if A and B are countable sets, then prove AxB is countable
To prove that the Cartesian product AxB of two countable sets A and B is also countable, we need to show that there exists a one-to-one correspondence between AxB and the set of natural numbers (N), which is known to be countable.
We can use the concept of a pairing function to establish this correspondence. A pairing function is a function that maps two natural numbers to a single natural number. One commonly used pairing function is the Cantor pairing function.
Here's a step-by-step explanation of how to prove that AxB is countable using the Cantor pairing function:
Step 1: Enumerate the elements of A and B
You need to create an enumeration or list of all the elements in A and B. Since A and B are countable sets, you can list their elements in a sequence like:
A = {a1, a2, a3, ...}
B = {b1, b2, b3, ...}
Step 2: Apply the Cantor pairing function to each pair (ai, bj)
For each pair (ai, bj) in AxB, apply the Cantor pairing function to generate a unique natural number. The Cantor pairing function is defined as follows:
f(ai, bj) = (ai + bj) * (ai + bj + 1) / 2 + ai
Step 3: Create a list of the resulting natural numbers
As you apply the Cantor pairing function to each pair (ai, bj), obtain a list of natural numbers:
f(a1, b1), f(a1, b2), f(a1, b3), ...
f(a2, b1), f(a2, b2), f(a2, b3), ...
f(a3, b1), f(a3, b2), f(a3, b3), ...
...
Step 4: Prove the one-to-one correspondence
To complete the proof, you need to show that there is a one-to-one correspondence between the resulting natural numbers and the elements of AxB.
To establish this correspondence, assume you have any natural number n. You can then use the Cantor pairing function in reverse to find the corresponding pair (ai, bj).
First, find the maximum value of ai that satisfies:
(ai + bj) * (ai + bj + 1) / 2 + ai <= n
Then, solve for bj using the equation:
bj = n - [(ai + bj) * (ai + bj + 1) / 2 + ai]
Thus, for every natural number n, you can find a unique pair (ai, bj) in AxB. This proves the one-to-one correspondence between AxB and N.
Therefore, AxB is countable.