# Are there continuous functions for which the epsilon-delta property doesn't hold? [duplicate]

The standard definition of continuity is as follows:

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

This may sound silly, but is the converse true? In other words, is there an example of a function which is continuous, but for which this property doesn't hold? I thought I remembered reading something about the exponential function being an example of a continuous function for which the epsilon-delta implication is false, since it gets arbitrarily steep as x approaches infinity, but I could be wrong. Thanks!

• Definitions work in both directions. The exponential function definitely satisfies the definition of continuity. Nov 7, 2016 at 23:52
• You are treating the definition as if it were a result. Because it's a definition, the word "continuous" means exactly your displayed bit.
– zhw.
Nov 7, 2016 at 23:58
• Yes this sounds silly but don't be afraid to ask silly questions! As long as you show some thinking behind it I believe people will be more than willing to help. Nov 8, 2016 at 0:02
• Since you mentioned a function that "gets arbitrarily steep as x approaches infinity", you may be confusing the ideas of continuity and uniform continuity. Or, if you haven't learned about uniform continuity yet, your intuition about what "continuity" means doesn't match its mathematical definition. Nov 8, 2016 at 1:47
• The definitions of continuous and uniformly continuous both use episilon-delta notation. The only difference between them is that two of the terms have switched places. Nov 8, 2016 at 9:37

The statement you made is the definition of continuity (in a metric space, where the notion of distance or $|x-x_0|$ makes sense). So it works both ways: If the function is continuous, it meets the definition and if it meets the definition then it is continuous.

You are encountering the idea of a function being "uniformly continuous." In the definition you wrote, there is nothing said about whether then same $\delta(\epsilon)$ can work irrespective of the point $x_0$. Your exponential example is indeed continuous, but it is not uniformly continuous, because $\delta$ has to depend both on $\epsilon$ and $x_0$.

In contrast, a function like $\frac{x^3}{1+x^2}$ is uniformly continuous, because for any given $\epsilon > 0$ you can find a $\delta$ such that that defining condition holds for every real $x_0$.

• Oh, uniform like uniform convergence sort of? Nov 8, 2016 at 21:32

Since it's a definition, your first "if" should really be read as an "if and only if". So the converse you ask about always holds.

The definition of a continuous function is that it satisfies the $\epsilon-\delta$ property. Definitions, by convention, work both ways. Hence, we would say that any function satisfying the $\epsilon-\delta$ property is continuous by definition.

I should add that the $\epsilon-\delta$ property, is something that would hold only in metric spaces i.e. where you have a distance function between points. In a general topological space (which need not have a distance function on it!), the definition of continuity of a function $f$ between topological spaces $X$ and $Y$ , is that the the pre-image of an open set in $Y$ must be open in $X$.

As it turns out, if $X$ and $Y$ are metric spaces, then this definition is equivalent to the $\epsilon-\delta$ property, which means that on metric spaces you can work with either definition and your job will be done.

Hence, continuity is equivalent to the $\epsilon-\delta$ property wherever this property is applicable.