Is there a norm on ${\Bbb R}^{\Bbb N}$ Let $E={\Bbb R}^{\Bbb N}$ be the real vector space of real sequences. 
1) Is there a norm on $E$?
2) Is there a norm $N$ on $E$ such that the restriction of $N$ to $\ell^2$ is finer than the standard norm $N_2$ of $\ell^2$ ?
 A: Pick a basis, and define the norm of a vector to be the sum of the absolute values of the coefficients of that vector in the basis.
For your extra requirement: Find a subspace $U$ of $\mathbb R^\mathbb N$ such that $U\oplus\ell^2=\mathbb R^\mathbb N$. Put on $\mathbb R^\mathbb N$ the norm which is the sup of the norm of $\ell^2$ and any norm (constructed as in my first paragraph) on $U$. This restricts to the norm of $\ell^2$, of course.
A: Assuming the axiom of choice, yes.
Using the axiom of choice/Zorn's lemma/etc. you can choose $B$ to be a basis for the vector space. Now we define for $v$, if $v=\sum_{b\in B}\alpha_bb$, then $$\|v\|=\sum_{b\in B}|\alpha_b|.$$
It is not hard to check that this is indeed a norm.
If we do not assume the axiom of choice, then this becomes an open problem, and there is a long standing thread about this on this very site as it is.
A: There is no norm on the entire $E=\Bbb R^{\Bbb N}$ (with the property that $||x||=0 \implies x=0$).
Suppose such norm exists. Let's denote the elements of $E$ as $a=(a_1,...,a_n,...)$ and suppose $||a_i||=\alpha_i>0$. Then form a vector $b=(b_1,...,b_n,...)$ such that $||b||=\infty$. 
Here's how you do that: start with $b_1=2^1\frac{a_1}{||a_1||}$, then choose between $b_2=\pm 2^2\frac{a_2}{||a_2||}$, depending for which $b_2$ the norm is greater, etc. This would make the norm of $b$ unbounded.
