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A scalar field $f : \mathbb{R}^n \to \mathbb{R}$ is said to be continuous at a point $\boldsymbol{a}$ if

$$ \lim_{\boldsymbol{x} \to \boldsymbol{a}} f(\boldsymbol{x}) = f(\boldsymbol{a}) $$

So in other words, $f$ has to be defined at $\boldsymbol{a}$ and also has to have a limit at $\boldsymbol{a}$. But isolated points are also defined to be continuous.

It seems to me like, given this definition, there isn't really an advantage to having a function continuous at a point, and continuity is only useful on an set or interval, because of the following:

  • $f$ can be continuous at a point but not differentiable (e.g. $f(\boldsymbol{x}) = \|\boldsymbol{x}\|$ at $\boldsymbol{0}$)
  • $f$ can be continuous at a point but the limit doesn't exist (e.g. $f(\boldsymbol{x}) = \sqrt{-\|\boldsymbol{x}\|}$ at $\boldsymbol{0}$ if $f(\boldsymbol{x}) \subset \mathbb{R}$)
  • $f$ can be continuous at a point but the first-order partials don't exist

So if $f$ is continuous at a point $\boldsymbol{a}$, is there anything we can say about $f$ at $\boldsymbol{a}$? Or is it just a nice-to-have?

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    $\begingroup$ I wouldn't call your second example continuous at $0$: if the limit doesn't exist, it can't be equal to $f(a)$. $\endgroup$
    – user7530
    Nov 2, 2013 at 20:27
  • $\begingroup$ @user7530 But it's an isolated point, no? $\endgroup$ Nov 2, 2013 at 20:30
  • $\begingroup$ Hmm, I suppose if you say the domain of $f$ is $\{0\}$, it is continuous at $0$... $\endgroup$
    – user7530
    Nov 2, 2013 at 20:30
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    $\begingroup$ @MasterOfBinary You seem to feel let down by that fact that continuous functions fail to be differentiable. All differentiable functions are continuous, but the converse is false. Consider the Weierstrass function. It is continuous everywhere and differentiable nowhere. en.wikipedia.org/wiki/Weierstrass_function $\endgroup$ Nov 2, 2013 at 20:36
  • $\begingroup$ @user7530 Maybe I should be more specific. $\boldsymbol{0}$ is the only place where that function is defined for real values. $\endgroup$ Nov 2, 2013 at 20:42

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Continuity is a very important property to have, in particular when studying topology. For analysis you usually want something more, for example you'd like the function to be $C^1$ or $C^\infty$ in order to differentiate.

Also, it is nice to have a definition of continuity at a point because it shows that continuity is a local property.

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  • $\begingroup$ +1 for the second paragraph. Continuity at a point is pretty useless as a property of functions, which is why the topological definition of continuity (inverse image of an open set is open) sees much more use. But it's one nice way of defining, without the topology, the property you are really after -- continuity on an interval. $\endgroup$
    – user7530
    Nov 2, 2013 at 20:33
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Let's start in $R^n$ which is where you are asking the question.

I think continuity at a point is the beginning of a process. Leaving out your suggestion of isolated points, once you have defined what you mean by continuous at a point (in any number of dimensions) you can now define what you mean by continuous on an interval (or ball in $R^n$, but I'm just going to use the word 'interval'),namely that the function is continuous at every point in the interval.

Now we have something to work with. For example, a function which is continuous on a closed, bounded interval (compact set) is bounded there and is guaranteed to have a maximum and minimum there. It is also guaranteed to be uniformly continuous, which is a stronger condition. It is guaranteed that there is a polynomial on the interval which approximates the function to any desired degree of accuracy. It is (Riemann and Lebesgue) integrable on that interval and has an anti-derivative in each variable, so you have some help with integration, double integration, etc. Even if you can't figure out what this antiderivative is, the fact that it exists is quite helpful for certain proofs.

Then the entire theory of derivatives is built on top of continuity. If the function is not continuous it cannot be differentiable. In $R^n$ differentiability (as opposed to having partial derivatives, which do not even imply continuity) tell us that the curve is smoother than a merely continuous one; and that in a small enough neighborhood of any point it is nearly linear (nearly can be defined exactly). This means that if you are asked to prove something and you can prove it for linear functions -- often quite easy -- you might be able to extend your proof to differentiable functions, the argument being that nearly linear is good enough.

The existence of each additional derivative (2nd, 3rd, etc) implies that the function is smoother and smoother. This is very helpful if you are trying to approximate a function with something easier to work with or show the existence of certain things, like the solution to a differential equation.

In the complex plane the existence of a derivative even at one point is an extremely strong statement, telling you the function is analytic in some vicinity of that point. Analytic functions have many, many properties that merely differentiable ones do not, and they are the cornerstone of many solutions to problems in physics.

Getting away from $R^n$ and into other spaces, continuity can still be defined, and again continuity on a compact set has many implications -- how many and what they are depends on what space you are talking about.

So you start with continuity at a point and build a huge structure on top of that concept. As they say, the longest journey starts with but a single step.

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  • $\begingroup$ It definitely seems nicer without including isolated points. $\endgroup$ Nov 2, 2013 at 21:13
  • $\begingroup$ Yes it is. The thing is mathematicians want to understand what the scope is of their definitions. Often a hypothesis seems to imply that certain things are true, and they may not be. To show this they construct something which meets the hyptheses and isn't true. Usually these things are horrible objects that are of no practical use, besides to teach you to be careful in your proofs. $\endgroup$
    – Betty Mock
    Nov 4, 2013 at 4:11
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Even though continuity doesn't imply differentiability, the fact that $f$ is not continuous at $\boldsymbol{a}$ means $f$ is not differentiable, and hence not continuously differentiable, at $\boldsymbol{a}$.

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  • $\begingroup$ Exactly the point. Without continuity, we aren't going to have differentiability or much of anything else. Practically all functions of any interest do have a continuous derivative at a. Those which do not fall into the category of my comment above: horrible objects which serve only to act as warnings not to be too glib. $\endgroup$
    – Betty Mock
    Nov 4, 2013 at 4:15

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