Can AxB be equal to the empty set if either A or B is empty?

[kathrynag]

Homework Statement

I just need to decide how to show this by contradiction.
If either A or B is the empty set then AxB=\oslash.

Homework Equations

The Attempt at a Solution

Here is how I started:
Assume either A or B is the empty set and AxB\neq\oslash

 

[statdad]

Reasonable start. What would A \times B \ne \emptyset mean?

 

[kathrynag]

Is that the correct way to do a proof by contradiciton?
AxB is defined as the set consisting of all ordered pairs (x,y) in which x is an element of A and y is an element of B. So, x and y exist?

 

[statdad]

Close: if you assume A \times B \ne \emptyset, there must be at least
one element (a,b) \in A \times B. If you think about the definition of cartesian products, this will lead to a contradiction - about what? (Hint: what did you assume about A \text{ and } B?)

 

[kathrynag]

Ok here's my idea for the proof.
Let A = null set and B be arbitrary. Then AxB= null set because of the definition of AxB. But there is no x which is an element of A. Therefore AxB=null set. Thus, contradiciton.

 

[statdad]

No - you can't assume A \times B = \emptyset and try to proceed with a proof by contradiction.

Assume A= \emptyset (B may or may not be empty: that is unimiportant).

If A \times B \ne \emptyset, then (by definition of the Cartesian Product and non-empty set)
you can find an element of the product, say (a,b) \in A \times B.

This means b \in B. From where do you get the object a?
Answering the second question gives the contradiction.

 

[kathrynag]

a is an element of A.
Oh, but then that mean A is nonempt and this a contradicition?

 

[statdad]

"a is an element of A.
Oh, but then that mean A is nonempt and this a contradicition?"

- yup - it contradicts A = \emptyset