$\{\emptyset\} \cup \{a, b\} = \{a, b\}$ is this true? I am really confused about the empty set. Does any set contain the empty set in general?
Is $\{\emptyset\} \cup \{a, b\} = \{\emptyset, a, b\}$? if so then it is not equal to $\{a, b\}$? then the statement is false? 
 A: The empty set is a set with nothing in it. It can be treated as an element, and may be an element in other sets, but is not in all of them. 
So $\{\emptyset\}$ is actually a set of size one, and the only item in it is the empty set. 
Now, $\{a, b\}$ does not contain the empty set (assuming $a,b \neq
 \emptyset$)
So $\{\emptyset\} \cup \{a, b\} = \{\emptyset, a, b\} $
A: You are correct.  The statement in your title is (usually) false.  It would be true only if $a = \varnothing$ or $b=\varnothing$ (or both).
A: Every set contains the empty set as a subset, not necessarily as an element.
For any set $A$, $\emptyset \subseteq A$ since for any element $x$ in the empty set (there is no such element), $x$ must be in $A$: this is just vacuously true.
The statement $\{\emptyset\} \cup \{a,b\} = \{\emptyset,a,b\}$ is correct. 
If $a$ and $b$ are not the empty set, then this union will not be $\{a,b\}$ since you have added a new element. The emptyset is a perfectly good candidate to be an element of a set (anything is, afterall).
A: Your title is not correct, nor is it true that the empty set is an element of every set.
Here, it's convienent to think of sets as "boxes" where we consider the boxes "the same" if they contain the same things (disregarding duplicates). If I have two things in a box, does it contain the same things as a box containing 1. an empty box and two other things? nope! 
