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location | ||
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visits | member for | 2 years, 4 months |
seen | Jan 5 '14 at 20:39 | |
stats | profile views | 152 |
Sep 8 |
awarded | Good Question |
Jul 2 |
awarded | Curious |
Jan 2 |
awarded | Yearling |
Oct 9 |
awarded | Constituent |
Oct 8 |
awarded | Caucus |
Sep 19 |
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Linear Algebra without Choice
you can still define a vector space to be finite dimensional if it is finitely generated. I think it is a basic theorem of linear algebra that such a vector space automatically has a (finite) basis, even without the axiom of choice. |
Jul 6 |
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When was the continuum hypothesis born?
what is Cantor's meaning of "Mannigfaltigkeiten" in this setting? surely it doesn't mean what we mean when we talk about manifolds. |
Jun 5 |
awarded | Popular Question |
Jun 4 |
asked | for which truth-operations f can f-membership in a prime ideal be represented by a polynomial? |
May 24 |
awarded | Nice Question |
May 20 |
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Objects which can't be defined without making choices but which end up independent of the choice
Nice. Is there a construction of this sort for $Tor$? |
May 20 |
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Objects which can't be defined without making choices but which end up independent of the choice
@Steven. actually in your example of Cauchy sequences there is no need to choose anything to define the sum of $A$ and $B$. You can just define the function $f\colon A\times B \to V$ defined by $f((x_n),(y_n))=(x_n+y_n)$. Then $A+B = Im(f)$ is nonempty if $A\times B$ is. |
May 20 |
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Objects which can't be defined without making choices but which end up independent of the choice
really? can't one just say $dim_k(V)$ is the smallest integer $n$ such that the set {$(v_1,\ldots,v_n)\in V^n\mid \langle v_1,\ldots, v_n\rangle=V$} is nonempty? |
May 20 |
asked | Objects which can't be defined without making choices but which end up independent of the choice |
May 16 |
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How to memorise (understand) Nakayama's lemma and its corollaries?
oh nice, thanks |
May 16 |
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How to memorise (understand) Nakayama's lemma and its corollaries?
I don't understand the last example. if you reduce that s.e.s. to $k$, you get an exact sequence $0\to 0\to k^n\to M\otimes k\to 0$. But we knew that already and you don't need flatness of M for that. How does it follow that in the original sequence $K$ vanishes? |
May 14 |
answered | Awfully sophisticated proof for simple facts |
May 12 |
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Magic trick based on deep mathematics
am I missing something or is $\sum_{g\in G}g$ always the unit element in the abelian group $(G,+)$? |
May 9 |
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A question in category theory
I would be interested whether there exists a counterexample if the isomorphism is not assumed natural. if C doesn't have to be abelian it is easy: Take C to be the category freely generated by two objects and two arrows between them (in different directions). Then for any $X$,$Y$ (possibly $X=Y$) in C we have $Hom(X,Y)\cong \aleph_0$, but the two objects in C are not isomorphic. |
Apr 8 |
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How long can this string of digits be extended?
Is there a good reason why N(b) can't be infinite? |