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I understand that there are functions that, by definition of continuity, can be continuous at only one point, such as

$$f(x)=\begin{cases} x,&\text{if }x\in\Bbb Q\\ 0,&\text{if }x\in\Bbb R\setminus\Bbb Q\;. \end{cases}$$

which is continuous only at $x=0$. But it is continuous because it satisfies the formal definition of continuity. Still, continuity at only-one-point sounds like an oxymoron to my mind. I understand that mathematical concepts are different than the standard meanings of the words in natural languages, so my question is this:

Does the classical definition of continuity fail to capture the intended concept of continuity for this pathological case? Has anybody attempted to modify the definition of continuity to make this pathological cases fail? I call it pathological because I imagine that, historically, the original concept of continuity attempted to capture the idea of "connected" line. But I might be wrong.

Michael Hardy
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Wolphram jonny
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  • I suppose you could require the graph of the function to be connected (in the precise topological sense). Or even arcwise connected. – Potato Oct 31 '14 at 02:51
  • what do you think the intended concept is? – azarel Oct 31 '14 at 02:58
  • a "connected" line? (even if it were of infinitesimal length) – Wolphram jonny Oct 31 '14 at 02:59
  • Here is a natural way of thinking about this. We first have the definition for a function to be "continuouse at a single point". We say a function is "continuous" if it is continuous at each single point in the domain. – Xiao Oct 31 '14 at 03:05
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    @julianfernandez Continuity is more about errors if you make a small error in the input you want a small error in the output. The function $\sin{1/x}\cup(0,0)$ is connected and not continuous. – azarel Oct 31 '14 at 03:13
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    In calculus, most professors go with "you can draw the graph of the function without picking up your pencil", but I like the "If you start close, you stay close" idea. – dannum Oct 31 '14 at 03:31
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    I don't consider this case pathological. If we removed the value $f(0)=0$ from the graph of your $f$, I think you'd still say that, based on how the rest of the graph looks, $f(0)=0$ is the most natural value (which is an informal interpretation of $f(x)=\lim_{a\to x} f(a)$). –  Oct 31 '14 at 04:47
  • Once a mathamatical thing is defined, then logic demands that the thing is the definition. I have spent way too many years watching students try to wrap their brains around $x^0 = 1$ to think otherwise. You need to figure out a way too look at it that makes sense, or else become powerful enough to create your own definition and rebuild mathamatics from that point on. – Steven Alexis Gregory Jun 16 '15 at 01:06

4 Answers4

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Sometimes, intuition is wrong, and we should learn from the math, rather than try to make the math fit the intuition.

That said, I bet what you are probably intuiting is not something that should be called "continuous at the point $P$" at all -- it should be called "continuous in a neighborhood of the point $P$". e.g. "there exists $a < P < b$ such that $f$ is continuous on $(a,b)$". So the math already fits the intuition, once you correctly translate between them.

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Via nonstandard analysis, there is a "infinitesimal segment" approach to continuity-at-a-point: a function $f$ is continuous at a point $a$ if, for any infinitesimal $h$, the difference $f(a+h)-f(a)$ (actually, ${}^*f(a+h)-{}^*f(a)$, to be precise; see below) is infinitesimal. That is, $f$ is continuous at $a$ if the graph of $f$ "infinitesmially near" $a$ varies only infinitesimally from a straight (horizontal, in fact) line. This definition is equivalent to the usual $\epsilon-\delta$ definition.

OK, so what is this "nonstandard analysis" I mention? Well, this is certainly too complicated for a short paragraph, but basically the idea is this: we start with the "real" universe of the real numbers $\mathbb{R}$ and all functions (continuous or not) on $\mathbb{R}$, and we consider a "blown up" version of the reals, called $^*\mathbb{R}$. $^*\mathbb{R}$ is an ordered field which contains the real numbers, and a lot more junk besides, including field elements which are less than every (real) positive real number; we call these infinitesimals. Moreover, each function $f$ on $\mathbb{R}$ has a version $^*f$ defined on $^*\mathbb{R}$, which agrees with $f$ on the actual reals. Statements in the $\epsilon-\delta$ language can be translated to (arguably) more intuitive and snappy definitions in terms of infinitesimals.

If this all seems suspicious to you, that's very reasonable; it takes a lot of work to set this up so it doesn't break. There are many good sources with actual details; I like http://homepages.math.uic.edu/~isaac/NSA%20notes.pdf, but I'm biased (I learned from it). See especially section 2, which explains why we can get away with this sort of silliness :D.

Noah Schweber
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Definition: For topological space $(X,\mathcal{T})$ and $(Y, \mathcal{S})$, a mapping $f: X\rightarrow Y$ is said to be continuous at the point $x_0$ in $X$ provided for any open neighborhood $O$ of $f(x_0)$, there is an open neighborhood $U$ for which $f(U) \subset O$.

Then we say $f$ is continuous provided it is continuous at each point in $X$.

This is the topological definition of continuity, even if you do not understand anything about topology, that is okay. The important thing to take away from this definition is that the idea of continuity at a single point comes first. Then if the function is continuous at each point in the domain, we abbreviate this by simply saying the function is continuous.

Xiao
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  • It's interesting you characterize the topological definition this way. At least when I was growing up, the topological definition was about continuity on subspaces or the entire topological space; later, we restricted this to a point. It was in the spirit of this approach that I wrote my comment above. – Simon S Oct 31 '14 at 04:10
  • I didn't characterize this, I am not a mathematician... : ) And normally after this definition, you would find an exercise asking to show $f$ is continuous if and only if the preimage of open sets is open. – Xiao Oct 31 '14 at 04:20
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One very intuitive idea about continuity is that $f $ should respect limits. With your function, if $x_n\to0$, then $f (x_n)\to0$; so we say that $f $ is continuous at $0$.

Martin Argerami
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