Continuity of a Function Around a Point

[quasar987]

Is it true that is f is continuous at t, then there exists an interval around t for each point of which f is continuous also?

Edit: In case it is false, is it true however on a set of measure zero?

P.S. Please just feed me the answer; I know nothing about measure except that a function is integrable on E iff E is of measure 0.

 

[dextercioby]

What's the definition of pointlike convergence (in a Banach space, for example) ? What is the definition of convergence on a domain (or a Banach space)...? Think straight and always start with the definitions.

Daniel.

 

[quasar987]

In a banach space, a function $\rho$ is pointlike convergant if $\forall \epsilon \leq \zeta(t_0)$ where t_0 of course are the points of uniform convergence of the generalised Laurent serie of the critical quaternion... every undergrad knows that Daniel.

 

[HallsofIvy]

Consider the function:

f(x)= 0 if x is irrational. If x is rational, x= m/n reduced to lowest terms, then
f(x)= 1/n. Where is that function continuous or discontinuous? (Oddly enough, the limit exists and is the same at every x.)

 

[quasar987]

(Oddly enough, the limit exists and is the same at every x.)

Soooo, this function is not a counter-exemple. So we're not more advanced.

 

[HallsofIvy]

Soooo, this function is not a counter-exemple. So we're not more advanced.

The question was not about limits, the question was about continuity. The example I gave has limit 0 at every point and so is continuous at each irrational number, discontinuous at each rational and thus not continuous on an interval about a point where it is continuous. Since the set of rational numbers has measure 0, the second question is still unanswered.

 

[Galileo]

Or, inspired by Halls, consider:

$$f(x)=\left\{ \begin{array}{cc}x^2 & \mbox{if x is rational}\\<br /> 0 & \mbox{if x is irrational}\end{array} \right.$$

This function is continuous only at x=0.

 

[Edwin]

You might be referring to continuity at a higher level of math than calculus, but continuity based on what was covered in our calculus class had something like the following definition:


A function f(x) is continuous at a point x = c if, and only if, the limit of f as x approaches c is equal to f(c). In order for a limit to exist, excluding limits that approach infinity, the limit of a function as x aproaches a number c from the left and from the right must exist and be equal. If you are dealing with a function like the piecewise defined function

f(x) = 1 if x is an integer, and f(x) = 0 if x is not an integer would be discontinuous at values of x that are integers because the limit the function as you approach any given integer from the left and right, does not equal 1.

Now I'm not sure about the continuity of the function you mentioned, f(x) = x^2 if x is rational, and 0 if x is irrational. Is there a continuous interval I of points in the domain of x that all consist of irrational numbers? If so, then you could argue that the function is continuous at least over the open interval of such a domain of the function.

I presume that there is another definition of continuity that you are going by, that is based in some higher form of math, like numerical analysis, or something of the like, that deals with discrete functions and continuity at a point. Is this so?

Inquisitively,

Edwin

 

[Galileo]

No. It is exactly the definition you stated. The function is continuous at x=0, because the limit of f(x) as x goes to 0 is 0. (There are otherways to state continuity, but these are generalizations with the equivalent meanings for this case).

You could use the definition of a limit to show it, or use the squeeze theorem, since $0\leq f(x) \leq x^2$.

 

[HallsofIvy]

Is there a continuous interval I of points in the domain of x that all consist of irrational numbers?

No, every interval of real numbers contains both rational and irrational numbers- that's the point. If a is irrational, then f(a)= 0 but within any interval, there exist rational numbers, x, close to a such that f(x)= x², not close to 0. If a is rational, but not 0, then f(a)= a² but within any interval, there exist irrational numbers, x, close to a such that f(x)= 0, not close to a². In neither of those cases does the limit exist- so the the function is not continuous at any non-zero number.

If a= 0, however, then f(a)= 0²= 0. If x is a rational number close to 0, then f(x)= x² is close to 0 and if x is an irrational number, f(x)= 0 so the limit, as x-> 0, exists and is equal to 0. The function is continuous only at x= 0.

The example I gave (I think it's "Dirichlet's function") is harder to show but, I think, "cuter" since the limit always exist but is equal to the value of the function only for a irrational.