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My question pertains to this answer Here.

Here it states that the "span of countably many real number over $\mathbb{Q}$ is countable" .

This is what I am having trouble proving. Can anyone provide me with an easy proof or some hint as to how I should prove this?.

Edit:-

If the basis is $\{a_{1},a_{2},....\}$

The map I have defined is $f:B_{m}\rightarrow \mathbb{Q}^{m}$ $$\sum_{i\in I} a_{i}q_{i}\rightarrow (q_{1},q_{2},...,q_{m})$$ where $I$ is any subset of $\mathbb{N}$ whose maximum element is $m$. Here $B_{m}=span\{a_{1},a_{2},...,a_{m}\}$.

$\bigcup_{m=1}^{\infty}B_{m}=\mathbb{R}$ .

PS. I am tagging this also to Real analysis as it involves a question about proof of uncountability.

Mr. Gandalf Sauron
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1 Answers1

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I think you can just argue as follows. Let $A=\{a_1,a_2,\ldots\}$ be your countably many real numbers, and set $A_n:=\{a_1,\ldots,a_n\}$. Certainly $\operatorname{Span}_{\mathbb Q}(A_n)$ is countable for each $n$ since it injects into $\mathbb Q^n$. Hence $$\operatorname{Span}_{\mathbb Q}(A)=\bigcup_n \operatorname{Span}_{\mathbb Q}(A_n)$$ is countable.

Dave
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  • Yes. I am trying to work out a proof. I have taken exactly that. The sum $$\sum_{i\in I} a_{i}q_{i}\rightarrow (q_{1},q_{2},....q_{n})$$. Is a bijection to a subset of $\mathbb{Q}^{m}$. Where I is any subset of $\mathbb{N}$ whose maximum element is $m$. Is that correct? – Mr. Gandalf Sauron Aug 27 '21 at 19:58
  • Yes, $\operatorname{Span}_{\mathbb Q}(A_n)$ bijects onto a subset of $\mathbb Q^n$. (I assume $n=m$ here) – Dave Aug 27 '21 at 20:00
  • Is my map axiomatically correct? – Mr. Gandalf Sauron Aug 27 '21 at 20:01
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    You need your map to well-defined, which it may not be if there is linear dependence amongst the $a_i$. You could instead take a basis for $\operatorname{Span}_{\mathbb Q}(A_n)$ and define your map on that basis (they key is that the basis has size $\leq n$). – Dave Aug 27 '21 at 20:11
  • Can you provide me with a correct mapping. The mapping is the exact thing I am having trouble to prove . I have edited in my effort in the question . Can you verify it please? – Mr. Gandalf Sauron Aug 27 '21 at 20:13
  • For a fixed $n$: you can fix a basis ${b_1,\ldots,b_m}$ of $\operatorname{Span}{\mathbb Q}(A_n)$; note that $m\leq n$ here. Then the map $\operatorname{Span}{\mathbb Q}(A_n)\to \mathbb Q^m$ is $$\sum_{i=1}^m q_ib_i\mapsto (q_1,\ldots,q_m)$$ where the $q_i$ are in $\mathbb Q$. This is good enough, but you can further inject $\mathbb Q^m$ into $\mathbb Q^n$ if you want. – Dave Aug 27 '21 at 20:16
  • Ah I get it . For the index set $I$ we can have different element in the range mapped to the same element in the domain. Thanks !!!! – Mr. Gandalf Sauron Aug 27 '21 at 20:17