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I’ve got some questions regarding set theory. I am struggling to find the right notation in order to express a number of conditions. I have a set named A that contains $N$ T-sized groups and each group is characterised by T$B_{x,y}$ elements. For example, consider the following case in which N=4 and T=3.

$$ A:=\{(B_{1,1}, B_{2,1}, B_{3,1}), (B_{3,1}, B_{2,1}, B_{5,1}), (B_{1,3}, B_{2,5}, B_{3,3}), (B_{7,3}, B_{4,5}, B_{3,3})\} $$

Each $A_k$ group (where $1≤k≤N$) represents a specific condition and $P$ is a property which can be found for each condition described by each $A_k$ group (for example, for ($B_{1,1}$,$B_{2,1}$,$B_{3,1}$), $P=5$).

What I want to express is the following conditions:

  1. Find all the P properties for all the $A_k$ groups. Each $P$ property is associated with one $A_k$ group.What I’ve got is: $∀1≤k≤N,P for A_k$
  2. Consider the $A_k$ group(s) which have the smallest value of $P$ property ($P^{min}$) among all P properties.
  3. Consider the $A_k$ groups which contain a specific $B_{x,y}$ element (e.g. $B_{2,1}$). For the above example, $B_{2,1}$ element should be in groups $A_1$ and $A_2$.
  4. I would also like to somehow express the “$P$ for $A_1$ is $5$”
  5. Return the $P$ for each of the $A_k$ groups which include the $B_{2,1}$ AND $B_{5,1}$ elements (for the aforementioned example, the $P$ of just $A_2$ group should be returned).

Thanks in advance.

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I don't understand your indexing scheme for the $B_{x,y}$s. It looks completely random -- and there are two $B_{3,3}$s. Shouldn't it be something like $$A=\{(B_{1,1},B_{1,2},B_{1,3}),(B_{2,1},B_{2,2},B_{2,3}),(B_{3,1},B_{3,2},B_{3,‌​3}),(B_{4,1},B_{4,2},B_{4,3})\}$$ And where do you get $P=5$ from? There doesn't seem to be any relevant $5$s in evidence already. –  Henning Makholm Oct 27 '12 at 18:49
    
Thanks for your reply. You're right, my $B_{x,y}$ indexing is misleading, yours makes much more sense. Each combination of $B_{x,y}$s which is declared by a corresponding $A_k$ group is statically mapped to a P value. For instance, the combination of $B_{x,y}$s denoted by $A_1$ is mapped to P=5, $A_2$ could be mapped to P=10, etc. I hope it makes more sense now. –  limp Oct 27 '12 at 19:08
    
One minor nitpick: the word "group" has a very technical meaning in mathematics. I think you mean "set" or perhaps "class" every time you use the word "group" in your question. –  Code-Guru Oct 28 '12 at 23:17
    
Also, can you give a more concrete example? –  Code-Guru Oct 28 '12 at 23:18

1 Answer 1

Even though all five expressions seem more in line with what one would expect for a computer science question, let me explain how they can be rendered in mathematical lingo. To do this in a natural way requires to scramble the order of your expressions a bit.

First of all, since each $A_k$ has a unique $P$ associated to it, we can say that $P$ is functional: each input (the $A_k$) determines a unique output. Therefore, we can use function notation for $P$: we write $P(A_k)$ for the $P$ associated to $A_k$.

Hence, we can write expression 4, "$P$ for $A_1$ is $5$", as: $P(A_1) = 5$.

Next, we arrive at set-builder notation. It is a way of describing a "set", that is, any collection of objects that we'd like to think about in unison. We write:

$$\{x: \text{what $x$ is, or should satisfy}\}$$

for the "collection"/"set" of those $x$ for which the part after the colon is true.

In particular, we write expression 1, "The $P$ properties of all $A_k$", as: $$\{P(A_k) : 1 \le k \le n\}$$

Also, for expression 2, we obtain: $$\{A_k: P(A_k) = P^{\rm min}\}$$

The last bit of notation we need is a symbol to denote elementhood. In mathematics, we use the symbol $\in$, and write:

$$x \in X$$

to signify that $x$ is an element of the set $X$.

Hence, for expression 3, we can write: $$\{A_k: B_{2,1} \in A_k\}$$ and for expression 5: $$\{A_k: B_{2,1} \in A_k \text{ and } B_{5,1} \in A_k\}$$

Because mathematicians are lazy, they commonly use the comma to abbreviate such similar statements as $B_{2,1} \in A_k$ and $B_{5,1} \in A_k$. Thus this is a shorter alternative for expression 5: $$\{A_k: B_{2,1},B_{5,1} \in A_k\}$$

I hope this is of some help to you.

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