Let $\mathbb{R}^{\infty}$ be the vector space of all real-valued sequences (over the field of real numbers), i.e., $$\mathbb{R}^{\infty}:= \{f: \mathbb{N}\rightarrow\mathbb{R}\}.$$
On the other hand, denote by $\mathbb{R}_0^{\infty}\subseteq\mathbb{R}^{\infty}$ the subspace consisting of all real-valued sequences with only finitely many components non-zero, i.e., $$\mathbb{R}_0^{\infty}:= \left\{f\in \mathbb{R}^{\infty}\,\vert\, f(n)\ne 0 \text{ for finitely many $n$}\right\}.$$
(Here, addition and scalar multiplication for these vector spaces is defined in the usual way.) Obviously both spaces are uncountable and $\mathbb{R}_0^{\infty}$ is a proper subspace of $\mathbb{R}^{\infty}$.
I am trying to prove that $\mathbb{R}_0^{\infty}$ and $\mathbb{R}^{\infty}$ are not isomorphic as vector spaces. I am not sure if going down a route of computing cardinalities is the way to go, or another way.
Best Answer
$\mathbb R_0^{\infty}$ has a countable (Hamel) basis but $\mathbb R^{\infty}$ doesn't. If $\mathbb R^{\infty}$ has a countable basis then this space would be a countable union of finite dimensional spaces. By introducing the topology defined by the metric $d(x,y)=\sum \frac 1 {2^{n}} \frac {|x_n-y_n|} {1+|x_n-y_n|}$ and applying Baire Category Theorem we get a contradiction since no proper subspace of $\mathbb R^{\infty}$ has an interior point.