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Suppose that $k$ is a field. Consider the vector space $k^{\times \mathbb{N}}:= \{(x_0, x_1, x_2,...) ∣ x_i \in k\}$ and its subspace $k^{\oplus N}:= \{(x_0, x_1, x_2, . . .) \in k^\mathbb{N}∣ x_i ≠ 0$ for finitely many $i\}$. Prove that $k^{\oplus \mathbb{N}}$ is not linear isomorphic to $k^{\times \mathbb{N}}$.

Attempt: The linear transformation from $k^{\otimes \mathbb{N}}$ to $k^{\times \mathbb{N}}$, for any fixed element in $k^{\otimes \mathbb{N}}$, is a finite-dimensional linear form. However the linear transformation from $k^{\times \mathbb{N}}$ to $k^{\otimes \mathbb{N}}$, for any fixed element in $k^{\times \mathbb{N}}$, is a infinite-dimensional linear form. Is there a theorem saying that such two spaces cannot be isomorphic?

Dasheng Wang
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Hint: $k^{\oplus\mathbb N}$ has a countable basis, which is not the case for $k^\mathbb N$.

Scientifica
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  • Why can't the bases of $k^\mathbb{N}$ be standard basis (1,0,0...), (0,1,0,...),..., which are bijective to natural numbers? – Dasheng Wang Apr 08 '20 at 12:18
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    @DashengWang Because the vector $(1,1,1,\ldots)$ is not in the span. – Servaes Apr 08 '20 at 12:24
  • @Servaes I meant $k^{\times \mathbb{N}}$. – Dasheng Wang Apr 08 '20 at 12:36
  • @DashengWang That doesn't that change the fact that $(1,1,1,\ldots)$ is not in the span. – Servaes Apr 08 '20 at 12:45
  • @Servaes Why not? Isn't $k^{\times \mathbb{N}}$ just a usual infinte-dimensional vector space? – Dasheng Wang Apr 08 '20 at 13:06
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    @DashengWang It is not a finite linear combination of these standard basis vectors. – Servaes Apr 08 '20 at 14:18
  • @DashengWang As Servaes pointed out, the span of a set of vectors is the set of finite linear combinations of the elements of that set. Every finite linear combination of elements of the family you gave (1,0,0,..), (0,1,0,...) is a sequence having all its terms 0 starting from a big enough term. – Scientifica Apr 08 '20 at 14:40
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    This being said, showing that $k^{\times \mathbb{N}}$ has no countable basis is not that trivial from a linear algebra point of view. The easiest solution is probably to look directly at the cardinals. – Captain Lama Apr 09 '20 at 11:55
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    May be this link is helpful – Peter Melech Apr 09 '20 at 12:28
  • @CaptainLama You're right. I should've pointed this out. – Scientifica Apr 09 '20 at 12:35
  • @PeterMelech Indeed the construction in the link does the job. – Scientifica Apr 09 '20 at 12:39
  • Thank you for letting me understand the hint. But is there a theorem saying a vector space with uncountable bases cannot be linear isomorphic to a vector space with countable bases? – Dasheng Wang Apr 10 '20 at 09:06
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    @DashengWang You're welcome. Yes. In fact two isomorphic vector spaces have bases of same cardinality (even if the bases are infinite), and this cardinality is the dimension of the vector space. To make things simple in this specific case, yes we can show that a vector space $V$ with no countable basis can't be isomorphic to a vector space $W$ with countable basis. Indeed, assume $f:W\to V$ is an isomorphism and $(w_n){n\in\mathbb N}$ a countable basis of $W$. You can then show that $(f(w_n)){n\in\mathbb N}$ is a countable basis of $V$. – Scientifica Apr 10 '20 at 09:19