How do we construct (in a vector space) a chain of countable dimensional subspaces that can only be bounded by an subspace of uncountable dimension? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:17:10Z http://mathoverflow.net/feeds/question/15297 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15297/how-do-we-construct-in-a-vector-space-a-chain-of-countable-dimensional-subspa How do we construct (in a vector space) a chain of countable dimensional subspaces that can only be bounded by an subspace of uncountable dimension? Z_3 2010-02-15T01:36:44Z 2010-02-15T02:21:58Z <p>In more rigorous language: " <strong>V</strong>: a vector space having an uncountable base <strong>S</strong>: The set of subspaces of <strong>V</strong> that have countable dimension. Can we construct explicitly a chain in the poset <strong>S</strong> (ordered by inclusion), such that this chain has NO upper bound in <strong>S</strong>? "</p> <p>Apparently, this chain must have uncountable terms. Also,because S doesn't satisfy Zorn's lemma, we know such chain must exist in <strong>S</strong>.</p> <p>But how do we construct it?</p> http://mathoverflow.net/questions/15297/how-do-we-construct-in-a-vector-space-a-chain-of-countable-dimensional-subspa/15299#15299 Answer by Qiaochu Yuan for How do we construct (in a vector space) a chain of countable dimensional subspaces that can only be bounded by an subspace of uncountable dimension? Qiaochu Yuan 2010-02-15T01:57:45Z 2010-02-15T01:57:45Z <p>Some subset of a basis of $V$ can be put into bijection with the first uncountable ordinal $\omega_1$. Consider the subspaces spanned by each initial segment of $\omega_1$, all of which have countable dimension.</p> http://mathoverflow.net/questions/15297/how-do-we-construct-in-a-vector-space-a-chain-of-countable-dimensional-subspa/15300#15300 Answer by James Pfeiffer for How do we construct (in a vector space) a chain of countable dimensional subspaces that can only be bounded by an subspace of uncountable dimension? James Pfeiffer 2010-02-15T02:01:10Z 2010-02-15T02:01:10Z <p>I'm not sure if this will satisfy your sense of "construct" but order your uncountable base in some well-ordering and let your chain be {first basis element}, {first two basis elements}, etc up through all countable ordinals. Then this will not have an upper bound.</p> http://mathoverflow.net/questions/15297/how-do-we-construct-in-a-vector-space-a-chain-of-countable-dimensional-subspa/15302#15302 Answer by Joel David Hamkins for How do we construct (in a vector space) a chain of countable dimensional subspaces that can only be bounded by an subspace of uncountable dimension? Joel David Hamkins 2010-02-15T02:21:58Z 2010-02-15T02:21:58Z <p>The other answers asked you first to well-order the whole vector space (or a basis for it), and those answers are perfectly correct, but perhaps you don't like well order the whole space. So let me describe a construction that appeals directly to the Axiom of Choice. </p> <p>Let V be your favorite vector space having uncountable dimension. For each countable dimension subpace W, let a<sub>W</sub> be an element of V that is not in W. Such a vector exists, since W is countable dimensional and V is not, and we choose such elements by the Axiom of Choice. </p> <p>Having made these choices, the rest of the construction is now completely determined. Namely, we construct a linearly ordered chain of countable dimensional spaces, whose union is uncountable dimension. Let V<sub>0</sub> be the trivial subspace. If V<sub>&alpha;</sub> is defined and countable dimensional, let V<sub>&alpha;+1</sub> be the space spanned by V<sub>&alpha;</sub> and the element a<sub>V<sub>&alpha;</sub></sub>. If &lambda; is a limit ordinal, let V<sub>&lambda;</sub> be the union of all earlier V<sub>&alpha;</sub>. It is easy to see that { a<sub>V<sub>&beta;</sub></sub> | &beta; &lt; &alpha;} is a basis for V<sub>&alpha;</sub>. Thus, the dimension of each V<sub>&alpha;</sub> is exactly the cardinality of &alpha;. In particular, if &omega;<sub>1</sub> is the first uncountable ordinal, then V<sub>&omega;<sub>1</sub></sub> will have uncountable dimension, yet be the union of all V<sub>&alpha;</sub> for &alpha; &lt; &omega;<sub>1</sub>, which all have countable dimension, as desired.</p> <p>If you forbid one to use the Axiom of Choice, then it is no longer true that every vector space has a basis (since it is consistent with ZF that some vector spaces have no basis), and the concept of dimension suffers in this case. But some interesting things happen. For example, it is consistent with the failure of AC that the reals are a countable union of countable sets. R = U A<sub>n</sub>, where each A<sub>n</sub> is countable. (The irritating difficulty is that although each A<sub>n</sub> is countable, one cannot choose the functions witnessing this uniformly, since of course R is uncountable.) But in any case, we may regard R as a vector space over Q, and if we let V<sub>n</sub> be the space spanned by A<sub>1</sub> U ... U A<sub>n</sub>, then we can still in each case make finitely many choices to witness the countability and conclude that each V<sub>n</sub> is countable dimensional, but the union of all V<sub>n</sub> is all of R, which is not countable dimensional.</p>