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It is well-known that if $\mathscr X$ is a finite-dimensional vector space, then any collection of vectors containing more elements than $\dim\mathscr X$ must be linearly dependent.

I am wondering if this result can be generalized to infinite-dimensional vector spaces. Specifically, suppose that $\mathscr X$ is an infinite-dimensional vector space (over $\mathbb R$) and let $\mathscr B\subseteq \mathscr X$ be a Hamel basis. Suppose that $\mathscr Y\subseteq\mathscr X$ is such a collection of vectors that $\#\mathscr Y>\#\mathscr B$ (where $\#$ denotes cardinality). Is it necessarily the case that the vectors in $\mathscr Y$ are linearly dependent? That is, do there exist $y_1,\ldots,y_n\in\mathscr Y$ and $\alpha_1,\ldots,\alpha_n\in\mathbb R$ for some $n\in\mathbb N$ such that $\sum_{m=1}^n\alpha_my_m=0$, yet not all of $(\alpha_m)_{m=1}^n$ vanish?

UPDATE #1: I have looked into this thread. According to a reference therein (Löwig, 1934), if $\#\mathscr X>\#\mathbb R$, then $\#\mathscr B=\#\mathscr X$, so no subcollection of vectors can have strictly greater cardinality than $\mathscr B$. My question is then relevant only for vector spaces whose cardinalities are between those of $\mathbb N$ and $\mathbb R$.

UPDATE #2: Having a closer look at the thread mentioned, if $\#\mathscr X\geq\#\mathbb N$ and $\mathscr X$ is a Banach space, then $\#\mathscr X=\#\mathscr B$, which renders the question trivial in this case, too. That leaves non-Banach spaces.

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triple_sec
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    Continuum Hypothesis aside, it is the case that any two bases of a vector space over a field have the same cardinality. – Nishant Nov 15 '14 at 03:48
  • @Nishant I see. That answers the question in the affirmative if $\mathscr Y$ spans $\mathscr X$. But what if $\mathscr Y$ does not span $\mathscr X$, despite the fact that it has greater cardinality than any basis? Is that possible? – triple_sec Nov 15 '14 at 03:50
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    Then unioning that with the basis gives a spanning set of the same cardinality (since cardinal addition is the same as the maximum). – Nishant Nov 15 '14 at 03:51
  • @Nishant Nice. Can you put that as an answer? – triple_sec Nov 15 '14 at 03:53
  • Wait a second... If $\mathscr Y$ and $\mathscr B$ are both linearly independent, does it follow that so is $\mathscr Y\cup\mathscr B$? – triple_sec Nov 15 '14 at 03:55
  • no, the union of independent sets need not be independent. – Ittay Weiss Nov 15 '14 at 03:58
  • No, but their union is a spanning set. – Nishant Nov 15 '14 at 04:34
  • @Nishant That's right, but then further work is needed to get the desired contradiction. $\mathscr Y\cup\mathscr B$ is a spanning set having greater cardinality than $\mathscr B$. However, if $\mathscr Y\cup\mathscr B$ is not independent then it is not a basis, so $#(\mathscr Y\cup\mathscr B)>#\mathscr B$ is not necessarily a contradiction to the fact that any two bases must have the same cardinality. – triple_sec Nov 15 '14 at 04:39
  • Yeah, you're right. I should've just stated that every independent set is included in a basis (by the same argument that leads to every vector space has a basis). – Nishant Nov 15 '14 at 05:06

2 Answers2

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Assuming the axiom of choice: 1) any two bases for a vector space have equal cardinality. 2) every set of independent vectors can be extended to a basis.

Thus if $B$ is a basis and $Y$ is an independent set of larger cardinality, then first extend $Y$ to a basis $Y'$, but then $Y'$ and $B$ must have the same cardinality, while clearly $|B|<|Y'|$.

So, yes, no set of larger cardinality than a basis can be linearly independent.

Ittay Weiss
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The proof below applies to spaces of cardinality $\# \mathbb{N}$, but generalizes easily to any space (replace $\mathbb{N}$ with whatever set of your choosing and countable with the cardinality of that set - it even works when we use $\mathbb{R}$, but it is, of course, vacuous since no larger subsets exist). In particular, all of these vector spaces are isomorphic to the space $V$ of real sequences with only finitely many non-zero terms. For any finite set $S\subset \mathbb{N}$, denote the subspace $$B(S)=\{v\in V:i\in \mathbb{N}\backslash S\rightarrow v_i=0 \}$$ of vectors $v$ which vanish on every coordinate not in $S$. Notice that $B(S)$ has dimension $|S|$. Thus, for any linearly independent set of vectors $L$, it holds that $|L\cap B(S)|\leq|S|$. However, it also holds, for any $l\in $L, that there exists a finite $S$ such that $l\in B(S)$. Therefore, we have that, where the union is taken over all finite subsets $S$ of $\mathbb{N}$: $$L=\bigcup_{S\subseteq \mathbb{N}}L\cap B(S).$$ However, there are only countably many finite subsets of $\mathbb{N}$ and each term $L\cap B(S)$ is finite - but a countable union of finite sets is countable, implying that any linearly independent set is countable as well.

Milo Brandt
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