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How do you show that the set of infinite sequences with entries in $\Bbb R$ is uncountably infinite-dimensional as a vector space over $\Bbb R$.

2 Answers 2

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Step 1: There is a family $\{A_r\mid r\in\Bbb R\}$ such that $A_r\subseteq\Bbb N$, and for $r\neq r'$ the intersection $A_r\cap A_{r'}$ is finite.

Step 2: Consider the characteristic functions of $A_r$ from the family of Step 1. Can they be linearly dependent?

(Another option is to use a diagonal argument to show that given any countable family of vectors, there is one not spanned by them. The above method has the benefit of showing that the dimension is as large as it can be, not just uncountable.)

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    I don't quite understand the family set. – 2017-02-06
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    What don't you understand about it? – 2017-02-06
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    How you know that there's an existence of such a family – 2017-02-06
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    Me? I proved that. Now you have to prove that also. You can find the answer on this site if you run into too much difficulties. Try thinking about branches in a binary tree, or sequences of rationals. – 2017-02-06
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    To the OP: Let $f:\mathbb Q\to \mathbb N$ be a bijection. For each $r\in \mathbb R$ let $(q_n(r))_n$ be a sequence of rationals converging to $r.$ Let $A_r=\{f(q_n(r))\}_n\;.$.......For distinct reals $r,r'$ the set $B(r,r')=\{q_n(r)\}_n\cap \{q_n(r')\}_n$ is finite .[.... Because, with $d =|r-r'|/2,$ if $n>n_0$ implies $|q_n(r)-r|n_0$ implies $ n\not \in B(r,r')$.....]. So $A_r\cap A_{r'}=\{f(q_n(r))\}_n\cap \{f(q_n(r'))\}_n$ is finite. – 2017-02-06
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Hint: For each $n \in \mathbb N$,can you see that $ \mathbb R ^n$ sits in your vector space?

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    Yes I can, but I don't see how it links to the question – 2017-02-06
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    So the dimension can't be finite. – 2017-02-06
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    So how do you prove that it's uncountably infinite? It could be countably infinite as well. – 2017-02-06
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    Use Baire´s category theorem – 2017-02-06
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    http://math.stackexchange.com/questions/1409963/a-banach-space-cannot-have-a-denumerable-basiswhy-is-it-true/1411478#1411478 – 2017-02-06
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    I have no idea what's that. Can we stick to linear algebra and basic set theory exclusively? – 2017-02-06
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    No, I don´t think so. You really need the Baire´s theorem,it´s a basic principle of functional analysis – 2017-02-06
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    @Peter: Why is $\Bbb{R^N}$ a Banach space? (I mean, it can be, but why is it naturally a Banach space? What is the definition of your norm?) – 2017-02-06
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    @Asaf Karagila It isn´t naturally but contains one,that by Baire can´t have a countable Hamel basis e.g.$\ell^{\infty}$, thus $\mathbb{R}^{\mathbb{N}}$ can´t have a countabel Hamel basis as it seems You found a way without Baire – 2017-02-06
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    @Peter: That's right, you can apply Baire's theorem on subspaces of $\Bbb{R^N}$. But nevertheless, this is not "the natural approach". See my answer. – 2017-02-06