User ostap chervak - MathOverflow

Estimate on radical of $2^n \pm 1$

2013-01-15T17:32:07Z

Not sure if this belongs to MO or not. Are there any lower bound on radical of $2^n \pm 1$ (Recall that radical of integer $rad(k)$ is a product of primes which divide integer $k$)?

As an example If abc-conjecture is true in the form $max(|a|,|b|,|c|) \leq rad(abc) ^2 $ then $$rad(2^n \pm 1) \geq 2^{n/2 - 1}$$ I wonder if this estimate is proven (or perhaps conjectured) by anyone? Are there any nontrivial results here?

Answer by Ostap Chervak for Old books still used

2012-12-31T10:37:01Z

R. Engelking (1977). General Topology.

Infinite domain with finite number of prime ideals(elements)

2012-06-14T13:51:34Z

While trying to prove one property of commutative rings with units I can't prove one fact without assuming existence of infinitely many different prime ideals or elements. I tried to test if it was the neccesary assumptions, but I failed, since I don't know any "toy"-examples of such rings.

I know only one example of this kind ($\mathbf{Q} [x]/ (x^2) $ ) but it's not a domain.

So,

Are there any infinite domains with finite number of prime ideals?

If no, then are there any infinite domains with finite (but nontrivial) number of prime elements?

I am interested in noncommutative examples as well. Sorry if this question is too elementary.

Answer by Ostap Chervak for closed meagre sets

2012-04-30T18:04:07Z

By the way, your question 1 is true for n-dimensional manifolds (since it is true for $\mathbb{R}^n$ by theorem of Menger and Urysohn).

Note that Q3 can also be answered noting that Cantor set $C$ is homeomorphic to $C^2$ and for every point $x$, {$x$} $\times C$ is a meagre subset of $C^2$. Note that this construction gives you Andreas's answer if x=0.

Answer by Ostap Chervak for Why are finiteness conditions important (and how to recognize them)?

2012-04-03T17:42:24Z

You can think of finiteness conditions from other point of view. Suppose you have some category (say, Set or Top) which have both "nice" objects and "pathological" objects. Then, it is natural to ask if there is a subcategory ("smaller" one) in which every set is "nice". That way, if your only goal is to study a single "nice" space you can study it's properties in subcategory and conclude something about larger one.

To study "smaller" category you need it to have some "nice" properties, like being cartesian-closed or something. That way Comp is "nice" subcategory of Top where you can use a lot of toopological constructions.

Now suppose, you want to study Von-Neumann universe of all sets. The only other, "smaller" von-neumann universe you can build is universe of heredetarily finite sets. That way $H_\omega$ is a "nice" sub-universe of $V$ where you can use almost all constructions fro mset theory (the only axiom which isn't true in $H_\omega$ is axiom of infinity).

But original question was stated not in the form "Why "niceness" properties are important" but "why finiteness conditions are important". Given that an answer to first question is much more understandable, we can say that finiteness conditions are important because all known "niceness" conditions are finiteness in nature.

So , for example following "niceness" condition in Top which doesn't look like finiteness condition is in fact equivalent to compactness:

$X$ is "nice" iff for every topological space Y projection $X\times Y\to Y$ is closed.

This situation ("niceness" conditions are hiddenly "finiteness") because almost all categories we study are set-like, so given a "nice" sub-universe in set-like category we can construct an induced "nice" sub-universe in Von-Neumann universe $V$ and the only nontrivial von-neumann universe is precisely $H_\omega$.

So, finiteness conditions are important because every regularity property in set-like category arises from a refularity property in universe of sets $V$, and the "best" regularity property in $V$ is a condition of being hereditarily finite so any good regularity property is esentially a finiteness condition.

Also note, that most of named finiteness conditions are actually "heredetarily finiteness" ones, they are usualy inherited by sum sub-objects.

Answer by Ostap Chervak for Linear Algebra - Find an N-Dimensional Vector Orthogonal to A Given Vector

2012-02-06T18:44:28Z

Actually you might use Gram Schmidt here.

Given a set of ortogonal vectors $x_1,x_2,\ldots,x_k$ you can use Gram-Shmidt algorithm for set of vectors ${x_1,x_2,...,x_k,e_i}$ adding basis vector to system of ortogonalysed vectors (note that you need use Gram Schmidt procedure only to find last vector since first k vectors are already orthogonal). Then (since vectors $e_1,e_2,\ldots,e_{k+1}$ are linearly independent) for some i between 1 and k+1 Gram Schmidt will give you non-zero vector which is ortogonal to given vectors $x_1,x_2,\ldots,x_k$

So to find a guess you simply need to use Gram Schmidt procedure several times (no more than k+1 for the first guess and no more then two times for next guesses).

To simplify this procedure you can do this only with first $k+1$ coordinates of vectors, so you will find a vector of form $(y_1,y_2,\ldots,y_{k+1},0,0,\ldots)$. Answers of Kapil and Klaus are actually equivalent to using this route.

Kronecker Approximation theorem and Fibonacci numbers

2011-03-10T22:53:24Z

There is a famous old theorem by Kronecker that for every positive real $\alpha$ and $\epsilon>0$ there exists a positive integer n such that $\alpha n$ is within $\epsilon$ of an integer.

Recently I found that the same result is true if we replace $\alpha n$ by $\alpha n^2$ or any polinomial p such that $p(0)=0$.

Could this result be generalised to other functions? Particularly I'm curious about sequences $\alpha 2^n$ and $\alpha F_n$ where by $F_n$ I denote n-th Fibonacci number.

Does anyone know anything about it?

Answer by Ostap Chervak for What are the most elegant proofs that you have learned from MO?

2011-04-15T15:37:09Z

I found several very nice proofs which I enjoyed:

1.Brilliant proof of fundamental theorem of algebra by Gian Maria Dall'Ara http://mathoverflow.net/questions/10535/ways-to-prove-the-fundamental-theorem-of-algebra/10684#10684

2.Some proofs of quadratic reciprocity: http://mathoverflow.net/questions/1420/whats-the-best-proof-of-quadratic-reciprocity (I especially liked that one: http://mathoverflow.net/questions/1420/whats-the-best-proof-of-quadratic-reciprocity/1431#1431)

3.Proof that $\mathbb{R}^{2n+1}$ does NOT have a square root (quite elementary and beatiful) http://mathoverflow.net/questions/60375/is-r3-the-square-of-some-topological-space/60389#60389

4.Nullstellensatz using model theory http://mathoverflow.net/questions/9667/what-are-some-results-in-mathematics-that-have-snappy-proofs-using-model-theory/9693#9693

5.If in ring R every countably generated ideal is principal than R is a PID http://mathoverflow.net/questions/8042/do-there-exist-non-pids-in-which-every-countably-generated-ideal-is-principal/8067#8067

6.An infinite dimensional vector space have smaller dimension than it's dual. http://mathoverflow.net/questions/13322/slick-proof-a-vector-space-has-the-same-dimension-as-its-dual-if-and-only-if-it/13372#13372

7.Topological proof that Z is a Bezout domain. http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts/64039#640397.

Answer by Ostap Chervak for Products of Baire spaces

2011-05-04T18:24:51Z

In addition to previous answers:

The easiest way of proving that space is Baire is using one of following theorems:

1)Any locally compact space is Baire 2)Any complete metric space is Baire

Actually there is a notion of Cech completeness which generalises both theorem. (A space is called Cech-complete if remainder of its Stone-Cech compactification $\beta X\setminus X$ is a $F_{\sigma}$ in Stone-Cech compactification, every locally compact is Cech-complete and every complete metric space is Cech-complete).

Then, while product of Baire spaces need not to be Baire, the product of ANY(even uncountable!) collection of Cech-complete spaces is Baire.

Answer by Ostap Chervak for Countable subgroups of compact groups

2011-04-24T20:11:49Z

By the way, $A_\omega ^{fin}$ (group of even permutation of countable set) can't be embedded into compact group (hence $S_\omega^{fin}$ also can't).

Proof(by a contradiction):

It is well-known (proof uses theory of representations) that every compact group is isomorphic to a subgroup of cartesian product of unitary groups, so WLOG we may assume that $A_\omega^{fin}$ is embedded in $\prod U({n_i})$. Now let $f_i$ be a homomorphism $f:A_\omega^{fin} \rightarrow U({n_i})$ such that $\prod f_i$ is an embedding in $\prod U({n_i})$. Obviously Ker $f_i$ can't be trivial for all i, so for some i $Ker(f_i) \neq A_\omega^{fin}$ , but since $A_\omega^{fin}$ is simple $Ker (f_i) =(0)$ and $f_i$ is injective.

So, we've just proved that if simple group can be embedded in compact group it can be embedded in $U(n)$ for some n.

Now, it's easy to finish the proof (there are lot's of methods, actually). For example note that there are infinitely many pairwise commuting different elements in $A_\omega^{fin}$ such that $x^2 = e $ but there are only finitely many such elements in $U(n)$.

Hope this helps.

Answer by Ostap Chervak for What are some reasonable-sounding statements that are independent of ZFC?

2011-03-25T19:48:27Z

My favourite one(in fact it is equivalent to continuum hypothesys, proving equivalency is a very nice exercise,btw):

Real line could be represented as a countable union of linearly independent(over $\mathbb{Q}$) subsets.

Comment by Ostap Chervak

2013-05-09T19:15:22Z

How can you do the second step? I don't see an imediate way of building a bijection between $\mathbb{R}$ and $\mathbb{R}\setminus\mathbb{Q}$

Comment by Ostap Chervak

2013-05-04T12:54:32Z

This prove shows that there exist a field with 83 elements

Comment by Ostap Chervak

2013-01-31T10:21:16Z

A city is compact if you can fire all except finite number of policemen patrolling the city

Comment by Ostap Chervak

2013-01-23T14:32:23Z

Note that this indeed is a generalization: original Euclid's proof uses $I=\emptyset$

Comment by Ostap Chervak

2013-01-19T18:47:55Z

Any hints on solving this problem?

Comment by Ostap Chervak

2013-01-19T11:42:53Z

"F_un mathematics" spin off exists, it's just isn't a website

Comment by Ostap Chervak

2013-01-17T18:59:30Z

And how to define paracompactness then? If you define it in terms of decompositions of unit you lose conection with compactness

Comment by Ostap Chervak

2013-01-15T18:23:11Z

$2^n + 1 = (2^n + 1) \rightarrow 2^n+1 \leq rad (1\times 2\times (2^n+1))^2 \rightarrow 2rad(2^n+1) \geq \sqrt{2^n+1} $ Thank you, quid, sorry for that

Comment by Ostap Chervak

2013-01-15T18:14:33Z

If space is compact you may use the word "remainder"

Comment by Ostap Chervak

2013-01-13T19:32:36Z

Note that In Solovay model of ZF+DC+Every Set is Lebesgue measurable Every Linear map is bounded, so you can't construct unbounded linear operator which is bounded on orthonormal basis without some form of choice (Hamel basis existence or Hahn-Banach which is on similar spirit). You may construct it by Hahn-Banach for linear functionals.Take $(e_i)$ to be orthonormal basis. Let $V_0 = L(e_0,e_1,...)$ $A_0=0$ then inductively choose $a_i\notin V_i$ and set $A_i (a_i) = n*a_i$, $A_i |_{V_{i-1}} = A_{i-1}$ $V_i=L(V_{i-1},a_i)$. Then extend operator $A$ defined on $\Cup L_i$ and you're done.

Comment by Ostap Chervak

2013-01-12T00:31:42Z

something may be clear while not being clearly clear

Comment by Ostap Chervak

2013-01-04T21:03:08Z

Irrartionality of $\pi$ uses infinitude of primes, so proofs by zeta function are all circular –

Comment by Ostap Chervak

2012-12-05T21:26:08Z

In fact $G/H$ is an ultrapower of $T$ but ultrapower of finite set is isomorphic to it.

Comment by Ostap Chervak

2012-12-04T06:39:44Z

Existence of non-measurables is equivalent to $\aleph_1 \leq |\mathbb {R}|$

Comment by Ostap Chervak

2012-05-17T08:53:41Z

I guess, you can make it more striking: "Using character theory, since any group of order 4 is abelian hence the only way to write 3 as a sum of squares is 3 =1^2 + 1^2+ 1^2" Right?