Re: Set thoery with denumerable.


"mina_world" <mina_world@xxxxxxxxxxx> wrote in message
news:fbgbio$jmn$1@xxxxxxxxxxxxxxxxxxx
Hello sir~

x_11 x_12 x_13 x_14 .......
x_21 x_22 x_23 x_24 .......
x_31 x_32 x_33 x_34 .......
x_41 x_42 x_43 x_44 .......
.............
.............

and then
x_11, x_12, x_21, x_13, x_22, x_31, x_14, x_23, x_32, x_41, .....
with diagonal order.

How do you show that this diagonal arrangement is denumerable
without the use of NxN~N ?

In fact, this is a part of proof .....
Let {A_n} be a family of denumerable sets.(A_n /\ A_m = empty for n=/=m)
Then U_{n=1 to 00} A_n is denumerable.

Arrange with x_11, x_12, x_21, x_13, x_22, x_31, x_14, x_23, x_32, x_41,
......

Let x_1, x_2, x_3, ..... in order.

Define f:N -> A, f(n) = x_n.

so f is bijection.

Thus, U_{n=1 to 00} A_n is denumerable.


This is a proof of my book.
Maybe, Author did prove this roughly.
This is a cause of my question.


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