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?

Solution 1:

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 $$ B = \{x \mid x \in B\} $$

Solution 2:

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.

Solution 3:

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 $A \cup \emptyset \subseteq A$.