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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.

Satana
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    The space $\mathbb{R}^{\infty}$ is the dual of $\mathbb{R}0^{\infty}$: the latter has a basis consisting of functions $\mathbf{f}_i$ with $\mathbf{f}_i(j) = \delta{ij}$. For infinite dimensional spaces, the dual is never isomorphic to the space – Arturo Magidin Apr 02 '19 at 21:15
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    Cardinality by itself won't do it; $|\mathbb{R}^{\infty}| = |\mathbb{R}|^{|\mathbb{N}|} = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\aleph_0} = 2^{\aleph_0}$; and $2^{\aleph_0}\leq |\mathbb{R}|\leq|\mathbb{R}_0^{\infty}|\leq |\mathbb{R}^{\infty}|= 2^{\aleph_0}$. So they have the same cardinality. – Arturo Magidin Apr 02 '19 at 21:16

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$\mathbb R_0^{\infty}$ 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.

Kavi Rama Murthy
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