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Given vector space $V$ over $\mathbb R$ such that the elements of $V$ are infinite-tuples. How to show that any basis of it is uncountable?

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How do you define "infinite $n$-tuple?" What is $n$? Do you just mean elements which are infinite sequences of real numbers? –  Thomas Andrews Oct 16 '12 at 17:25
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Then you don't mean $n$-tuple, which means a sequence $n$ real numbers. –  Thomas Andrews Oct 16 '12 at 17:29
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My bad. Let me clarify it then: V contains all possible infinite tuples. –  p.s Oct 16 '12 at 17:40
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@Pilot As several people have mentioned, it's probably best to not use the term "tuple". I would just stick with "infinite sequence of real numbers". –  EuYu Oct 16 '12 at 17:43
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I found an older question, which is a generalization of this one: Vector dimension of a set of functions –  Martin Sleziak Oct 16 '12 at 20:26

2 Answers 2

Take any almost disjoint family $\mathcal A$ of infinite subsets of $\mathbb N$ with cardinality $2^{\aleph_0}$. Construction of such set is given here.

I.e. for any two set $A,B\in\mathcal A$ the intersection $A\cap B$ is finite.

Notice that $$\{\chi_A; A\in\mathcal A\}$$ is a subset of $\mathbb R^{\mathbb N}$ which has cardinality $2^{\aleph_0}$.

We will show that this set is linearly independent. This implies that the base must have cardinality at least $2^{\aleph_0}$. (Since every independent set is contained in a basis - this can be shown using Zorn lemma. You can find the proof in many places, for example these notes on applications of Zorn lemma by Keith Conrad.)

Suppose that, on the contrary, $$\chi_A=\sum_{i\in F} c_i\chi_{A_i}$$ for some finite set $F$ and $A,A_i\in\mathcal A$ (where $A_i\ne A$ for $i\in F$). The set $P=A\setminus \bigcup\limits_{i\in F}(A\cap A_i)$ is infinite. For any $n\in P$ we have $\chi_A(n)=1$ and $\sum\limits_{i\in F} c_i\chi_{A_i}(n)=0$. So the above equality cannot hold.


You can find a proof about dimension of the space $\mathbb R^{\mathbb R}$ (together with some basic facts about Hamel bases) here: Does $\mathbb{R}^\mathbb{R}$ have a basis?

In fact, it can be shown that already smaller spaces must have dimension $2^{\aleph_0}$, see Cardinality of a Hamel basis.

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Thank you Martin.It is such a beautiful proof! –  p.s Oct 16 '12 at 18:23
    
The idea of using AD families comes from Lacey's paper mentioned in the related thread. (Of course, he shows a different result there, but AD-families are one of the ingredients of his proof.) –  Martin Sleziak Oct 16 '12 at 18:26

I like the following solution (a colleague of mine told me about it):

For any $a\in\mathbb R$ we have a sequence $\widehat a=(1,a,a^2,\dots,a^k,\dots)$.

The set $\{\widehat a; a\in\mathbb R\}$ is linearly independent. (To see this just notice that if you choose $n$ sequences from this set then the first $n$ coordinates of these sequences form Vandermonde matrix.)

Thus we have a linearly independent of cardinality $\mathfrak c$. Hence cardinality of any Hamel basis is at least $\mathfrak c$.

At the same time, the cardinality of the whole space is $|\mathbb R^{\mathbb N}|=\mathfrak c^{\aleph_0}=\mathfrak c$, so the basis cannot have more than $\mathfrak c$ elements. Thus the Hamel dimension of this space is $\mathfrak c$.

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