Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose I have a set $\Omega = \{\omega_1,\omega_2,\omega_3,\ldots,\omega_k,\omega_n\}$ of $n$ elements. I ordered $\Omega$ with two different criteria by defining two new sets $A$ and $B$ of $k$ elements where $k$ is a fixed quantity $k<n$. Therefore $A$ and $B$ are two subsets. Suppose I want to define $C$ as a combination of the two subsets according a weight $\lambda \in [0,1]$, where if $\lambda=0$, $C=B$ and when $\lambda=1$, $C=A$. what is a proper definition of the set C?

Roughly explanation: What I want to do is the same reasoning behind a linear convex combination of two points $a$ and $b$ where $\lambda \in [0,1]$ , so $c = \lambda a + (1-\lambda)b$.

Can you help me? Thank you very much.

share|improve this question
1  
Not every question involving sets has to do with set theory, and even less with the [set-theory] tag. –  Asaf Karagila May 10 '12 at 21:24

1 Answer 1

up vote 0 down vote accepted

If $A=\{{\,a_1,a_2,\dots,a_k\,\}}$ and $B=\{{\,b_1,b_2,\dots,b_k\,\}}$ you could define $$C_{\lambda}=\{{\,\lambda a_1+(1-\lambda)b_1,\lambda a_2+(1-\lambda)b_2,\dots,\lambda a_k+(1-\lambda)b_k\,\}}$$ This assumes that when you wrote "I ordered $\Omega$ with two different criteria" you were using the word "ordered" the way I would use it, so that $A$ and $B$ are actually lists, rather than sets, and it makes sense to say which element of the one corresponds to any given element of the other.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.