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Let us consider space $R = \mathbb{R}^{\mathbb{N}}$ of infinite sequences of real numbers. I have to prove two claims:

  1. Let $a_i=(0,0,\ldots,1,0,0,\ldots)$; $1$ at i-th place. Then $B=\cup_{i=1}^{\infty}a_i$ is not a basis of $R$.
  2. It can proved with the Zorn lemma that there is basis in $R$. Prove that there is no countable basis in $R$.

My attempt:

  1. Let us consider sequence $A=(1, 1, \ldots)$ (all $1$). Since linear combination is by definition finite then it is clear that we can not express $A$ as finite linear combination of sequences from $B$, because any linear combination of sequences from $B$ would have only finite number of nonzero elements.

  2. Well, honestly I do not know to solve this. I think we should do something like this - assume that there is countable basis $B'$. Then we should construct a sequence that does not belong to $\operatorname{span}(B')$ explicitly or use some set-theoretic theorems about cardinalities that $\operatorname{span}(B')$ has smaller size compared to $R$. However while I am pretty sure about the approach we should take, I do not see how to execute it.

So my questions are:

  1. Is $1$ is correct/make sense?
  2. How to approach $2$?

Thanks a lot for your time!

William Elliot
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Hedgehog
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  • Spanning seems the more obvious counter example. Take a look at the proof that the reals are uncountable. I believe the trick you need is in there. – Chris C Oct 31 '17 at 04:42
  • For number 2, do you know the Baire Category Theorem? – Mark Schultz-Wu Oct 31 '17 at 04:46
  • @Mark I will learn it now and see what I can do. – Hedgehog Oct 31 '17 at 04:58
  • By basis, do you mean a vector space basis or a topological basis? – William Elliot Oct 31 '17 at 06:47
  • @WilliamElliot Vector space basis. – Hedgehog Oct 31 '17 at 06:53
  • I don't know what $\bigcup_{i=1}^\infty a_i$ means, seeing as the $a_i$ are not sets, they are sequences. – bof Oct 31 '17 at 08:54
  • For number 2) I think you could argue as follows: if there were a countable basis, then there would also be a countable dense set, so it suffices to show that there can't be one. To show that there can't be a countable dense set, given the particular nature of $\mathbb{R}^\mathbb{N}$, you can use a diagonal Cantor-style argument. (I'm not sure about this, I didn't check the details. But if you don't want to use Baire's theorem could be an alternative approach) – Lucio Oct 31 '17 at 10:20

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