Continuous Nowhere, Continuous Everywhere: Tricky Function Example [0,1] Domain

[Nebula]

I was asked to find a function that is continuous nowhere but its square is continuous everywhere but 0. The domain of the function is [0,1].

I can't come up with an example...
Any help is appreaciated. Thanks

 

[StatusX]

if your allowed to define it piecewise:

f(x)=
1 for x rational
-1 for x irrational

and if it absolutely needs to be discontinuous at 0

0 for x=0

 

[Tide]

if your allowed to define it piecewise:

f(x)=
1 for x rational
-1 for x irrational

and if it absolutely needs to be discontinuous at 0

0 for x=0

Hmm. That raises the question: If p is a number in [0, 1], what is the "next number" after p? Is it rational or irrational? It seems rather difficult to answer for a continuum! :-)

 

[cogito²]

One of the properties of real numbers is that there is no next number.

 

[matt grime]

Hmm. That raises the question: If p is a number in [0, 1], what is the "next number" after p? Is it rational or irrational? It seems rather difficult to answer for a continuum! :-)

How on Earth did you reach that conclusion?

 

[HallsofIvy]

Hmm. That raises the question: If p is a number in [0, 1], what is the "next number" after p? Is it rational or irrational? It seems rather difficult to answer for a continuum! :-)

On the contrary, it's very easy to answer: there is NO "next number" in the set of real numbers or the set of rational numbers. If there were, every subset would be "well ordered" and that's a property of integers, not rational or real numbers.

 

[Tide]

How on Earth did you reach that conclusion?

Actually, it wasn't a conclusion. It was a question for StatusX to ponder in which he proposes a "solution" to Nebula's query by, in effect, making "every other number" change the sign of a function. The answer, of course, is that there is no "next number" in the continuum of reals. No matter how close two numbers are you can always find other numbers in between.

 

[StatusX]

he wanted a function that was nowhere continous, so that's what i gave. i never said that rationals and irrationals alternate, if that's what you mean, which doesn't even make sense. all i mean is that the limit of f(x) as x goes to c does not exist for any c, and so does not equal f(c), so the function isn't continuous anywhere. maybe that's not the right way to define continuity, but i still think its obviously discontinous.

 

[matt grime]

You did have a conclusion, tide: that the example 'raised the question'... how or why does it make you think of this question? The example does not make any such 'alternating sign every other number' argument. It is just +1 and -1 for two different dense subsets. Any such function is a solution, and there's no need to put quotation marks around the word as if it weren't.

There's no next number in the rationals either, and that isn't a continuum, but the question in my mind is still, 'what has this got to do with anything?'

 

[jcsd]

By the defintion of continuity and the fact that the rationals and irrationals are both dense in R the function is obviously discontinious as at every point.