Question #230848
State and prove Baire' s category theorem.
1
Expert's answer
2021-08-30T17:34:37-0400

Definition 1. A topological space X is said to be a Baire space, if for any given countable collection AnA_{n} of closed sets with empty interior in X, their union n=1An\bigcup\limits_{n=1}^{\infty} A_n also has empty interior in X.

In terms of open sets this property can be reformulated as follows:

Definition 2. A topological space X is said to be a Baire space, if for any given countable collection GnG_{n} of open dense subsets of X, their intersection n=1Gn\bigcap\limits_{n=1}^{\infty} G_n is dense.

Baire' s category theorem. Every complete metric space is a Bair space.

Proof. The set n=1Gn\bigcap\limits_{n=1}^{\infty} G_n is dense iff for every open set U, n=1GnU\bigcap\limits_{n=1}^{\infty} G_n\cap U\ne \empty. We will construct a monotone sequence of closed balls Bˉ(xk+1,rk+1)Bˉ(xk,rk)\bar B(x_{k+1},r_{k+1})\subset \bar B(x_{k},r_{k}) such that Bˉ(xk,rk)n=1kGnU\bar B(x_{k},r_{k})\subset \bigcap\limits_{n=1}^{k} G_n\cap U and rk<2kr_k<2^{-k}. Then the sequence {xn}\{x_n\} will be a Cauchy sequence. Indeed, for all ε>0\varepsilon>0 let NNN\in\mathbb{N} be arbitrary such that 2N<ε/22^{-N}<\varepsilon/2 and n,m>Nn,m>N. Since xnBˉ(xn,rn)Bˉ(xN,rN)x_n\in \bar B(x_{n},r_{n})\in \bar B(x_{N},r_{N}) and xmBˉ(xm,rm)Bˉ(xN,rN)x_m\in \bar B(x_{m},r_{m})\in \bar B(x_{N},r_{N}) then dist(xn,xm)<2rN<22N<εdist(x_n,x_m)<2r_N<2\cdot 2^{-N}<\varepsilon. Therefore, the sequence {xn}\{x_n\} is a Cauchy sequence and has a limit xn=1Bˉ(xn,rn)n=1GnUx\in\bigcap\limits_{n=1}^{\infty}\bar B(x_{n},r_{n})\subset \bigcap\limits_{n=1}^{\infty} G_n\cap U. Therefore, n=1GnU\bigcap\limits_{n=1}^{\infty} G_n\cap U\ne \empty.

A monotone sequence of closed balls Bˉ(xk+1,rk+1)B(xk,rk)\bar B(x_{k+1},r_{k+1})\subset B(x_{k},r_{k}) can be constructed by induction, since for every kNk\in\mathbb{N} the set Gk+1G_{k+1} is dense and open, and hence, has a non-empty intersection with non-empty open set B(xk,rk)B(x_{k},r_{k}). Therefore, it contains some closed ball Bˉ(xk+1,rk+1)Bˉ(xk,rk)\bar B(x_{k+1},r_{k+1})\subset \bar B(x_{k},r_{k}).

The proof is completed.


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