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Please note that this is a homework question, I've been thinking about this for a couple days and I haven't had much luck! Thought I'd get some other input.

Let $V$ be the vector space of all infinite sequences with entries in an arbitrary field $\mathbb{F}$. (The goal is to show it doesn't have a countable basis)

First show that if $V$ admits a countable basis then there is a nested sequence of subspaces $W_{1} \subset W_{2} \subset ...$ such that $dim(W_{j}) = j$ and $V = \cup W_{j}$.

This proof was fairly obvious because you are given the fact that there is a countable basis so you can just take spans. However, then ext part says

We can represent an infinite sequence as a concatenation of sequences $(s_{2}, ... )$ where $s_{j}$ is a sequence of length $j$. For each $j \geq 2$ show that there exists a sequence $S_{j}$ with length $j$ and the property that

$s_{j} = S_{j} \rightarrow (s_{2}, ... S_{j}, .. ) \not\in W_{j-1}$

For me what this is trying to get at that given any element in your vector space you can find a sequence of length $j$ such that if you shove it in the $j$th spot of the sequence representation of the element in the vector space than it isn't in $W_{j-1}$.

My first idea was thinking about picking our "countable basis"(which we know doesn't exist) to be the element whose ith entry is 1 and the rest is 0. Then we could just pick up a diagonal argument to ensure that it wouldn't be in $W_{j-1}$ but I can't quite wrap my head about how to apply the same idea to an arbitrary countable basis where we don't know what the elements look like!

Lastly it is to show that there's an element which doesn't belong to any of the $W_{j}$ which follows from the above just by picking all those finite length sequences and putting them together.

So what gives, I feel as though I'm getting too focused around the idea that I know this not to be the case that I haven't been able to quite get it.

Thanks!

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  • $\begingroup$ Can you clarify what is meant by $s_{j} = S_{j} \rightarrow (s_{2}, ... S_{j}, .. )$? $\endgroup$ – wckronholm Mar 20 '14 at 1:11
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Let $E^i = (E^i_j)$ be a countable subset of $V$. We define an element $a_j$ such that $a_j$ is not a finite linear combination of the $E^i$.

Let $a_1=E^1_1$. Let $a_2\neq E^1_2$. Then any element $(a_1,a_2,\dots)$ is not in the span of $E^1$.

Suppose we have found $a_1,\dots a_m$ such that no element $(a_1,\dots,a_m,\dots)$ is in the span of $E^1,\dots,E^{n-1}$. We define the next $n+1$ terms. Consider $(E^i_{m+1},\dots,E^i_{m+n+1}) \subset F^{n+1}$ for $i=1,\dots,n$. These elements span at most an $n$ dimensional subspace of $F^{n+1}$. Hence we may choose $(b_1,\dots,b_{n+1})\in F^{n+1}$ which is not in the span. Let $a_{m+j}=b_{j}$ for $j=1,\dots,n+1$. Then any element $(a_1,\dots,a_{m+n+1},\dots)$ is not in the span of $E^1,\dots,E^n$.

By induction we have $a_j$ which is not in the span of any finite subset of $E^i$ and hence not in the span of $E^i$.

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  • $\begingroup$ That looks about like what I was trying to figure out in my head for figuring out how to show it. Thanks so much, no questions! $\endgroup$ – Nick R Mar 20 '14 at 11:11
  • $\begingroup$ You are welcome! $\endgroup$ – Seth Mar 20 '14 at 11:18
  • $\begingroup$ Actually why do we know that there is a unique linear combination of $E^{1} ... E^{n-1}$ such that the first $n-1$ terms are hit? $\endgroup$ – Nick R Mar 20 '14 at 11:38
  • $\begingroup$ You are right, that part was incorrect. I believe I have fixed the proof and also I made it direct. No more contradiction. Please let me know if you agree that it is correct. $\endgroup$ – Seth Mar 20 '14 at 19:15
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Given a field $\mathbb K$, we can show that any basis of $\mathbb K^\mathbb N$ has size at least $\mathfrak c=|\mathbb R|$. I'll prove this as a consequence of the following combinatorial statement:

Lemma. There is a collection of infinite subsets of $\mathbb Q$ of size $\mathfrak c$ with the property that any two of them have finite intersection.

Such sets are called almost disjoint. To see the lemma, note that to each real we can associate (the range of) a strictly increasing sequence of rationals converging at the real. Any two such sequences have only finitely many terms in common.

Given an almost disjoint family of subsets of $\mathbb Q$, we obtain an almost disjoint family $\mathcal F$ of subsets of $\mathbb N$ of the same size simply because there is a bijection between $\mathbb Q$ and $\mathbb N$. Now, given such a family $\mathcal F$, consider its characteristic functions as vectors in $\mathbb K^\mathbb N$. More precisely, if $A\subseteq \mathbb N$, we associate to $A$ the vector $v_A=(a_1,a_2,\dots)$ where each $a_i$ is $0$ or $1$, and it is $1$ iff $i\in A$.

We are done if we can prove that these vectors $v_A$, $A\in\mathcal F$, are linearly independent. To see this, consider any finitely many of them, say $v_{A_1},\dots,v_{A_n}$. Note that (since $A_1,\dots,A_n$ are almost disjoint) for each $j$ with $1\le j\le n$ there is an $\displaystyle i\in A_j\setminus\bigcup_{\substack{k=1\\ k\ne j}}^n A_k$ since $A_j$ is infinite, but each intersection $A_j\cap A_k$, $k\ne j$, is finite. This means that $v_{A_j}$ has a $1$ at the $i$th position, while the other $v_{A_k}$ have a $0$ there. But then, any combination of the $v_{A_k}$, $k\ne j$, also has a $0$ at the $i$th position, and cannot equal $v_{A_j}$.


By the way, note that the same argument shows that the $v_A$, $A\in\mathcal F$, in spite of how many there are, do not form a basis. For instance, none of the standard vectors $e_i$ is a combination of them.

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