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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!

zyx
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Bobby Mack
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    Definitions work in both directions. The exponential function definitely satisfies the definition of continuity. – Justin Young Nov 07 '16 at 23:52
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    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 07 '16 at 23:58
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    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. – BigbearZzz Nov 08 '16 at 00:02
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    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. – alephzero Nov 08 '16 at 01:47
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    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. – joeytwiddle Nov 08 '16 at 09:37
  • This is like asking if there are three-sided polygons that are not triangles. – Matthew Leingang Nov 08 '16 at 11:28
  • Your expression does not include the quantification over $x_0$. Depending on whether $\delta$ is allowed to depend on $x_0$, what you wrote can be the definition of a uniformly continuous function, which the exponential function is not, or the definition of a continuous function, which the exponential function is. – Sasho Nikolov Nov 08 '16 at 14:22
  • What would a continuous function be, if not one satisfiying the condition you wrote out? – anomaly Nov 09 '16 at 00:25
  • I voted to close, even though the questions are slightly different, they are essentially on the same thing and the accepted answer at the target is exactly the correct answer in this case: when reading a definition, an "if" should be read as "if and only if". (There might be better duplicate targets of this nature, this is the first one I found.) – Asaf Karagila Nov 09 '16 at 06:06

4 Answers4

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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 $x_0$.

Mark Fischler
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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.

aduh
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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.

Martin Sleziak
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Sarvesh Ravichandran Iyer
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This article from wikipedia may also look interesting. Especially this part:

In mathematics, a definition is used to give a precise meaning to a new term, instead of describing a pre-existing term. Definitions and axioms are the basis on which all of mathematics is constructed.

And:

An Intensional definition, also called a connotative definition, specifies the necessary and sufficient conditions for a thing being a member of a specific set. Any definition that attempts to set out the essence of something, such as that by genus and differentia, is an intensional definition.

polfosol
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