Continuity of function

[Leviathan]

This is one problem from the ETS Practice Book:



Let g be the function defined on the set of all real number by

g(x) = 1 if x is rational.

g(x) = e^x if x is irrational.

Then the set of numbers at which g is continuous is


(A) the empty set
(B) {0}
(C) {1}
(D) the set of rational numbers
(E) the set of irrational numbers

Answer:

SPOILER: B

Can anyone prove why the answer is so? I have no idea how to tackle such problem. Thank you.

[WRusin]

Any real number can be approximated by a series of elements from Q and R\Q. If we take a real number different from 0 and want to approximate it with such series we realize that the limits of the function at those series are different. For example if x = 2 than \limit_{x \rightarrow 2, x \in Q} = 2 and \limit_{x \rightarrow 2, x \in R\Q} = e^2. x=0 is the only argument for which those limits coincide. That is the reason for which the right answer is (B).

[Leviathan]

I get it now. Thanks WRusin!!

[lmtuan]

Answer: B

[lmtuan]

It is easy to so that:
* g is continuous at x=0
* at x_0 not equal to 0, we have:
Limit of g(x) as x-->x_0 does not exist. Thus, g is not continuous at all points, but 0.
Answer: B