I think I remember reading somewhere that the Baire Category Theorem is supposedly quite powerful. Whether that is true or not, it’s my favourite theorem (so far) and I’d love to see some applications that confirm its neatness and/or power.

Here’s the theorem (with proof) and two applications:

(Baire) A non-empty complete metric space X is not a countable union of nowhere dense sets.

Proof: Let X=⋃Ui where ¯Ui˚. We construct a Cauchy sequence as follows: Let x_1 be any point in (\overline{U_1})^c. We can find such a point because (\overline{U_1})^c \subset X and X contains at least one non-empty open set (if nothing else, itself) but \mathring{\overline{U_1}} = \varnothing which is the same as saying that \overline{U_1} does not contain any open sets hence the open set contained in X is contained in \overline{U_1}^c. Hence we can pick x_1 and \varepsilon_1 > 0 such that B(x_1, \varepsilon_1) \subset (\overline{U_1})^c \subset U_1^c.

Next we make a similar observation about U_2 so that we can find x_2 and \varepsilon_2 > 0 such that B(x_2, \varepsilon_2) \subset \overline{U_2}^c \cap B(x_1, \frac{\varepsilon_1}{2}). We repeat this process to get a sequence of balls such that B_{k+1} \subset B_k and a sequence (x_k) that is Cauchy. By completeness of X, \lim x_k =: x is in X. But x is in B_k for every k hence not in any of the U_i and hence not in \bigcup U_i = X. Contradiction. \Box

Here is one application (taken from here):

Claim: [0,1] contains uncountably many elements.

Proof: Assume that it contains countably many. Then [0,1] = \bigcup_{x \in (0,1)} \{x\} and since \{x\} are nowhere dense sets, X is a countable union of nowhere dense sets. But [0,1] is complete, so we have a contradiction. Hence X has to be uncountable.

And here is another one (taken from here):

Claim: The linear space of all polynomials in one variable is not a Banach space in any norm.

Proof: “The subspace of polynomials of degree \leq n is closed in any norm because it is finite-dimensional. Hence the space of all polynomials can be written as countable union of closed nowhere dense sets. If there were a complete norm this would contradict the Baire Category Theorem.”

**Answer**

If P is an infinitely differentiable function such that for each x, there is an n with P^{(n)}(x)=0, then P is a polynomial. (Note n depends on x.) See the discussion in Math Overflow.

**Attribution***Source : Link , Question Author : Community , Answer Author :
4 revs, 2 users 50%*