Symmetric difference is one of forms which are infrequently used in elementary level. Given a set $A$ and $B$, it is denoted by $A\vartriangle B$ .
By using preceding terms it means that
$A\vartriangle B=(A-B)\cup (B-A)=(A\cap B^c)\cup (A^c\cap B)$ or
$A\vartriangle B=\left\{ x | (x\in A, x\notin B)\cup (x\notin A, x\in B) \right\}$ .
For example, if $A=\left\{ 1,2,3,4 \right\}$ and $B=\left\{ 3,4,5,6 \right\}$ ,
$A\vartriangle B=\left\{ 1,2,5,6 \right\}$ .
Therefore, these below are true.
$A\vartriangle B=B\vartriangle A$ (Commutative law is satisfied)
$(A\vartriangle B)\vartriangle C=A\vartriangle (B\vartriangle C)$ (Associative law is satisfied)
$A\cap (B\vartriangle C)=(A\cap B)\vartriangle (A\cap C)$ (Distributive law in intersection is satisfied)
$A\vartriangle \phi=A$
$A\vartriangle A=\phi$
2014/02/23
2014/02/09
power sets
If the element of a set is also a set, the set whose elements are a set is called a power set.
Given a set $X=\left\{1, 2, 3 \right\}$ , the number of all subsets of $X$ is eight,
$\phi, \left\{1 \right\}, \left\{2 \right\}, \left\{3 \right\}, $
$ \left\{1,2 \right\}, \left\{2,3 \right\}, \left\{1,3 \right\}, \left\{1,2,3 \right\}$ .
Therefore, the power set of $X$ is the set whose elements are all above.
$\left\{\phi, \left\{1 \right\}, \left\{2 \right\}, \left\{3 \right\}, \left\{1,2 \right\}, \left\{2,3 \right\}, \left\{1,3 \right\}, \left\{1,2,3 \right\} \right\}$
In general, if the number of the elements of a finite set is $n$ , the number of the elements of the power set becomes $2^n$ . It will be understood easily.
As this sample is very easy, you should visualize the power set of $[0,1]$
(the interval of real numbers). You will find it is impossible to imagine or write the result.
But it exists definitely.
If the set $X$ is the empty set $\phi$ , the power set of $X$ is $\left\{\phi \right\}$ .
Strictly, this power set of $X$ is not the empty set. That is to say,
$\left\{\phi \right\}\ne \phi$ .
Because the power set of $X$ has one element, but $X$ has no element.
Given a set $X=\left\{1, 2, 3 \right\}$ , the number of all subsets of $X$ is eight,
$\phi, \left\{1 \right\}, \left\{2 \right\}, \left\{3 \right\}, $
$ \left\{1,2 \right\}, \left\{2,3 \right\}, \left\{1,3 \right\}, \left\{1,2,3 \right\}$ .
Therefore, the power set of $X$ is the set whose elements are all above.
$\left\{\phi, \left\{1 \right\}, \left\{2 \right\}, \left\{3 \right\}, \left\{1,2 \right\}, \left\{2,3 \right\}, \left\{1,3 \right\}, \left\{1,2,3 \right\} \right\}$
In general, if the number of the elements of a finite set is $n$ , the number of the elements of the power set becomes $2^n$ . It will be understood easily.
As this sample is very easy, you should visualize the power set of $[0,1]$
(the interval of real numbers). You will find it is impossible to imagine or write the result.
But it exists definitely.
If the set $X$ is the empty set $\phi$ , the power set of $X$ is $\left\{\phi \right\}$ .
Strictly, this power set of $X$ is not the empty set. That is to say,
$\left\{\phi \right\}\ne \phi$ .
Because the power set of $X$ has one element, but $X$ has no element.
2014/02/02
complementary sets
If every sets which we are addressing in a problem is a subset of a big set, the big set which contains all sets is called a universal set. Given the universal set $X$ and a object set $A$ in the problem, a difference set $X-A$ is able to be defined. We usually express it by $A^c$ .
$A^c$ is also called a complementary set of $A$ . By connotative form,
$A^c=\left\{x | x\notin A, x\in X \right\}$
In general as $x\in X$ has been understood and omitted, $A^c$ becomes $\left\{x | x\notin A\right\}$ .
When the universal set is implicit, it is very convenient. Of course,
$(A^c)^c=A$
$\phi^c=X$ and $X^c=\phi$ ($\phi$ is the empty set)
The expressions of complementary sets is very famous in De Morgan's law.
$(A\cup B)^c=A^c\cap B^c$
$(A\cap B)^c=A^c\cup B^c$
If one universal set for $A$ and $B$ exists, these are true. Please try proofs. However, if the universal set of $A$ is not equal the universal set of $B$ , these are not true.
$A^c$ is also called a complementary set of $A$ . By connotative form,
$A^c=\left\{x | x\notin A, x\in X \right\}$
In general as $x\in X$ has been understood and omitted, $A^c$ becomes $\left\{x | x\notin A\right\}$ .
When the universal set is implicit, it is very convenient. Of course,
$(A^c)^c=A$
$\phi^c=X$ and $X^c=\phi$ ($\phi$ is the empty set)
The expressions of complementary sets is very famous in De Morgan's law.
$(A\cup B)^c=A^c\cap B^c$
$(A\cap B)^c=A^c\cup B^c$
If one universal set for $A$ and $B$ exists, these are true. Please try proofs. However, if the universal set of $A$ is not equal the universal set of $B$ , these are not true.