# Cardinality of the set of all complex sequences converging to zero.

I was asked to show that the set of all complex sequences converging to zero has the same cardinality as the set [0,1].

This is the only hole in a proof that I am working on. I need to show there exists a bijection between these two sets. If I can show these two sets have the same cardinality then the bijection exists and the proof I am currently working on is finished.

I have very little experience with finding Cardinality of uncountable sets. I have no idea where to even start. Any help is appreciated.

• Your hole is... the whole (of the problem). – JP McCarthy Feb 12 '15 at 21:02
• Possibly helpful: math.stackexchange.com/questions/218959/… – Joe Johnson 126 Feb 12 '15 at 21:07
• Well, it was a sub portion of a problem I was working on. As stated yes, it is the whole problem sorry for the confusion. @ Jp MaCarthy – Ben Feb 12 '15 at 21:13

$|\mathbb{C}| = |\mathbb{R} \times \mathbb{R}| = |\mathbb{R}| = 2^{\aleph_0}$. So the set of all complex sequences equals $|\mathbb{C}^\mathbb{N}| = |2^{\aleph_0}|^{\aleph_0} = 2^{\aleph_0 \times \aleph_0} = 2^{\aleph_0}$ as well. So the sets have equal size.
First you need to show $|\mathbb{C}| = |[0,1]|$. For this you can use the $\tan$ and $\arctan$ functions to show that $\mathbb{R}$ and a bounded interval have the same cardinality. Then to show $|\mathbb{C}| = |[0,1]|$ you can take the even indexed digits and the odd indexed digits of an arbitrary real number in $[0,1]$ to get the 2-dimensional coordinates of an arbitrary pair of numbers in $[0,1]^2$. Then to show that the set of all complex valued sequences has the same cardinality as $[0,1]$, take each number in $[0,1]$ and map it to the countably infinite sequence of numbers in $[0,1]$ where the $k$th number in the sequence is defined by the sequence of digits in the original number that are indexed by the powers of the $k$th prime number.