# Cardinality of the continuum

• Sep 21st 2010, 03:42 PM
anlys
Cardinality of the continuum
Hello,
I need help with the following problem:

Prove that Card(l^p) = c (cardinality of the continuum), p >= 1.

I am not quite sure how to start the proof. I'd appreciate anyone's help. Thank you.
• Sep 21st 2010, 04:28 PM
DontKnoMaff
unable to understand
I would like to help you but I am unsure what your notation means. You are trying to prove that the cardinality of some set I to the pth power is equal to c? I am sorry if this is a dumb question, this is my first post onto this website.
• Sep 21st 2010, 04:40 PM
anlys
Hi, that's okay. Actually, that is not an I, it's a lower case for L. The notation l^p refers to the metric space with || x ||_p = (summation |x_i|^p from i = 1 to infinity)^(1/p). That's all I was given in the problem.
• Sep 22nd 2010, 12:10 AM
Tikoloshe
I think the additional structure on $\ell^p(\mathbb{N})$ is a red herring. All you need is that $\ell^p\subset\mathbb{R}^\mathbb{N}$, and that $\lvert\mathbb{R}\rvert=\lvert\{0,1\}^\mathbb{N}\rv ert$.

Then, $\lvert\mathbb{R}^\mathbb{N}\rvert=\lvert(\{0,1\}^\ mathbb{N})^\mathbb{N}\rvert=\lvert\{0,1\}^{\mathbb {N}\times\mathbb{N}}\rvert=\lvert\{0,1\}^\mathbb{N }\rvert=\lvert\mathbb{R}\rvert$.