The idea of continuity in (real) analysis—and indeed everywhere else it is used: topology, etc.—seems to me superfluous. For if, as I learnt from Stewart (Concepts of Modern Mathematics), it is based on the idea of excising jumps in so-called ‘functions’, then upon considering it carefully one sees that all functions that are called discontinuous at some point are not really functions at all; the problem, it seems, has only stuck because of the non-rigorous way analysis developed, in which Newton, Euler and the other early guys called everything that produced values a function, like the unrestricted square-root ‘function’, for example.

Now that the function concept has been placed on rigorous footing by set theory, one sees that the concept of continuous functions is just what we properly today call a mapping, and all the fuss is just linguistical. Thus, all definitions of continuity are just alternative definitions, in analytical guise, of the modern concept of mapping; the difference being just that in analysis, the sets are more specified than general—they’re usually continua, e.g.—but the idea is the same! Thus, continuity just requires that all points in the domain get mapped to exactly one point—i.e., not neither or more than one. Thus, for all x in the domain of some mapping f, there’s one and only one f(x). That, as clear as day, is the modern definition of ‘mapping’.

Or, if this is not so, please enlighten me: Does the idea of continuity capture more than what the modern set-theoretic definition of mapping does? If so, what exactly is that extra stuff it covers? Thanks.

PS. Of course I’m aware that there are theories of multi-valued functions, but that’s clearly not what I’m interested in; I’m talking about mappings as traditionally envisaged (i.e., single-valued).


closed as unclear what you're asking by Alex M., Tim Raczkowski, TravisJ, yoknapatawpha, Jyrki Lahtonen Sep 30 '15 at 18:39

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    $\begingroup$ If $f(x)=0$ for $x$ irrational, and $f(x)=1$ for $x$ rational, well, that's a perfectly good mapping, but it's not continuous. $\endgroup$ – Gerry Myerson Sep 30 '15 at 13:04
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    $\begingroup$ You are wrong. There are plenty of functions that are not continuous. $\endgroup$ – AD. Sep 30 '15 at 13:05
  • $\begingroup$ I think his concern has something to do with for example the functions defined on the natural numbers $\mathbb N$. All of them are continuous, or are they ? $\endgroup$ – Svetoslav Sep 30 '15 at 13:11
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    $\begingroup$ Where did you get the idea that continuity is equivalent to there being one and only one value of $f(x)$? For instance, consider $f(x)=0$ for $x<0$, $f(x)=1/2$ for $x=0$, $f(x)=1$ for $x>0$. This has one value at every real number but it is not continuous. Why does it "seem" continuous to you? $\endgroup$ – Ian Sep 30 '15 at 13:13
  • $\begingroup$ @Ian Yes, I know, but this was referring more to him. $\endgroup$ – Svetoslav Sep 30 '15 at 13:13

Your argument seem to be around the terminology that has been chosen. At the end of the day there exists constructs (we call now continuous functions) that have different characteristics than other (we now call non-continuous functions). These characteristics of the former have been noted as being useful in some areas and therefore deserve to be distinguished from the latter.

So whether or not we call them continuous function or something else the concept as such is not redundant.

(note that your point about there's one and only one f(x) doesn't imply continuity, even non-continuous function show this characteristic)


"Thus, for all x in the domain of some mapping f, there’s one and only one f(x). That, as clear as day, is the modern definition of ‘mapping’."

Exactly. And this holds for any mapping from any set $X$ to any set $Y$.

But continuity is about the quality of such mappings, when $X$ and $Y$ are not just sets, but spaces. In (metric or topological) spaces some notion of nearness or neighborhood is defined. A map $f:\>X\to Y$, well defined in the set theoretical sense, is continuous at $x_0\in X$ if, whenever $x$ is sufficiently near $x_0$ then $f(x)$ is near $f(x_0)$, whereby the tolerance may be fixed in advance. (I'm sure you have been told the exact definition.)


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