Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

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.

share|cite|improve this answer

Your Answer


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.