Let f:(0, 1) --> R be continuous in every irrational x. Then

[nikolany]

Let f0,1)-->R continuous in every irrational number x. Then f is continuous at every x.

Is that true? Do I need to find an example?
Thx

 

[Deleted member 4993]

Let f0,1)-->R continuous in every irrational number x. Then f is continuous at every x.

Is that true? Do I need to find an example?
Thx

One - even thousand examples - does not prove "truth" of a statement.

However. one counter-example is enough to prove a statement to be false.

 

[pka]

Let f0,1)-->R continuous in every irrational number x. Then f is continuous at every x.

I think that you mean \(\displaystyle f\) is continuous at each irrational.
Consider \(\displaystyle \left\{ {\lambda _1 ,\lambda _2 ,\lambda _3 , \cdots } \right\}\) to a listing of all the rational numbers in \(\displaystyle (0,1)\).
Define a function \(\displaystyle f(x) = \sum\limits_{\lambda _n < x} {\frac{1}{{2^n }}}\).
That is if a rational is less that x we add one half to its subscripted value.
See if that is a counterexample.

 

[PAULK]

Let f0,1)-->R continuous in every irrational number x. Then f is continuous at every x.

Is that true? Do I need to find an example?
Thx

It is not so. It is possible to construct f(x) continuous only at the irrationals. (You will find, in your real analysis course, that the 'reverse' is not so -- you cannot construct a function continuous at only the rationals.)

SO: Consider this (admittedly disgusting) function:

f(x) = 0, if x is irrational (easy part)

f(x) = 1/q, if x is rational and x = p/q in lowest terms.

Chew on that a bit and I think you will find (i.e. prove) it is continuous at the irrationals only.