Most among none - Set Theory and Logic

Question

Let's say I have a collection of particles with $0$ mass.

Is it true to say that any particle is the heaviest in the set?

And, what if the collection has $0$ particles?

My motivation for posting these questions is to familiarize new students of advanced math with a specific kind of subtle logic which can show up but which seems unusual at the beginning.

The first is a test of a good understanding of implications and the second is a test of a good understanding of how sets work.

Maybe we can get a wiki going here of different kinds of examples of such subtle logic?

Students who are studying advanced math dont need these logical cul de sacs. — Rene Schipperus, Dec 15 '17 at 18:54
I don't think that this is a logic or set theory question. It looks more like a word problem on a precalc level. — Asaf Karagila, Dec 15 '17 at 18:54
If you have theoretical physicists in the class you might get a answer that there is a particle with negative mass. — kingW3, Dec 18 '17 at 13:01
@ThisIsNotAnId: What is "what"? (Also, https://math.stackexchange.com/a/1452437/622 (also, what do you think is a set?)) — Asaf Karagila, Dec 18 '17 at 17:41
@AsafKaragila I don't know. I was just going by what my elders taught me >.< — ThisIsNotAnId, Dec 18 '17 at 22:41

score 2 · Accepted Answer · answered Dec 15 '17 at 19:18

Heaviest may be defined in two ways

$x\in A$ is the heaviest element of a set $A$ if all other elements in that set are less heavy.
- $\forall a \in A:x>a$, where $a > b$ iff $a$ is heavier than $b$.
$x\in A$ is the heaviest element of a set $A$ if for all other elements in that set $x$ is at least as heavy as they are.
- $\forall a \in A: x \ge a$, where $a\ge b$ iff $a$ is at least as heavy as $b$.

Now based which definition of heaviest (I think definition 1 is more natural and common) you will find answers to be trivial.

By definition 1 no particle is the heaviest if all particles are the same weight. By definition 2 all particles are the heaviest if they are all the same weight.

To move on the the second part of the question, if there are no particles then either statement is valid and they are equivalent in this case, as all particles would be equivalent to no particles.

The second definition is more commonly used in physics – ℋolo Dec 15 '17 at 19:23 — ℋolo, Dec 15 '17 at 19:23
A rigorous answer to what I had in mind. Thank you. – ThisIsNotAnId Dec 18 '17 at 22:40 — ThisIsNotAnId, Dec 18 '17 at 22:40

Most among none - Set Theory and Logic

1 Answers1