Question on Empty set and Power set Does set A equal to set B ? where
$$ A= \{ \emptyset, a,b,c \} $$ $$ B=\{ a,b,c \}  $$
second question: I know the order of the elements does not matter in a Set.
But i once read about the order of elements(or subset) matters in some case, and i can't remember what it is. 
Some help from you guys would be nice.
Thanks
 A: The order of the items in the set does not matter.  So these three sets are equal:
$$B = \{a,b,c\} = \{c,b,a\} = \{b,a,c\}$$
A set may have another set as a member.  So a set containing the empty set and another element is distinct from a set containing just one element.  For example, the $A$ and $B$ in your question are distinct sets.
The order of items in ordered pairs contained in sets matter.  So, the following two sets, $C$ and $D$, are distinct because they contain different ordered pairs.
$$C = \{ (a,b) , (c,d) \} $$
$$D = \{ (b,a) , (d,c) \} $$ 
A: No. $A$ and $B$ are different as they have three elements in common, but $A$ also has $\emptyset$. You might think it is like adding zero, but it is not. The order matters when you have a Cartesian product or ordered $n$-tuples in general, hence the name.
A: *

*Assuming $a,b,c \not = \emptyset$, we have $A-B=\lbrace \emptyset \rbrace\not =\emptyset,$ so $A\not =B$. 

*Perhaps the time you saw that the order was of importance was when studying sequences?
A: *

*If $A=B$, it must satisfy $A⊆B$, and $B⊆A$. In your case, $A$ has an element $∅$ that does not appear in $B$. Therefore, $A⊄B$,  they are not equal~

*Oder matters in Pair Set. For instance $A=${(1,2)} and B={(2,1)}$ $,then$A≠B$ .

