let A and B be sets. Prove that AXB=0 iff A=0 or B=0
To prove that AXB = 0 if and only if A = 0 or B = 0, we need to show two separate implications:
1. If AXB = 0, then A = 0 or B = 0.
2. If A = 0 or B = 0, then AXB = 0.
Let's prove each implication separately:
1. If AXB = 0, then A = 0 or B = 0.
To prove this implication, we can use the fact that the Cartesian product of two non-empty sets is never empty.
Assume that neither A nor B is empty. This means that there exists at least one element in both A and B. Let's call these elements a in A and b in B.
Now, since (a,b) is an element of AXB, this implies that AXB is not empty. However, we assumed that AXB = 0, which is a contradiction. Therefore, our assumption that neither A nor B is empty must be false.
Hence, if AXB = 0, then A = 0 or B = 0.
2. If A = 0 or B = 0, then AXB = 0.
To prove this implication, we need to show that if A = 0 or B = 0, then AXB is empty.
If A = 0, it means that A is an empty set. In this case, AXB will be empty since there are no elements in A to form the Cartesian product with any element of B.
Similarly, if B = 0, it means that B is an empty set. Again, AXB will be empty since there are no elements in B to form the Cartesian product with any element of A.
Thus, if A = 0 or B = 0, then AXB = 0.
Therefore, we have shown both implications, and hence, we can conclude that AXB = 0 if and only if A = 0 or B = 0.