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Continuity is defined in my book Basic Analysis (by Lebl pg 86) like this:

Let $S \subset \mathbb R$, $f : S\rightarrow \mathbb R$ be a function, and let $c \in S$ be a number. We say that $f$ is continuous at c if for every $\epsilon > 0$ there is a $\delta > 0$ such that whenever $x \in S$ and $|x − c| < \delta$ , then $| f (x) − f (c)| < \epsilon$.

When $f : S \rightarrow \mathbb R$ is continuous at all $c \in S$, then we simply say that f is a continuous function.

But how does this definition ensure continuity?

If, for $c=0,\epsilon = 10$ there exists $\delta = 5$ for which $|x-0| = |x| < 5, |f(x)-f(c)| < 10$ , and for another $\epsilon=30$, there exists $\delta = 1$ for which $|x| < 1, |f(x)-f(c)| < 30$, wouldnt this increase chances of a function that wildly jumping between values, and perhaps indicating a noncontinous function? I just picked these values out of a hat of course, but the definition does not say anything about how values are structured, if they become smaller while approaching $c$ for example.

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This should hold for every (positive) epsilon, in particular for positive epsilons close to zero (note that if this holds for a given epsilon, it also holds ipso facto for every larger epsilon). – Did Jan 17 '12 at 12:22
If I had a discontinuous function that for $x<5$ was $f(x) = 2x$, but after $x \ge 5$ it was $f(x) = 3x$, wouldn't this function pass the test for the definition above? It simply says "there exists" placing certain bounds, but I have a feeling it would miss the jump points. – BB_ML Jan 17 '12 at 12:29
This function would be continuous at c=0 hence it should definitely pass the test of continuity at c=0 (and it does, take delta=min(5,epsilon/2)). And it should fail the test at c=5 (and it does, take epsilon=4). – Did Jan 17 '12 at 12:34
That is not the graph of the function you wrote down... – Thomas Rot Jan 17 '12 at 15:20
Indeed, more like the graph of the function $g$ defined by $g(x)=2x$ if $x\lt5$ and $g(x)=3x-5$ if $x\geqslant5$. And $g$ is continuous at $c=5$ (choose $\delta=\frac13\varepsilon$). – Did Jan 17 '12 at 16:05
up vote 1 down vote accepted

Think of $\epsilon$ as a margin of error. If we are close enough to $c$, within $\delta$, the function value $f(x)$ is within the margin of error $\epsilon$. As Didier says it is important to let $\epsilon$ be arbitrarily small. For each margin of error $\epsilon$ there is a closeness $\delta$, such that all points within a distance $\delta$ of $c$, all the values of these points are within the margin of error $\epsilon$. Thus $f(x)$ is close to $f(c)$ if $x$ is close to $c$. This is the idea of continuity. A nice picture is at the wikipedia page.

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@Didier: Thanks – Thomas Rot Jan 17 '12 at 12:48

The definition says that no matter how small a ‘target’ you set around $f(c)$, you can guarantee that $f(x)$ is inside that target whenever $x$ is sufficiently close to $c$. Do you want to guarantee that $f(x)$ is closer than $0.001$ to $f(c)$? Take $\epsilon=0.001$, and the definition of continuity tells you that there’s some positive $\delta$, possibly very small, such that $f(x)$ is less than $0.001$ away from $f(c)$ provided that $x$ is less than $\delta$ away from $c$. Do you want to be sure that $f(x)$ is less than $10^{-10}$ away from $f(c)$? Apply the definition with $\epsilon=10^{-10}$: there is some $\delta$ such that as long as you choose $x$ between $c-\delta$ and $c+\delta$, $f(x)$ will be less than $10^{-10}$ away from $f(c)$. In short, it’s really the small values of $\epsilon$ that you should be interested in.

This doesn’t prevent $f$ from oscillating more and more frequently as $x$ approaches $c$, but it does force the magnitude of the oscillations to approach $0$. Take a look, for instance, at the graph of the function

$$f(x)=\begin{cases} x\sin\frac1x,&\text{if }x\ne 0\\ 0,&\text{if }x=0\;, \end{cases}$$

which is shown here. As you can see, they oscillate infinitely often as $x$ approaches $0$, but because the amplitude of the oscillations approaches $0$, both functions are continuous at $0$. In fact, no matter what $\epsilon>0$ you pick as a ‘target’, the same number will work for $\delta$: it’s pretty straightforward to check that if $|x|<\epsilon$, then $|f(x)<\epsilon$ as well, since $|\sin(1/x)|\le 1$ for all $x$.

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