Answer to Question #284163 in Real Analysis for Amit

Question

Let a and b be two cardinal numbers. Modify Cantor’s definition of a < b to define a ≤ b. (Hint: Examine what happens if you drop condition (a) from Cantor’s definition of a < b.) 2. Prove that a ≤ a. 3. Prove that if a ≤ b and b ≤ c, then a ≤ c. 4. Do you think that a ≤ b and b ≤ a imply

a = b? Explain your reasoning. (Hint: This is not as trivial as it might look.)

Answer 1

1.

If for two aggregates M and N with the cardinal numbers a and b , with the condition:

There is a part N1 of N, such that N1 ∼ M,

is fulfilled, it is obvious that this condition still hold if in them M and N are replaced by two equivalent aggregates M₀ and N₀ . Thus it express a definite relation of the cardinal numbers a and b to one another.

2.

Since a = a, then a = a or a < a

so, $a\le a$

3.

if a ≤ b, then $b=a+k_1,k_1\ge 0$

if b ≤ c, then $c=b+k_2,k_2\ge 0$

so, $c=a+k_1+k_2\implies a\le c$

4

if a ≤ b, then $b=a+k_1,k_1\ge 0$

if b ≤ a, then $a=b+k_2,k_2\ge 0$

so, $b=b+k_1+k_2\implies k_1+k_2=0$

$\implies k_1=k_2=0\implies a=b$