Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with vector-spaces
Search options not deleted
user 22989
5
votes
Accepted
Linearly independent family of sequences of rationals with a cardinal equal to the continuum
Choose a bijection $\alpha:\mathbb{N}\to\mathbb{Q}$, and for each $x\in\mathbb{R}$ let
$$a(x)_i=\begin{cases}0&\mbox{ if $\alpha(i)<x$}\\1&\mbox{ if $\alpha(i)\geq x$}\end{cases}.$$
Then the set of s …
- 35.2k
11
votes
Properties of vector spaces without AC
I think vector spaces must still be flat (tensor product is exact). I don't think any of the steps in the following proof use choice, although it's quite possible I'm mistaken:
Finite dimensional vec …
- 35.2k
4
votes
Accepted
Maximal commutative subrings of the endomorphism ring of a vector space
Even for a $2$-dimensional vector space (so we're looking at subrings of the ring $M_2(\mathbb{F})$ of $2\times2$ matrices over $\mathbb{F}$) there are nonisomorphic maximal commutative subrings.
Bo …
- 35.2k
46
votes
Accepted
Do vector spaces without choice satisfy Cantor-Schroeder-Bernstein?
Without the axiom of choice, it is possible that there is a vector space $U\neq 0$ over a field $k$ with no nonzero linear functionals.
Let $V$ be the direct sum of countably many copies of $U$, and $ …
- 35.2k