Proof of Union and Intersection with nothing

[Seydlitz]

Homework Statement

Prove:
$A \cup \varnothing = A$
$A \cap \varnothing = \varnothing$

The Attempt at a Solution

Intuitively both are true. The first is true because union with nothing will eventually return the original set. The second is true because there is no element that can be in a set and at the same time belongs to nothing.

First proof:
$A \cup \varnothing = A$
Suppose $x \in (A \cup \varnothing)$, then $x \in A$ or $x \in \varnothing$, the only true implication under such condition is, $x \in A$. Hence $x \in A$ for all $x$.

$A \cap \varnothing = \varnothing$
Suppose $x \in (A \cap \varnothing)$, then $x \in A$ and $x \in \varnothing$. It can be seen that there is no $x$ that can satisfy the definition, hence the intersection will be equal to $\varnothing$.

The thing I'm worried about the last proof is by using the fact that $x \in A$ and $x \in \varnothing$ which will be always false, you can use that statement to prove anything you want. Say, like this:

Suppose $x \in (A \cap \varnothing)$, then $x \in A$ and $x \in \varnothing$, hence $x \in A$.

But it will be always vacuously true?

 

[Dick]

Homework Statement

Prove:
$A \cup \varnothing = A$
$A \cap \varnothing = \varnothing$

The Attempt at a Solution

Intuitively both are true. The first is true because union with nothing will eventually return the original set. The second is true because there is no element that can be in a set and at the same time belongs to nothing.

First proof:
$A \cup \varnothing = A$
Suppose $x \in (A \cup \varnothing)$, then $x \in A$ or $x \in \varnothing$, the only true implication under such condition is, $x \in A$. Hence $x \in A$ for all $x$.

$A \cap \varnothing = \varnothing$
Suppose $x \in (A \cap \varnothing)$, then $x \in A$ and $x \in \varnothing$, hence $x \in \varnothing$.

The thing I'm worried about the last proof is by using the fact that $x \in A$ and $x \in \varnothing$ which will be always false, you can use that statement to prove anything you want. Say, like this:

Suppose $x \in (A \cap \varnothing)$, then $x \in A$ and $x \in \varnothing$, hence $x \in A$.

But it will be always vacuously true?

There is NO x such that $x \in \varnothing$, so you can't conclude $x \in A$.

 

[Seydlitz]

There is NO x such that $x \in \varnothing$, so you can't conclude $x \in A$.

The proof will not be correct yes. But will the statement still stand logically if you follow the truth table of implication? The antecedent is certainly false, but then I can add new consequence which is true. This will cause the whole implication to be true also.

On top of that I just edited the first post to add slight revision to the proof.

 

[Dick]

The proof will not be correct yes. But will the statement still stand logically if you follow the truth table of implication? The antecedent is certainly false, but then I can add new consequence which is true. This will cause the whole implication to be true also.

On top of that I just edited the first post to add slight revision to the proof.

Yes, the antecedent is always false. So the implication is a true statement. But that doesn't prove the consequent. If p->q and p is always false, p->q is always true. But that gives you no information about whether q is true or false.

 

[Seydlitz]

Yes, the antecedent is always false. So the implication is a true statement. But that doesn't prove the consequent. If p->q and p is always false, p->q is always true. But that gives you no information about whether q is true or false.

Ok I get your point for the latter part, but do you have any problem with the union proof? All of the options for 'or' will be false except if x belongs to A. So it is a must if we want to follow the definition, that the union will always be A. Is this sort of reasoning correct?

 

[Dick]

Ok I get your point for the latter part, but do you have any problem with the union proof? All of the options for 'or' will be false except if x belongs to A. So it is a must if we want to follow the definition, that the union will always be A. Is this sort of reasoning correct?

It looks fine to me. If a or b is true, and b is false, then a must be true.

 

[vela]

You've shown that $A \cup \emptyset \subset A$. You still need to show the opposite direction before you can conclude equality.

 

[Seydlitz]

You've shown that $A \cup \emptyset \subset A$. You still need to show the opposite direction before you can conclude equality.

Let $x \in A$, hence $x \in A$ or $x \in \varnothing$, thus $x \in A \cup \varnothing$. So $A \subset (A \cup \varnothing)$

Is that sufficient?

 

[vela]

Yup.