proof of Baire category theorem

Let $(X,d)$ be a complete metric space, and $U_{k}$ a countablecollection of dense, open subsets. Let $x_{0}\\in X$ and $\\epsilon_{0}>0$ begiven. 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 ofradius $\\epsilon_{1}$ about $x_{1}$ lies entirelyin $U_{1}$ . 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 ballof radius $\\epsilon_{2}$ about $x_{2}$ liesentirely in $U_{2}$ . We continue by induction, and construct a sequenceof 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 openball 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 thesequence $x_{k},\\;k=1,2,\\ldots$ is Cauchy, and converges by hypothesisto 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