$\newcommand{\defeq}{\mathrel{\mathop:}=}$

## 2007/08/15

### 亂排

\input amssym

\def\proof{\noindent{\bf Proof.} }

\proclaim Definition (open covering). An open covering $\scr F$ of a
set $S$ is a family of open sets such that $x \in S$ implies $x \in O$
for some $O \in {\scr F}$.

\proclaim Definition (compactness). A set $S$ is called compact if
every open covering of $S$ has a finite subcovering.

\proclaim Theorem (Heine-Borel). Every open covering of\/ $[a, b] \subset {\Bbb R}$ has a finite subcovering.

\proof Assume the contrary is true. Then there exists an open covering
$\scr F$ of $[a, b]$ which doesn't have a finite subcovering. Divide
$[a, b]$ into two equal-length parts $[a, {a+b \over 2}]$ and
$[{a+b \over 2}, b]$. It must be the case that one of them cannot be
covered by finite subcovering of $\scr F$; let it be
$I_1 = [a_1, b_1]$, whose length is $(b-a) \mathbin{/} 2$. Applying
the above argument iteratively yields $I_2$, $I_3$, $\ldots$ and every
$I_n$ cannot be covered by finite subcovering of $\scr F$ and has
length $(b-a) \mathbin{/} 2^n$. Thus
$$I_1 \supset I_2 \supset I_3 \supset \cdots \qquad {\rm and} \qquad |I_n| = {b - a \over 2^n} \to 0 \quad {\rm as} \quad n \to \infty.$$
By the theorem of nested intervals, there uniquely exists a real
number $x_0 \in [a, b]$ such that
$$\bigcap_{n = 1}^\infty I_n = \{x_0\}.$$
$\scr F$ is an open covering of $[a, b]$ and $x_0 \in [a, b]$, so
there exists an open set $O \in {\scr F}$ covers $x_0$. $O$ is
open'' implies there exists $B_\delta(x_0) \subset O$. $I_n$
shrinks'' to $x_0$ as $n \to \infty$, so eventually $I_n$ would be
covered by $B_\delta(x_0) \subset O$ for some $n$. But this
contradicts with the fact that $I_n$ cannot be covered by
finite subcovering of $\scr F$.

\proclaim Corollary. $[a, b] \subset {\Bbb R}$ is compact.

\vfill\eject
\end


Blogger 的圖片上傳目前似乎有點問題，所以截圖暫略 XD。

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yen38/15/2007 3:53 pm 說：