Hi i cans show that the below problem is an equivalence class no problem, but i am finding it difficult to describe its equivalence classes.

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- Jan 20th 2010, 09:25 AM1234567equivalence classes
Hi i cans show that the below problem is an equivalence class no problem, but i am finding it difficult to describe its equivalence classes.

Attachment 14917 - Jan 20th 2010, 09:30 AMPlato
The post says "whenever $\displaystyle |A|-|B|$." What does that mean?

That is not a relation. Please reply with a complete question. - Jan 20th 2010, 09:33 AMJhevon
- Jan 20th 2010, 09:41 AMPlato
- Jan 20th 2010, 09:44 AMJhevon
- Jan 20th 2010, 09:47 AMPlato
- Jan 20th 2010, 10:08 AMJhevon
Maybe. Or at least insist that questions are not posted in image files, unless there are accompanying diagrams or something.

I don't really see a better way to describe the classes other than to reuse the language of the problem. Something like,

For $\displaystyle A \in \mathcal P (\mathbb N),~ [A] = \{ B \in \mathcal P (\mathbb N) ~:~ |A| = |B| \}$ - Jan 20th 2010, 12:44 PMDrexel28
- Jan 20th 2010, 12:53 PMJhevon
- Jan 20th 2010, 12:55 PMDrexel28
- Jan 20th 2010, 11:48 PMShanks
A~B iff A and B have the same cardinal.

therefore, there are countable many equivalent classes:

1-class: the collection of all sets that contains only one element.

2-class: the collection of all sets that contains two elements.

...

n-class: the collection of all sets that contains n elements.

...

infinite-class:the collection of all sets that contains countable infinitely many elements.

A further question: What is P(N)/~? I leave it for you to solve it. Solve it, and you will understand the cardinality better.