Complexity Theory – Polynomial Close Closure

Having trouble generalizing when a class of complexity $ D $ is closed under polynomial reduction.

For example, consider the following examples:

  1. $ bigcup_ {c ge 1} mathsf {DTime} (2 ^ {cn ^ 5}) $
  2. $ bigcup_ {c ge 1} mathsf {DTime} (5 ^ {cn}) $
  3. $ bigcup_ {c $ ge 1} mathsf {DTime} (n ^ {c log ^ 5n}) $

$ P $ is contained in all these classes, so if we are given a problem $ A $ who belongs to one of these classes, and a reduction of another problem $ B $ at $ A $the polynomial reduction will not affect the overall execution time of the decision. Can it be generalized to a higher class than $ P $?

Transitive closure vs. accessibility in graphics

I am faced with the most curious situation with [my current information of] transitive closure algorithms. Specifically, what follows is not an algorithm to determine the transitive closure of a graph G (V, E) and the time complexity is not O (| V | + | E |):

Use Kosaraju to calculate SCCs (strongly connected components) – O (| V | + | E |)

Create a DAG from SCCs as (super) nodes – O (| V | + | E |)

In a single super-node, the reachability of all the nodes is identical and equal to the union of the nodes in the containing supernode, plus the supernodes accessible in the DAG-O (| V | + | E |) to compute the in the DAG and O (| V | * | V |) total to compute the accessibility sets for all nodes

These accessibility sets make to constitute the transitive closure of the graph altogether, no?
But there are not any O (| V | + | E |) transitive closure algorithms around.
What do I miss?

Commutative algebra – What is the difference between total integral closure and integral closure?

I have been advised here to make this question a new question:

What is the difference between total integral closure and integral closure (geometrically, in the context of rigid analytical geometry)? I have read in several places that the difference is essentially between rank 1 and higher rank evaluations, but how can we clarify that?

It is surprisingly difficult even to find a definition of total integral closure …

Closed convex cone – equivalence of the definition via closure and via infinite sums

I have a set $ P $ of points in a Banach space. Consider the following two cones:

  • Closing the set of all non negative linear (negative) combinations of $ P $. (I.e., topological closure of $ { sum_ {i = 1} ^ n a_ip_i: a_i geq0, p_i in P } $.)
  • The set of all non-negative infinite linear combinations of $ P $. (That is to say., $ { sum_i a_ip_i: a_i geq0, p_i in P } $ or $ i $ can go on infinite sets, and we only consider sums that converge absolutely.)

Are these sets equal?

Prove that $ K = k (t) $ is the fractional body of R and $ k[t]$ is the full closure of $ R $ in $ K $

assume $ k is a field and leaves $ t = overline {x} / overline {y} $ in the field of integral domain fractions $ R = k[x,y]/ (x ^ 2-y ^ 3) $. Prove it $ K = k (t) $ is the fraction field of R and k $

angular – Is there a way to hide a closure container in a model?

I may cache DOM child / grandchild items with * ngIf, but the parent and grandparent elements have formatting that remains even when the child / grandchildren have been removed from the DOM (the limits remain). Is it possible to hide parent / grandparent elements to completely remove formatting when children / grandchildren are deleted?

  

Find the closure and derived set of $ A $ in some topologies.

Let $ X =[0,1] subset mathbb {R} $, let $ mathscr {B} $ to be a base for the topology $ mathscr {T} _ {1} $ sure $ X $ given by $$ mathscr {B} = { {0 }, {1 } } cup {(0,2 ^ {- k}) ,: , k in mathbb {Z} _ { ge0} }, $$and let $ mathscr {T} _ {2} $ to be a topology on $ X $ given by $$ mathscr {T} _ {2} = left {G subset X ,: , frac {1} {2} not in G right } cup {G subset X ,: , (0,1) subset G }. $$Yes $ A = left { left ( frac {1} {4}, frac {1} {2} right) right $ is a subset of the product space $ X ^ { ast} = (X, mathscr {T} _ {1}) times (X, mathscr {T} _ {2}) $, find the closure $ overline {A} $and the derived set $ A $ in $ X ^ { ast} $.

Find $ A $, I've watched the $ A = left { frac {1} {4} right } times { frac {1} {2} $ and using the formula for $ A $ in the product space like $ A = left ( left { frac {1} {4} right } times { frac {1} {2} } right) & # 39; = left (, overline { left { frac {1} {4} right }} times { frac {1} {2} } & # 39; right) cup left ( left { frac {1} {4} right } times overline { { frac {1} {2} }} , right) = left ( left[Frac{1}{4}1right)times{frac{1}{2}}right)cupleft(left(frac{1}{4}1right)times{frac{1}{2}}right)=left[Frac{1}{4}1right)times{frac{1}{2}}$[Frac{1}{4}1right)times{frac{1}{2}}right)cupleft(left(frac{1}{4}1right)times{frac{1}{2}}right)=left[Frac{1}{4}1right)times{frac{1}{2}}$[frac{1}{4}1droite)times{frac{1}{2}}droite)cupleft(left(frac{1}{4}1droite)times{frac{1}{2}}right)=left[frac{1}{4}1droite)times{frac{1}{2}}$[frac{1}{4}1right)times{frac{1}{2}}right)cupleft(left(frac{1}{4}1right)times{frac{1}{2}}right)=left[frac{1}{4}1right)times{frac{1}{2}}$

But the solution is $ A = left ( frac {1} {4}, 1 right) times { frac {1} {2} $.

Where am I wrong? or can not use this formula for a subset of a point in the product space?

ct.category Theory – Closure of presentable objects within finite limits

In a locally presentable category $ cal E $, there are arbitrarily large regular cardinals $ lambda $ such as $ lambda $-presentable (a.k.a. $ lambda $-compact) are closed under withdrawals. Namely, the withdrawal functor $ { cal E} ^ {( to leftarrow)} to cal E $ is a right assistant, so accessible. So, he keeps $ lambda $-presentable objects for arbitrarily large $ lambda $so just check that the $ lambda $objects -presentable in $ { cal E} ^ {( to leftarrow)} $ are the ones that are punctual in $ cal E $. (A version of this argument is given in this answer in the case of finished products.)

Of course, "arbitrarily big" means that for any cardinal $ mu $ there is a regular cardinal $ lambda> mu $ with this property. A stronger statement would be that this is true for all big enough regular cardinals $ lambda $, that is, invert the quantifiers and say that there is a $ mu $ such as all regular cardinals $ lambda> mu $ have this property $ lambda $-The presentable objects are closed under withdrawals). Is this stronger affirmation true?

Note that this is certainly not true all regular cardinals $ lambda $ have this property; counterexamples can be found here.

abstract algebra – For $ Omega $ an algebraic closure of k $, why is $ k[x]/ (f) otimes_k Omega cong Omega ^ m $?

Fix a field k $ and an algebraic closure $ Omega $ of k $. For a monic, irreducible, separable polynomial $ f in k[x]$, let $ f $ to divide $ Omega $ as

$$ f (x) = prod_ {i = 1} ^ m (x – a_i). $$

It was said that the following holds:

$$ frac {k[x]} {(f)} otimes_k Omega con prod_ {i = 1} ^ m frac { Omega[x]} {(x-a_i)} cong Omega ^ m. $$

I can see the Chinese rest theorem here, but what is not clear to me is the resolution of the tensor because I am not yet comfortable working with them. Why is the first isomorphism true?

I like Google Plus, but its closure is imminent. Sigh

Google Plus is one of my favorites because it sells your privacy, but they are out of business soon. Sigh…