Proof of intersection and union of Set A with Empty Set I need to prove the following:
Prove that $A\cup \!\, \varnothing \!\,=A$ and $A\cap \!\, \varnothing \!\,=\varnothing \!\,$
It's my understanding that to prove equality, I must prove that both are subsets of each other.
So to prove $A\cup \!\, \varnothing \!\,=A$, we need to prove that $A\cup \!\, \varnothing \!\,\subseteq \!\,A$ and $A\subseteq \!\,A\cup \!\, \varnothing \!\,$.
However, I found an example proof for $A \cup \!\, A$ in my book and I adapted it and got this:
$A\cup \!\, \varnothing \!\,=$ {$x:x\in \!\, A  \ \text{or} \ x\in \!\, \varnothing \!\,$}
= {$x:x\in \!\, A$} = A
$A\cap \!\,  \varnothing \!\,=$ {$x:x\in \!\, A  \ \text{and} \ x\in \!\, \varnothing \!\,$}
= {$x:x\in \!\, \varnothing \!\,$} = $\varnothing \!\,$
Do my proofs look ok?
 A: Yes. The solution works, although I'd express the second last step slightly differently.
$\begin{align}
A\cup \varnothing & =  \{x:x\in A \vee x\in\varnothing \} & \text{definition of union}
\\ &= \{x:x\in A \} & \neg\exists x~(x\in \varnothing)
\\ & = A
\\[2ex]
A\cap\varnothing & =  \{x:x\in A \wedge x\in \varnothing \} & \text{definition of intersection}
\\ & = \{\} & \neg\exists x~(x\in \varnothing \wedge x\in A) 
\\ & = \varnothing
\end{align}$ 

$A\cup \varnothing = A$ because, as there are no elements in the empty set to include in the union therefore all the elements in $A$ are all the elements in the union.  Hence the union of any set with an empty set is the set.
$A\cap \varnothing = \varnothing$ because, as there are no elements in the empty set, none of the elements in $A$ are also in the empty set, so the intersection is empty.  Hence the intersection of any set and an empty set is an empty set.
A: This looks fine, but you could point out a few more details. For instance, $x\in \varnothing$ is always false. Therefore
$x \in A \text{ or } x\in \varnothing
$
is logically equivalent to
$
x \in A
$
and therefore the two set descriptions
$$
\{x \mid x \in A \text{ or } x \in \varnothing\},\quad \{x\mid x \in A\} 
$$
must describe the same set, since the conditions are true for exactly the same elements $x$.
Similarily, because $x \in \varnothing$ is trivially false, the condition $x \in A \text{ and } x \in \varnothing$ will always be false, so the two set descriptions
$$
\{x \mid x \in A \text{ and } x \in \varnothing\},\quad \{x\mid x \in \varnothing \} 
$$
must describe the same set.
Of course, for any set $B$ we have
$$
B = \{x \mid x \in B\}
$$
A: I think your proofs are okay, but could use a little more detail when moving from equality to equality.  I like to stay away from set-builder notation personally. You could also show $A \cap \emptyset = \emptyset$ by showing for every $a \in A$, $a \notin \emptyset$. That proof is pretty straightforward. For showing $A\cup \emptyset = A$ I like the double-containment argument. As a freebie you get $A \subseteq A\cup \emptyset$, so all you have to do is show $A \cup \emptyset \subseteq A$.
A: we need to proof that A U phi=A,
it can be written as,
A U PHI={$X$:$X \in A$ OR $X \in \phi$}
Apparently no $X$ satisfies the second part of the conjunction. Or, in a simplest way I will use a example,
write in roaster form
A={1,2,3}
PHI={}
WHEN YOU WRITE THE UNION,you just write all elements in the sets in a single set. So it COMES OUT TO BE {1,2,3}
THEREFORE $A\cup \phi=A$
