proof of Baire category theorem

Let $(X,d)$ be a complete metric space, and $U_{k}$ a countable collection of dense, open subsets. Let $x_{0}\in X$ and $\epsilon_{0}>0$ be given. We must show that there exists a $x\in\bigcap_{k}U_{k}$ such that

d(x_{0},x)<\epsilon_{0}.

Since $U_{1}$ is dense and open, we may choose an $\epsilon_{1}>0$ and an $x_{1}\in U_{1}$ such that

d(x_{0},x_{1})<\frac{\epsilon_{0}}{2},\quad\epsilon_{1}<\frac{\epsilon_{0}}{2},

and such that the open ball of radius $\epsilon_{1}$ about $x_{1}$ lies entirely in $U_{1}$ . We then choose an $\epsilon_{2}>0$ and a $x_{2}\in U_{2}$ such that

d(x_{1},x_{2})<\frac{\epsilon_{1}}{2},\quad\epsilon_{2}<\frac{\epsilon_{1}}{2},

and such that the open ball of radius $\epsilon_{2}$ about $x_{2}$ lies entirely in $U_{2}$ . We continue by induction, and construct a sequence of points $x_{k}\in U_{k}$ and positive $\epsilon_{k}$ such that

d(x_{k-1},x_{k})<\frac{\epsilon_{k-1}}{2},\quad\epsilon_{k}<\frac{\epsilon_{k-% 1}}{2},

and such that the open ball of radius $\epsilon_{k}$ lies entirely in $U_{k}$ .

By construction, for $0\leq j<k$ we have

d(x_{j},x_{k})<\epsilon_{j}\left(\frac{1}{2}+\cdots+\frac{1}{2^{k-j}}\right)<% \epsilon_{j}\leq\frac{\epsilon_{0}}{2^{j}}.

Hence the sequence $x_{k},\;k=1,2,\ldots$ is Cauchy, and converges by hypothesis to some $x\in X$ . It is clear that for every $k$ we have

d(x,x_{k})\leq\epsilon_{k}.

Moreover it follows that

d(x,x_{k})\leq d(x,x_{k+1})+d(x_{k},x_{k+1})<\epsilon_{k+1}+\frac{\epsilon_{k}% }{2},

and hence a fortiori

d(x,x_{k})<\epsilon_{k}

for every $k$ . By construction then, $x\in U_{k}$ for all $k=1,2,\ldots$ , as well. QED

Title	proof of Baire category theorem
Canonical name	ProofOfBaireCategoryTheorem
Date of creation	2013-03-22 13:06:55
Last modified on	2013-03-22 13:06:55
Owner	rmilson (146)
Last modified by	rmilson (146)
Numerical id	10
Author	rmilson (146)
Entry type	Proof
Classification	msc 54E52