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Let $V=\{(a_1,a_2,...,a_n,0,0,...)|n\in \mathbb{N}, a_i\in \mathbb{R}\}$ be the vector space of eventually zero sequences in $\mathbb{R}^\mathbb{N}$ with addition and scalar multiplication point-wise. I am trying to show that this is $not$ isomorphic to the vector space $\mathbb{R}^\mathbb{N}$. One approach I tried was to compare cardinalities. So I calculated

$$|V|=\aleph_1+\aleph_1^2+\aleph_1^3+...=\aleph_1+\aleph_1+\aleph_1+...=\aleph_1*\aleph_0=\aleph_1$$ and $$|\mathbb{R}^\mathbb{N}|=\aleph_1^{\aleph_0}$$

So I am wondering whether it is known whether or not $$\aleph_1^{\aleph_0}>\aleph_1$$

Asaf Karagila
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Elie Bergman
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  • As Asaf Karagila wrote, the notion of dimension (over $\mathbb{R}$) is crucial here. Two vector spaces over some field are isomorphic iff they have the same dimension over said field. The dimension of a vector space may not be easy to compute however. – nombre Mar 03 '17 at 18:36

1 Answers1

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There are two issues here:

  1. $\Bbb R$ is a set of cardinality $2^{\aleph_0}$, which may or may not be equal to $\aleph_1$.

  2. $(2^{\aleph_0})^{\aleph_0}=\aleph_1^{\aleph_0}=\aleph_0^{\aleph_0}=2^{\aleph_0}$.

The first one is in fact the continuum hypothesis and the fact it is neither provable nor disprovable from the axioms of $\sf ZFC$ requires a lot of technical work. The latter can be proved using simple cardinal arithmetic and the Cantor–Bernstein theorem.

Comparing cardinalities won't do you much good here. But you can prove that exactly one of these spaces has a countable dimension. (See this for more details.)

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Asaf Karagila
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