

  1. $ \forall \varepsilon > 0\ \exists \delta > 0\ \forall x \big[ 0< |f(x) - f(x_0)| < \varepsilon \implies |x - x_0| < \delta \big] $

This would mean that the constant zero function, $f(x)=0$ would not be continuous anywhere, nor would the sine or cosine functions. There would never be any $\delta$ large enough to capture all of the places where the function's value is close to $f(x_0)$.

On the other hand, option 4 would still allow something like $$ f(x) = \begin{cases} x+1 & x\le 0 \\ x-1 & x> 0 \end{cases} $$ to count as "continuous" everywhere.


Henning Makholm has explained why the first three are unsatisfactory: The first and third would imply that no functions are continuous, because you bar $x = x_0$ for $f(x) = f(x_0)$, yet clearly $f(x_0) = f(x_0)$; and the second is unsatisfactory because any bounded function would be considered continuous (along with many other functions).

The fourth definition is essentially continuity of the inverse function. As such, we would expect any function that has intervals over which $f$ is constant to fail this definition of "continuity." And indeed, if we have $f(x) = 1$ and $x_0 = 0$, then we have for all $x$, $|f(x)-f(x_0)| < \varepsilon$ for whatever value of $\varepsilon$ you care to choose, but none of those is sufficient to imply that $|x-x_0| < \delta$ for any finite $\delta$.


Taking your four alternatives in turn:

  1. Consider $f(x)=\sin(x)$, which usually thought to be continuous though $|f(n\pi)-f(0)|=0$ for all integer $n$
  2. Consider $f(x) = \text{sign}(x)$, with $\epsilon = 3$ bigger than the jump at $0$, which usually thought to be discontinuous at $0$
  3. Same issue as 1
  4. Consider $f(x)=x$ for positive $x$, and $0$ otherwise, which is usually thought to be continuous for all $x$ though would not satisfy this definition for $x_0 \lt -\delta$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .