bio  website  

location  sqrt(1)  
age  
visits  member for  4 years, 5 months 
seen  7 hours ago  
stats  profile views  3,178 
two queens
mathematics and death
not cheerful not sad
guide me through the land
pat me on my head
they live
in my one man nation
i follow them
to my destination
wh,
(long ago)
polynomials over finite Galois field
spread so evenly across their finite affine space
i wish for a network of friends
to count on
wh,
19960305/06
1d

comment 
On independent sets of graph
$\alpha(G)\ $ and $\ N(G)\ $ have to be balanced to get an objectively good inequality, however possibly not as good as you'd like itit has to be seen. 
1d

comment 
On independent sets of graph
Yes, as an upper bound. These are trivially equivalent (sorry to inertially waste time on my talking). 
1d

comment 
On independent sets of graph
I think that only my accidental conclusion was wrong (a result of a mistaken thinking at that momentmy concentration gave up when I mixed the general definition and the peculiarities of my construction). Another equivalent formulation: $\ m:=M(G)\ $ is the largest integer such that there exists $\ W\subseteq V\ $ such that $\ W=\alpha(G)\cdot m\ $ and W is a union of $\ m\ $ maximal independent sets (i.e. od independent sets $\ J\subseteq V\ $ such that $\ J=\alpha(G)$). 
1d

comment 
On independent sets of graph
I slightly and equivalently reformulated you definition of $\ M(G)\ $ ok. In my comment I got confused only about its conclusion. Of course, as you've written, $\ M(G)\le\frac n{\alpha(G)}$. 
1d

comment 
On independent sets of graph
The last portion of the general part was wrong (the two preceding portions, and the small example, were fine). 
1d

revised 
On independent sets of graph
The last traces of the error removed. 
1d

comment 
On independent sets of graph
Thank you. But at least what I called my small EXAMPLE was fine, I was not confused at that stage. 
1d

revised 
On independent sets of graph
An unfortunate Error removed. 
1d

comment 
On independent sets of graph
Turbo, about def. of $\ M(G).\ $ Is $\ M(G)\ $ the maximal cardinality of a family of pairwise disjoint independent sets of cardinality $\ \alpha(G)\ $?  so we would have $\ M(G)\ \le\ \binom n{\alpha(G)}\ $ (where $\ n\ $ is the number of vertices). 
1d

revised 
On independent sets of graph
mth typo (x2) 
2d

comment 
On independent sets of graph
Your notion $\ f(\alpha(G)\ N(G))\ :=\ \max(\alpha(G)\ N(G)),\ $ and similar functions $\ f(\alpha(G)\ N(G))\ $ introduce an interesting internal pressure to the graphs, and it should lead to a whole subtopic. 
2d

comment 
On independent sets of graph
Perhaps. This and variations were on my mind, but I wanted to finish first step to answer properly on your question. Just to establish the $\ \LaTeX\ $ phrases was hard to me (I am slow). Now wider possibilities are open. 
2d

comment 
On independent sets of graph
No yet, but there should be more fun ahead of us :) 
Jan 25 
comment 
On independent sets of graph
This general (:) example is done; it's time for a more dramatic one, hey! 
Jan 25 
revised 
On independent sets of graph
Corollary 
Jan 25 
revised 
On independent sets of graph
More complete (THEOREM) 
Jan 25 
revised 
On independent sets of graph
smoother 
Jan 25 
comment 
On independent sets of graph
I fixed $\ M(G)$now it's perfect! (every ksubset is used). In particulat the small example got improved as the result. 
Jan 25 
revised 
On independent sets of graph
A typo in Remark 2. 
Jan 25 
revised 
On independent sets of graph
REMARK 0 added 