Let $L$ be the vector space without other structures (topology and so on) over some field. We consider the chains (i.e. linearly ordered sets) of vector subspaces in $L$ (such chain is called a flag). The question is the following: is it true, that the cardinal number of the maximal chain (that can be built with Zorn's lemma) is $\dim(L)$? The answer is obviously 'yes' if $L$ is finite dimensional. What about the infinite case?
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- $\begingroup$ I think that this question may have been more suited for math.stackexchange.com $\endgroup$Asaf Karagila– Asaf Karagila ♦2013-04-25 08:34:26 +00:00Commented Apr 25, 2013 at 8:34
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2 The way you ask your question is quite confusing: typically "chain" means a well ordered set. In this case the answer is ${\bf YES}$. More precisely - you don't have the "maximal chain" but still each chain has cardinality not exceeding $dim(V)$: having a chain $V_\alpha$ you obtain a linearly independent sequence $v_\alpha\in V_{\alpha+1}\setminus V_\alpha$.
However you wrote "linearly ordered sets". In this case the answer is ${\bf NO}$: Let $V$ be a verctor space with basis $\mathbb{Q}$ and for every $r\in\mathbb{R}$ take $V_r=\mathop{\rm span} \{q \mid q< r \}$.
- 2$\begingroup$ Typically a subset of a partially ordered set is a chain if any two elements are comparable (i.e. it is linearly ordered). Inside any chain you can always find a cofinal well-ordered chain, but that´s another story. $\endgroup$Ramiro de la Vega– Ramiro de la Vega2013-04-25 13:37:28 +00:00Commented Apr 25, 2013 at 13:37
- $\begingroup$ Yes Ramiro - chain is any linearly ordered subset. I was misled by a previous answer (already deleted) which assumed well ordering of the subspaces. $\endgroup$Adam Przeździecki– Adam Przeździecki2013-04-25 19:09:57 +00:00Commented Apr 25, 2013 at 19:09