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A PhD student of mathematics at Uniwersytet Wrocławski.


3h
comment Number of group actions
@UNM: It isn't. So how many homomorphisms are there?
3h
answered Number of group actions
5h
comment Define a projection homomorphism and find the kernel
In general, there is no nontrivial homomorphism ${\bf Z}_n\times {\bf Z}_m\to {\bf Z}_l$.
14h
comment Robinson's Consistency Theorem for first order languages
I think it's enough to have $M_n\subseteq N_n\subseteq M_{n+1}$. But it's prettier that way. ;-)
14h
answered Robinson's Consistency Theorem for first order languages
15h
comment Robinson's Consistency Theorem for first order languages
It doesn't just require background knowledge, it also requires some large cardinal assumptions (for existence of saturated models).
2d
answered Model complete theories without quantifier elimination
Dec
16
comment Is indefinite integration non-linear?
But the indefinite integral can be seen as a function into the quotient space of functions from reals to reals modulo constants. As such, it is perfectly linear.
Dec
16
comment A question regarding non-(Lebesgue)-measurable sets in models of ZFC+$2^{\aleph_0}$=$\aleph_2$
That sounds a little too broad to me. Do you have something more specific in mind?
Dec
14
comment Why are every structures I study based on Real number?
Another way to generalise normed vector spaces is when the underlying field is a normed field. This allows us to define a $k$-norm as a function into reals satisfying $\lVert av\rVert=\lvert a\rvert\cdot\lVert v\rVert$ (in addition to obvious axioms). If $k$'s norm is onto ${\bf R}_+$, you can then do normalisation (although it is non-unique, and most of the time this condition implies that $k$ is wildly inseparable, unless it is actually a subfield of the complex numbers with standard norm). Of course, this does not get us very far from reals in some ways. But then there are valued fields...
Dec
13
comment What's the difference between a bijection and an isomorphism?
@JiK: The question is general. In general, an isomorphism in a category is a morphism which has a two-sided inverse. If morphisms are functions, then isomorphisms are obviously bijections, but that is not always the case. Moreover, even if morphisms are functions, in general not every bijective morphism is an isomorphism (because the inverse may fail to be a morphism), for instance the mapping $x\mapsto x^p$ gives a bijective morphism from the affine line into itself (in the category of algebraic varieties over field of characteristic $p$), but it is not an automorphism if it is not identity.
Dec
12
comment What's the difference between a bijection and an isomorphism?
Note that an isomorphism is not always a bijection. For example, an isomorphism in the category of topological spaces up to homotopy equivalence can be very far from it, as can be an isomorphism in the category of metric spaces up to a quasi-isometry. Not to mention some more abstract categories, where morphisms are not even (equivalence classes of) functions.
Dec
12
comment Why is such an ideal ambiguous?
What do you mean by ambiguous?
Dec
11
answered The probability of modulo a prime
Dec
9
answered Order of abelian groups
Dec
9
comment No norm consistent with given topology
You really ought to mention what space this is.
Dec
8
awarded  Caucus
Dec
7
answered Orthonormal Hamel Basis is equivalent to finite dimension
Dec
7
comment Extend concept of weak derivatives?
Seems okay to me, but my pde are a bit rusty, so I might be overlooking something.
Dec
7
comment Extend concept of weak derivatives?
Why would that be strange? Most of the time, you want the set of test functions to be as small as possible (while having the properties you want it to have), not large.