Is $A=B$ if $A=\{a,b\}$ and $B=\{a,b,∅\}$? I know that $A∩∅=∅$. But what about $A∩\{∅\}$?
Shouldn't  $A∩\{∅\}=\{∅\}$.
If $A=\{a,b\}$ and $B=\{a,b,∅\}$ then  we can conclude $A=B$?
 A: It's really simple:
$$A\cap\{\varnothing\} = \left\{\begin{array}{ll}
\varnothing &\text{if }\varnothing\notin A,\\
\{\varnothing\} &\text{if }\varnothing \in A.
\end{array}\right.$$
If $A=\{a,b\}$ and $B=\{a,b,\varnothing\}$, then $A=B$ if and only if $a=\varnothing$ or $b=\varnothing$.
(Informally: an empty bag is not the same thing as a bag that contains nothing except for an empty bag)
A: I will try to give a more informal and intuitive clarification. There is an important distinction between the set that contains nothing (which is the empty set specified as $\varnothing$) and the concept we call nothing (which doesn't have a symbol in set theory). To clarify that, we can rewrite $\varnothing$ as $\{\}$. You can then clearly see that $\{\}$ is not the same as nothing; it is something containing nothing. An analogy that is often made in mathematical textbooks is the following: The empty set is analogous to an empty shoebox. Therefore, a set containing the empty set, i.e. $\{\{\}\}$ or $\{\varnothing \}$ is a shoebox containing an empty shoebox. Thus, the empty set is not nothing, it is the set that holds nothing.
To come back to your question: $A \cap \varnothing = \varnothing $. That is true. But why is that? What does a set containing something (i.e. $A$) have in common with a set containing nothing? Well, nothing, regardless of how we specify $A$. And because the intersection of sets is always a set, we get a set containing nothing. And what is a set containing nothing? It is the empty set! Now why is that different to $A \cap \{\varnothing\} = \{\varnothing \}$? Because now you are comparing two sets that hold something. One set holds something unspecified (i.e. $A$), while the other set holds a set containing nothing (i.e. the empty set). And what do both sets have in common? Well, that depends on the specification of $A$. If $A$ contains the empty set, i.e. if $\varnothing \in A$, then we get $A \cap \{\varnothing\} = \{\varnothing \}$. If not, then we get $A \cap \{\varnothing\} = \varnothing$. And, as we said before, $\varnothing \neq \{ \varnothing \}$.
I hope that gives you a more intuitive explanation to a distinction that deludes a lot of people starting with elementary set theory.
A: For Solution of this, You just need to  understand $ \{\phi\} \neq \phi  $  ... (Assuming $ a\neq \phi, b\neq \phi $) $ \{ \phi \}  $ has one element in set
As $ \{ \phi \} \neq \phi $ , Its clear that set $ A=\{a,b\} $ has two elements while set $ B=\{a,b,\phi\} $ has three elements..
So $ A\neq B $
