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I have a question regarding set theory when using forall on the same set twice, in the same equation. More specifically, this is the equation I want to write:

$ \forall u_i \in U : u_i \; mutual_{reach} \; \forall u_k \in U $

What this should mean is: If any given user in the set U of users have a mutual reachability with every other user, the group of users have a mutual reachability. I have not seen this way of iterating through every element of the same set twice before and I am therefore not sure if this is the correct notation for it, or if it even makes sense mathematically. Does anyone know if this is correct, or if I am completely off track?

Andrés E. Caicedo
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Bouet
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1 Answers1

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Let me rephrase your statement: You say exactly that " If any given user in the set U of users have a mutual reachability with every other user, the group of users have a mutual reachability".

That is to say: "the group of users has a mutual reachability if and only if every pair of users in the group have a mutual reachability with each other".

Let's rewrite this: Suppose our group is $U$ (it's a set, rather than a group, but anyway). For every pair of users, you take the Cartesian product of $U$ with itself to form $U^2$, and then we say they are mutually reachable.

Note the following: A group having a mutual reachability is different from two users having rechability, although the word "mutual reachability" is used for both. So we need to have separate notations to take care of both.

We should invent notation for ourselves. Let's denote $u_1 \sim u_2$ for $u_1$ has mutual reachability with $u_2$, and keep the other one fixed.

Now we can write: $$ U \text{ has mutual reachability} \iff \forall (u_1,u_2) \in U^2, u_1 \sim u_2 $$

So that reduces two $\forall$ statements to one, and all we did was take the Cartesian product and write our statement more mathematically.

Sarvesh Ravichandran Iyer
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