Countable Cartesian Product

Let be countable sets, and let their Cartesian product x x x ..... be defined to be the set of all sequences where , is an element of . Prove that the Cartesian product is uncountable. Show that the same conclusion holds if each of the sets has at least two elements.

Re: Countable Cartesian Product

Hi, kalyanram.

Have you ever seen the proof that (0,1) (the open interval in R) is uncountable? I think you could do something like that here and get the result you want.

Re: Countable Cartesian Product

Quote:

Originally Posted by

**kalyanram** Let

be countable sets, and let their Cartesian product

x

x

x ..... be defined to be the set of all sequences

where

, is an element of

. Prove that the Cartesian product is uncountable. Show that the same conclusion holds if each of the sets

has at least two elements.

Consider modifying the diagonal argument.

This condition "each of the sets has at least two elements" makes that possible

Re: Countable Cartesian Product

Ok here goes my argument.

Consider that every has at least two elements and consider that the = x x x....... is countable and has elements say .....

now consider the element in the i-th co-ordinate. By the construction of from the very construction of . Hence is an element of the Cartesian product that a contradiction.

Thanks for the help.

~Kalyan.

Re: Countable Cartesian Product

Quote:

Originally Posted by

**kalyanram** Ok here goes my argument.

Consider that every

has at least two elements and consider that the

=

x

x

x....... is countable and has elements say

.....

now consider the element

in the i-th co-ordinate. By the construction of

from the very construction of

. Hence

is an element of the Cartesian product that

a contradiction.

The idea is correct. But I don't follow your reasons.

For each , where .

Define . How do we know that exists?

Is it possible that for some

Re: Countable Cartesian Product