The Baire category theorem says that any compact Hausdorff space or a complete metric space is a Baire space.

A Baire space is when you take the union of a countable collection of close sets in a space X, each of whose interior is non empty in X, also has an empty interior is X.

I am trying to use this theorem to prove the continuous function $f:[0,1] \to \mathbb{R}$ is nowhere differentiable.

I understand how to do the first two steps which involve defining your sets and showing they are closed.

But I do not understand the general argument when trying to show your set is nowhere dense. How do we do this?

  • $\begingroup$ "the continuous function $f$"? Do you mean you're trying to use Baire Category to prove that there exists a nowhere differentiable continuous function? That "most" continuous functions are nowhere differentiable? $\endgroup$ Aug 5, 2016 at 20:05
  • $\begingroup$ @tedshifrin I am trying to prove there exists a continuous function f which is nowhere differentiable $\endgroup$
    – Al jabra
    Aug 5, 2016 at 20:10
  • $\begingroup$ Have you tried googling it? Look here: homepages.math.uic.edu/~marker/math414/fs.pdf $\endgroup$ Aug 5, 2016 at 20:27

1 Answer 1


The Baire Category Theorem can be used to proof the following theorem by Banach:

Let $D_+$ be the collection of all $f \in C[0,1]$ for which there is a point $x_f \in [0,1)$ at which $f$ has a finite right-hand derivative. Then $D_+$ is of the first category in $C[0,1]$.

It is from this statement that one can deduce that "a "typical" $f \in C[0,1]$ is not differentiable anywhere". The above statement is Theorem 1.5.5 in Megginson's An Introduction to Banach Space Theory, and I refer to this book for the proof as it is quite lengthy.

  • $\begingroup$ For those interested, Banach's actual original proof in German (1931) is very closely followed (almost an English translation) somewhere in Ralph Henstock's 1967 book Linear Analysis. And for more than you'd probably ever want to know about this topic, see my answer to Generic Elements of a Set.. $\endgroup$ Aug 5, 2016 at 21:17

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