We shall summarize the definitions of compact sets.
At first glance, you may not think these are equivalent.
However, these present a same concept and show the properties of compact sets.
(1)A set is compact if and only if any open cover of the set has a finite open sub cover.
It depends on the theorem of Heine-Borel. Please note "finite".
(2)A set in $\mathbb{R}$ is compact if and only if the set is complete and bounded.
Hence, a closed interval in $\mathbb{R}$ is compact. It is understood easily.
(3)A metric space is compact if and only if any infinite sequences has a convergent sub sequence.
It is called the theorem of Bolzano-Weierstrass.
Note that these theorems would not be true if the existing conditions were changed even if only slightly.
2014/05/25
2014/05/18
compact sets
We will again give the definition of the compactness.
If a open cover of $X$ always contains a finite sub cover, the set $X$ is compact.
The reasons why the compactness is more stronger than the completeness depend
on two propositions as follow.
(1)A compact space becomes complete.
(2)A compact set becomes bounded.
We leave out the proofs. However, as a open cover of $X$ always contains a finite sub cover,
no one will have a feeling of strangeness.
Therefore, a set becomes compact if and only if a set is complete and has a finite open cover.
Then we reach the theorem of Heine Borel because a complete subset in a complete space is a closed set. (a finite open cover means bounded.)
2014/05/10
some definitions related to open sets 5 (compact)
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.
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.
2014/05/03
some definitions related to open sets 4 (a interval is connected)
We reached in the preceding post to a proposition which insisted that a connected set on the real number line was a interval. So, we shall give a brief proof of the proposition.
Suppose that $I$ is a connected set on the real number line
and $I=A\cup B$ , where $A,B\subset \mathbb{R}$ , $A\cap B=\phi$
and both $A$ and $B$ are closed and not empty.
We are able to choose two points $a_1$ from $A$ and $b_1$ from $B$ , and put $a_1<b_1$ .
Dividing the new interval $(a_1,b_1)$ in half, the one must contain points of $A$ and $B$ .
Let the one be a small interval $(a_2,b_2)$ and we repeat the same operation.
Then, we get the sequence of the interval $(a_n,b_n)$ where $a_n\in A$ and $b_n\in B$
and $(a_n,b_n)\supset (a_{n+1},b_{n+1})$ .
If the operation is repeated infinitely, as the new interval becomes smaller and smaller, the real number sequence $a_n$ and $b_n$ have a same limit point $c$ .
As $A$ and $B$ are closed, $c$ is a intersection point of two sets. However, it is in contradiction to the precondition.
Suppose that $I$ is a connected set on the real number line
and $I=A\cup B$ , where $A,B\subset \mathbb{R}$ , $A\cap B=\phi$
and both $A$ and $B$ are closed and not empty.
We are able to choose two points $a_1$ from $A$ and $b_1$ from $B$ , and put $a_1<b_1$ .
Dividing the new interval $(a_1,b_1)$ in half, the one must contain points of $A$ and $B$ .
Let the one be a small interval $(a_2,b_2)$ and we repeat the same operation.
Then, we get the sequence of the interval $(a_n,b_n)$ where $a_n\in A$ and $b_n\in B$
and $(a_n,b_n)\supset (a_{n+1},b_{n+1})$ .
If the operation is repeated infinitely, as the new interval becomes smaller and smaller, the real number sequence $a_n$ and $b_n$ have a same limit point $c$ .
As $A$ and $B$ are closed, $c$ is a intersection point of two sets. However, it is in contradiction to the precondition.