Why power set of integer is uncountable?

Asked Jun 20 '20 at 05:06

Active Jun 20 '20 at 05:09

Viewed 191 times

Z+ = {1,2,3,4,...}

if we constuct the P(Z+) like that,

0={}

1={1}

2={2},{1,1}

3={3},{1,2}

4={4},{1,3}

5={5} {1,4} {2,3}

6={6} {1,5} {2,4} {1,2,3}

7={7} {1,6} {2,5} {3,4} {1,2,4}

8={8} {1,7} {2,6} {3,5} {1,2,5} {1,3,4}

...... Then we use hilbert hotel to move the "guest" to ensure they are one-to-one response.

edited Jun 20 '20 at 05:09

asked Jun 20 '20 at 05:06

toby

3

For example the subset ${1,3,5,7,9,\ldots}$ is still without a room. The set of finite subsets of $\Bbb{Z}_+$ on the other hand is countable. – Jyrki Lahtonen Jun 20 '20 at 05:09
1

Or this. LInks to many other related threads there. – Jyrki Lahtonen Jun 20 '20 at 05:15
3

Does this answer your question? Is the power set of the natural numbers countable? – Kumar Jun 20 '20 at 05:18
Thanks ,It solve my confusion. – toby Jun 20 '20 at 05:23

0 Answers0