Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

enter image description here

enter image description here

share|improve this question
1  
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
1  
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
1  
That is not the graph of the function you wrote down... –  Thomas Rot Jan 17 '12 at 15:20
1  
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

2 Answers 2

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.

share|improve this answer
    
@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 $x$.

share|improve this answer

Your Answer

 
discard

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.