Timeline for If $V$ is a vector space with a basis. $W\subseteq V$ has to have a basis too?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 14, 2011 at 6:43 | vote | accept | Asaf Karagila♦ | ||
| Nov 14, 2011 at 1:31 | comment | added | Goldstern | One could try, but it would not work, or at least contradict my claim below. The set $W$ below is the kernel of the linear map from $V$ to $3^{<\omega}$, defined by $f(a) = (0,0,..., 1)\in 3^n$ for both $a\in S_n$. So in this case, both $V$ and $V/W$ have a basis, but $W$ does not. | |
| Nov 13, 2011 at 12:10 | comment | added | François Brunault | Another remark about the question is that if $W$ has finite codimension in $V$ then $W$ admits a basis. One could then try to generalize this by proving that if both $V$ and $V/W$ admit a basis, then so does $W$. | |
| Nov 13, 2011 at 12:08 | comment | added | François Brunault | @Asaf : I just meant that if $V$ is a vector space that doesn't admit a basis (such a vector space exists under $\lnot AC$), then you can cover it by the vector space $\oplus_{v \in V} \mathbb{F}$, which admits a basis. One could then try to produce a counterexample to your question by looking at the kernel of the canonical map $\oplus_{v \in V} \mathbb{F} \to V$. | |
| Nov 13, 2011 at 7:10 | comment | added | Asaf Karagila♦ | Francois, I am not sure that I understand the remark. | |
| Nov 12, 2011 at 21:41 | answer | added | Goldstern | timeline score: 17 | |
| Nov 12, 2011 at 21:38 | comment | added | François Brunault | Minor remark : the corresponding problem for surjections admits easy counterexamples, namely for every vector space $W$ there is a surjection $V \to W$ with $V$ admitting a basis (one can take $V=k^{(W)}$). One can then ask whether the kernel of this surjection admits a basis. | |
| Nov 12, 2011 at 21:37 | comment | added | Martin Brandenburg | Let me just remark that $V$ is the direct sum of copies of $\mathbb{F}$ indexed by $A$ and that a vector space has a basis iff it arises as such a direct sum; these standard arguments don't use AC. So the question really is: Can we conclude that in (ZF) every subspace of a direct sum $\oplus_{i \in I} \mathbb{F}$ of copies of $\mathbb{F}$ is again a direct sum of copies of $\mathbb{F}$? The answer is yes if there is a well-ordering on $I$, but probably no in general; somebody here will force you a counterexample :). | |
| Nov 12, 2011 at 18:02 | history | asked | Asaf Karagila♦ | CC BY-SA 3.0 |