- #1

Seydlitz

- 263

- 4

## 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?

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