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{\bf 4.4.1.} {\sc The Stone Theorem.} {\it Every open cover of a metrizable space has an open refinement
which is both locally finite and $\sigma$-discrete.}
{\sc Proof.} Let $\{U_s\}_{s\in S}$ be an open cover of a metrizable space $X$; take metric $\rho$ on the
space $X$ and a well-ordering relation $<$ on the set $S$. Define inductive families ${\cal V}_i = \{V_{s,i}\}_{s\in S}$
of subsets of $X$ by letting
$$V_{s,i} = \bigcup B(c,1/2^i),$$
where the union is taken over all points $c\in X$ satisfying the following conditions:
$$\hbox{$s$ is the smallest element of $S$ such that $c\in U_s$.}\leqno(1)$$
$$\hbox{$c\not\in V_{t,j}$ for $j < i$ and $t\in S$.}\leqno(2)$$
$$\hbox{$B(c,3/2^i)\subset U_s$.}\leqno(3)$$
It follows from the definition that the sets $V_{s,i}$ are open, and (3) implies that $V_{s,i}\subset U_s$.
Let $x$ be a point of $X$; take the smallest $s\in S$ such that $x\in U_s$ and a natural number $i$ such
that $B(x,3/2^i)\subset U_s$. Clearly, we either have $x\in V_{t,j}$ for a $j < i$ and a $t\in S$ or $x\in V_{s,i}$.
Hence, the union ${\cal V} = \bigcup_{i=1}^\infty {\cal V}_i$ is an open refinement of the cover $\{U_s\}_{s\in S}$.
We shall prove that for every $i$
$$\hbox{if $x_1\in V_{s_1,i}$, $x_2\in V_{s_2,i}$ and $s_1\ne s_2$, then $\rho(x_1,x_2) > 1/2^i$,}\leqno(4)$$
and this will show that the families ${\cal V}_i$ are discrete, because every $1/2^{i+1}$-ball meets at most
one member of ${\cal V}_i$.
Let us assume that $s_1 < s_2$. By the definition of $V_{s_1,i}$ and $V_{s_2,i}$ there exist points $c_1,c_2$
satisfying (1)-(3) such that $x_k\in B(c_k,1/2^i)\subset V_{s_k,i}$ for $k = 1,2$. From (3) it follows that
$B(c_1,3/2^i)\subset U_{s_1}$ and from (1) we see that $c_2\not\in U_{s_1}$, so that $\rho(c_1,c_2)\ge 3/2^i$. Hence,
$$\rho(x_1,x_2)\ge\rho(c_1,c_2)-\rho(c_1,x_1)-\rho(c_2,x_2) > 1/2^i,$$
which proves (4).
To conclude the proof of the theorem it suffices to show that for every $t\in S$ and any
pair $k,j$ of natural numbers
$$\hbox{if $B(x,1/2^k)\subset V_{t,j}$, then $B(x,1/2^{i+k})\cup V_{s,i} = \emptyset$ for $i\ge j+k$ and $s\in S$,}\leqno(5)$$
because for every $x\in X$ there exist $k,j$ and $t$ such that $B(x,1/2^k)\subset V_{t,j}$ and thus the ball
$B(x,1/2^{j+k})$ meets at most $j+k-1$ members of ${\cal V}$.
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It follows from (2) that the points $c$ in the definition of $V_{s,i}$ do not belong to $V_t,j$ whenever
$i\ge j+k$; since $B(x,1/2^k)\subset V_{t,j}$, we have $\rho(x,c)\ge 1/2^k$ for any such $c$. The inequalities
$j+k\ge k+1$ and $i\ge k+1$ imply that $B(x,1/2^{j+k})\cup B(c,1/2^i) = \emptyset$, and this yields (5). $\diamondsuit$
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