PlanetMath: proof of Baire category theorem

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proof of Baire category theorem

(Proof)

Let be a complete metric space, and 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

$\displaystyle d(x_0,x)<\epsilon_0.$

Since

is dense and open, we may choose an $\epsilon_1>0$ and an $x_1\in U_1$ such that

$\displaystyle d(x_0,x_1)<\frac{\epsilon_0}{2},\quad \epsilon_1<\frac{\epsilon_0}{2},$

and such that the open ball of radius $\epsilon_1$ about

lies entirely in

. We then choose an $\epsilon_2>0$ and a $x_2\in U_2$ such that

$\displaystyle d(x_1,x_2)<\frac{\epsilon_1}{2},\quad \epsilon_2<\frac{\epsilon_1}{2},$

and such that the open ball of radius $\epsilon_2$ about

lies entirely in

. We continue by induction, and construct a sequence of points $x_k\in U_k$ and positive $\epsilon_k$ such that

$\displaystyle d(x_{k-1},x_k)<\frac{\epsilon_{k-1}}{2},\quad \epsilon_k<\frac{\epsilon_{k-1}}{2},$

and such that the open ball of radius $\epsilon_k$ lies entirely in

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

$\displaystyle d(x_j,x_k) < \epsilon_j \left(\frac{1}{2} + \cdots + \frac{1}{2^{k-j}}\right) < \epsilon_j \leq \frac{\epsilon_0}{2^j}.$

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

we have

$\displaystyle d(x,x_k) \leq \epsilon_k.$

Moreover it follows that

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

and hence a fortiori

$\displaystyle d(x,x_k)<\epsilon_k$

for every

. By construction then, $x\in U_k$ for all $k=1,2,\ldots$ , as well. QED

"proof of Baire category theorem" is owned by rmilson. [ full author list (2) ]

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54E52 (General topology :: Spaces with richer structures :: Baire category, Baire spaces)

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