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Feb
29 |
awarded | Enlightened |
Feb
29 |
awarded | Nice Answer |
Feb
9 |
awarded | Commentator |
Feb
9 |
comment |
Additional condition to the Bollobas theorem (Sperner's therorem) in extremal set theory
I'm not sure what you're asking here. Clearly the old proof still goes through if you have the additional assumption (since you don't need to actually use the assumption), and I think the standard example $A_i = \{1, \ldots, n\} \setminus \{i\}$, $B_i = \{i\}$ satisfies your extra assumption and shows that the upper bound is still best possible. |
Feb
8 |
revised |
Blinking graphs
added 47 characters in body |
Feb
8 |
revised |
Blinking graphs
added 4 characters in body |
Feb
8 |
answered | Blinking graphs |
Feb
6 |
awarded | Yearling |
Feb
5 |
awarded | Editor |
Feb
5 |
revised |
How many uniquely colored degree two vertices in 3-coloring of subcubic graph?
added 23 characters in body |
Feb
5 |
answered | How many uniquely colored degree two vertices in 3-coloring of subcubic graph? |
Dec
7 |
comment |
“Let” versus “for all”
@quid - the page also says, of the Let/Then construction, "The second sentence is not a sentence, since the implicative sense of "then" plays the role of a conjunction". I'm not sure why you glossed over that part, since it's exactly the explanation you're asking for. Note that temporal "then" is not being used as a conjunction, but as an adverb, so the observation that starting a sentence with temporal "then" is common English usage isn't an argument in favor of the correctness of starting a sentence with implicative "then". |
Jun
4 |
comment |
Provability of unprovability
Thanks for pointing that out -- as much fun as I had fussing around with modal logic for the nonconstructive proof, it's certainly better to have an explicit example. (Is it obvious that this example works? It's been a while since I've thought about these things.) |
Jun
4 |
answered | Provability of unprovability |
Apr
17 |
comment |
Is there a Degenerate Dependency Local Lemma?
It seems to me that you can cash this out formally in terms of the assignment version of the problem by taking all $x(E_i)$ and $x(A)$ equal to $1/2$, and taking the ordering so that the center vertex is minimal. Now to satisfy the hypothesis in the question we just need all $P[E_i] \leq 1/4$ and $P[A] \leq 1/2$, which as you say is easy to arrange. |
Nov
24 |
awarded | Citizen Patrol |
Nov
22 |
awarded | Critic |
Nov
21 |
awarded | Yearling |
Nov
13 |
answered | Combinatorial Databases |
Nov
3 |
answered | Deciding whether a given graph has an f-factor or not! |