In preceding posts, a definition of the completeness of real numbers was given.
Here we shall give the similar definition of the completeness of a metric space.
In a metric space if any Cauchy sequences is always convergent, the metric space is complete.
Therefore, closed intervals on the real number line is complete. That is to say,
a closed subset in a complete metric space is complete.
The compactness is more stronger than the completeness. However the definition is most
difficult and abstract. A open cover must be prepared at first.
Given a set $X\subset S$ , $S=\cup A_i$ and all $A_i$ is open, $S$ is called a open cover of $X$ .
$[0,1)\subset \cup_{i=1}^{\infty}\left( -\frac{1}{i},1-\frac{1}{i} \right)$
Hence, $\cup_{i=1}^{\infty}\left( -\frac{1}{i},1-\frac{1}{i} \right)$ is a open cover of $[0,1)$
In general, there are many open covers of $X$ .
If $S'$ which is made by joining some selected $A_i$ also becomes a open cover of $X$ ,
$S'$ is called a sub cover. If $i$ is finite, $S$ is called a finite open cover.
If a open cover of $X$ always contains a finite open sub cover, the set $X$ is compact.
The famous Heine Borel theorem is as follows.
A set $X\subset \mathbb{R}$ is compact if and only if $X$ is closed and bounded.
It is not easy to understand the essence which the theorem insists.
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