Topology of real numbers


	Topology of real numbers

Topology on a set X is defined by a family of its subsets that are called open.The family should satisfy the following requirements.

Open sets

Union of any family of open sets is an open set. In other words, if $U_\alpha \in \tau$ for any $\alpha \in A$ then

$\bigcup_{\alpha \in A } U_\alpha \;\;\in \;\tau$

where A is the set of indices.

Intersection of a finite number of open sets is an open set. If $U_i \in \tau$ for $i=1,\;2,\;\dots \;n$ then

$\bigcap_{ i=1 }^n U_i \;\;\in \;\tau$

Empty set $\emptyset \in \tau$ is open and so is X itself;

$\emptyset \in \tau \;\;\mbox{ and } X\;\in \;\tau$

As soon as you have topology defined on X you immediately obtain the family of subsets $\bar \tau$ from X that are complimentary to open sets. They are called closed. That means

$M \;\in\;\bar \tau \;\;\mbox{ if, and only if, } \;\;\;X \setminus M \in \;\tau$

A set M is closed if, and only if, its complimentary set $X \setminus M$ is open. Closed sets $\bar \tau$ have the following properties.

Closed sets

Union of a finite number of closed sets is closed.

$\bigcup_{i=1}^n U_i \; \in \; \bar \tau$

under the condition that

$\forall \;\;i\;\;U_i \in \bar \tau$

Intersection of an arbitrary family of closed sets is closed.

$\bigcap_{\alpha \in A } U_\alpha \;\;\in \;\bar \tau$

as long as,

$\forall\;\;\alpha \in \;A\;\;\;U_\alpha \in \bar \tau$

where A is the set of indices.

The empty set

$\emptyset \mbox{ and } X$

are both closed.

Real numbers

Topology of real numbers (real line $\mathbb{R}$ ) is defined by open intervals

$(a,\;b)=\{a \;<\; x \;< \; b\}$

A subset

$M\subset \mathbb{R}$

is called open if it is the union of open intervals;

$M = \bigcup_\alpha (a_\alpha,\; b_\alpha )$

The set of open intervals is called the basis for the topology of real numbers.

Limit points

A point

$p \in \mathbb{R}$

is called a limit point for a set

$S \subset \mathbb{R}$

$\forall \;\varepsilon > 0\;\;\exists\;\;x \in S \;\;\mbox{ such that } \;\;x\in\{p- \varepsilon < x < p\;\mbox{ or } p<\;x<\; p + \varepsilon \}$

The set S taken together with all its limit points is closed. It is called closure of S and denoted by $\bar S$ .

Compact set

A set

$S \subset \mathbb{R}$

is called compact if any open cover of S:

$S \subset \bigcup_\alpha (a_\alpha ,\; b_\alpha )$

permits a finite subcover

$S \subset \bigcup_{i=1}^n (a_{\alpha_i} ,\; b_{\alpha_i} )$

where each of the intervals from

$\{(a_{\alpha_i} ,\; b_{\alpha_i} ) \}_{i=1}^n$

coincides with one of the intervals from

$\{(a_{\alpha} ,\; b_{\alpha} ) \}_\alpha$

Now we present an easy verifiable criterion of compactness.

      
Theorem (Heine, Bolzano, Borel, Lebeg) A set is compact if, and only if, it is closed and bounded.

Proof.

Let

$S \subset \mathbb{R}$

be a compact set. Our goal is to show that it is closed and bounded.

Let us first show that it is bounded. Consider an open cover

$S\subset \bigcup_n (-n,\;n)$

Compactness implies the existence of a finite subcover. Hence, there exists a natural number N such that

$S\subset (-N,\;N)$

Therefore, the set is bounded.

Now we show that the set is closed. Consider a limit point p for S. If p does not belong to S then one can not choose a finite subcover for S from

$S \subset \bigcup_n J_n$

where

$J_n = \{x <\;p-\frac{1}{n}\;\;\mbox{ and } p+\frac{1}{n} < x \}$

Hence, all limit points for S are in S. That means S is closed.

It remains to prove that any closed bounded set S is compact. In other words any open cover for S permits a finite subcover. We conduct the proof by reductio ad absurdum.

Let us assume that S closed and bounded. However, there exists an open cover

$S \subset \bigcup_\alpha U_\alpha$

that does not permit a finite subcover. Since S is bounded there exists a closed interval

$[a,\;b]=\{a \le x \le b \}$

such that

$S \subset [a,\;b]$

We divide the interval into two parts at choose the part that does not permit a finite subcover. Then again we divide the chosen part into two pieces and again choose the one that does not permit a finite subcover. We continue in the manner and obtain the sequence of closed intervals shrinking to a point p from S that does not belong to the open cover

$\bigcup_\alpha U_\alpha$

That contradicts to the assumption that

$S \subset \bigcup_\alpha U_\alpha$

The obtained contradiction completes the proof.

Cantor set

Consider the unit interval

$[0,\; 1 ] = \{0 \le \;x \le 1 \}$

We divide this interval into three equal parts and throw away the middle piece. As result we are left with the following set

$A_1 = [0, \; \frac{1}{3} ] \bigcup [\frac{2}{3},\; 1 ]$

Then we divide each interval from A₁ into three equal parts and again get rid of the middle pieces. The resulted set is

$A_2 = [0, \; \frac{1}{9}] \bigcup [\frac{2}{9},\frac{1}{3} ] \bigcup [\frac{2}{3},\; \frac{7}{9}]\bigcup [\frac{8}{9},\; 1]\;$

We continue in this fashion, each time, in order to construct A_{n + 1} we divide intervals from A_n into three equal parts and throw away middle pieces.

     The Cantor set is the intersection