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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