Characteristic function on I is uncountable

Question

I studied characteristic function in real analysis in Richard R goldberg book for methods of real analysis and there is an exercise problem asking to prive that characteristic function on I is uncountable.

My attempt:if we composite every function whose range is I with the characteristic function on I we would still get a characteristic function on I. ie., if f is a function whose range is I and χ is the characteristic function on I then

χ○f is a characteristic function on I

But my doubt is, Is the set of all functions whose range is I is uncountable?

I saw some proof using the help of power set of I. I didn't understand it. Please help me with the proof.

Answer 1

You have to consider two cases:

1. $I$ is finite.
2. $I$ is not finite.

Since the set of characteristic functions on $I$ is in bijection with its power set $\mathcal{P}(I)$ we can answer this very quickly since $\mathcal{P}(I)$ is countable off $I$ is finite… thus either it is finite or uncountable!

Beware: by countable I mean $\sharp I \leq \aleph_0$ si finite sets are also countable with my definition!