Can a Real Function Be Continuous Only at a Single Point?

[glance]

Hi
The question is the following: is it possible for a (say) real function to be continuous at a certain point internal to its domain, and be discontinuous in some neighborhood of that point?
I am not talking about a function defined at a single point or things like that, but of a function defined on the entire \mathbb{R} (or some interval in it, whatever).

Now, i have also came up with an answer: a function f defined as f(0)=0, f(x)=x for every rational x, and f(x) = 2x for every irrational x. Such a function would be (seems to me) continuous at x=0 and discontinuous for any other x. I am not completely certain of this, though, and for that reason i would like some feedback on this.

I am also asking this question because strangely enough I have never heard of the concept of an isolated continuity point, while for example the "opposite" (that of an isolated singularity) is quite common, and I would like to know if it's just me or if it is just a "useless" pathological concept.

Thank you in advance.
Bye

 

[gopher_p]

Is it possible for a (say) real function to be continuous at a certain point internal to its domain, and be discontinuous in some neighborhood of that point?

Yes.

In fact, there is a function defined on all of $\mathbb{R}$ which is continuous at a single point.

 

[micromass]

Now, i have also came up with an answer: a function f defined as f(0)=0, f(x)=x for every rational x, and f(x) = 2x for every irrational x. Such a function would be (seems to me) continuous at x=0 and discontinuous for any other x.

Yes, that is a correct example.

Now, can you come up with an example of a function defined on entire $\mathbb{R}$ that is differentiable only in one point?

 

[pasmith]

Hi
The question is the following: is it possible for a (say) real function to be continuous at a certain point internal to its domain, and be discontinuous in some neighborhood of that point?
I am not talking about a function defined at a single point or things like that, but of a function defined on the entire \mathbb{R} (or some interval in it, whatever).

If f : \mathbb{R} \to \mathbb{R} and g : \mathbb{R} \to \mathbb{R} are continuous such that there exists a unique a \in \mathbb{R} such that f(a) = g(a), then the function <br /> h : \mathbb{R} \to \mathbb{R} : x \mapsto \begin{cases} f(x) &amp; x \in \mathbb{Q} \\<br /> g(x) &amp; x \in \mathbb{R} \setminus \mathbb{Q} \end{cases} is discontinuous on \mathbb{R} \setminus \{a\} and continuous at a.

 

[glance]

If f : \mathbb{R} \to \mathbb{R} and g : \mathbb{R} \to \mathbb{R} are continuous such that there exists a unique a \in \mathbb{R} such that f(a) = g(a), then the function <br /> h : \mathbb{R} \to \mathbb{R} : x \mapsto \begin{cases} f(x) &amp; x \in \mathbb{Q} \\<br /> g(x) &amp; x \in \mathbb{R} \setminus \mathbb{Q} \end{cases} is discontinuous on \mathbb{R} \setminus \{a\} and continuous at a.

That is a very interesting example, thank you.

Now, can you come up with an example of a function defined on entire $\mathbb{R}$ that is differentiable only in one point?

That seems to be tricky! I did some research and stumbled upon this discussion of that matter, in which that question is very well explained.

Now however I wonder if it is accidental that in all of these examples the functions are constructed using rational and irrational numbers. I think that the important point is to have one subset which is dense in the other. Is there some example of functions having this kind of "pathologies" NOT using rational/irrational numbers in the definition?
Even better, is it possible to find a function of this kind NOT using at all dense subsets of the real numbers in the definition?

Thanks