Cardinality of the set of all finite subsets of [0,1]

[Damidami]

Hello, I was wondering this, what is the cardinality of the set of all finite subsets of the real interval [0,1]

It somehow confuses me because the interval is nonnumerable (cardinality of the continuos \mathfrak{c}), while the subsets are less than numerable (finite). It is clear that it has to be equal or greater than \mathfrak{c} because one can consider subsets of only one element and there you got one set for each real number in the interval. It is equal to it, isn't?

Thanks.

 

[PatrickPowers]

Hello, I was wondering this, what is the cardinality of the set of all finite subsets of the real interval [0,1]

It somehow confuses me because the interval is nonnumerable (cardinality of the continuos \mathfrak{c}), while the subsets are less than numerable (finite). It is clear that it has to be equal or greater than \mathfrak{c} because one can consider subsets of only one element and there you got one set for each real number in the interval. It is equal to it, isn't?

Thanks.

You are correct that the cardinality is at least as much as the cardinality of [0.1]. The power set of an infinite set always of greater cardinality than the base set. I don't know about the finite subsets.

 

[micromass]

Hello, I was wondering this, what is the cardinality of the set of all finite subsets of the real interval [0,1]

It somehow confuses me because the interval is nonnumerable (cardinality of the continuos \mathfrak{c}), while the subsets are less than numerable (finite). It is clear that it has to be equal or greater than \mathfrak{c} because one can consider subsets of only one element and there you got one set for each real number in the interval. It is equal to it, isn't?

Thanks.

Yes. Equality holds. To see this, try to figure out the cardinality of

\mathcal{A}_n=\{A\subseteq [0,1]~\vert~|A|=n\}

Then the set you are looking for is

\bigcup_n \mathcal{A}_n

a countable union.