What is the meaning of ∅ used in the context of a relation?

[StatOnTheSide]

Hi. I am reading Halmos's Naive Set Theory book. I have the following doubts.

1. In chapter 7. on relations, he says "The least exciting relation is the empty one. (To prove that ∅ is a set of ordered pairs, look for an element of ∅ that is not an ordered pair)
". Whenever someone talks about ∅ used in the context of a relation, it becomes very confusing. Formally, is there a way to logically explain as to why ∅ is a relation?

2. At the end of chapter 8., there are two excercises namely "(i) Y^∅ has exactly one element, namely ∅, whether Y is empty or not, and (ii) if X is not empty, then ∅^X is empty". What do these statements mean in the first place? How do you prove them?

I have been able to understand almost everything in Halmos's book except few things which are really confusing. I kindly request you to help me understand them. I really like Halmos's style but I get struck at places and I feel very helpless during those times.

 

[SteveL27]

Hi. I am reading Halmos's Naive Set Theory book. I have the following doubts.

1. In chapter 7. on relations, he says "The least exciting relation is the empty one.

The set of exciting relations is well-ordered, so there must be a least exciting one.

Just kidding, could't resist. You'll get used to these empty set arguments after a while. To see that the empty set is a relation, write down the definition of relation (set of ordered pairs such and so) and try to see how the empty set could possibly fail to satisfy the definition. You'll see that you can't make the empty set fail to be a relation.

In other words: there is no way to logically falsify the statement, "The empty set is a relation." So it must be true.

 

[920118]

2. At the end of chapter 8., there are two excercises namely "(i) Y^∅ has exactly one element, namely ∅, whether Y is empty or not, and (ii) if X is not empty, then ∅^X is empty". What do these statements mean in the first place? How do you prove them?

X^Y is most likely the set of (total) functions from Y to X. They shouldn't be that hard now that you know what it means.

 

[StatOnTheSide]

Thanks for the reply.

Steve: Halmos actually says that "To prove that ∅ IS a set of ordered pairs, look for an element of ∅ that IS NOT an ordered pair". What if I say "To prove that ∅ IS NOT a set of ordered pairs, look for an element of ∅ that IS an ordered pair".

In general, "to prove a Property about ∅ is True, look for elements in ∅ which do not satisfy the Property" but then, if I use the reverse logic that "to prove a Property about ∅ is False, look for elements in ∅ which do satisfy the Property". SInce the answer to both the statements is "no element in ∅ satisfies the property" or "no element in ∅ satisfies the negation of the property", you can prove anything about ∅ by saying there are no elements in ∅ which satisfies/do not satisfy a property.

Given that, how is it possible to conclude one way or another?

920118: What does it even mean when you say all functions from Y to ∅ or ∅ to X? There are no functions right? Is that the reason why the answer to that is ∅?

 

[SteveL27]

Thanks for the reply.

Steve: Halmos actually says that "To prove that ∅ IS a set of ordered pairs, look for an element of ∅ that IS NOT an ordered pair". What if I say "To prove that ∅ IS NOT a set of ordered pairs, look for an element of ∅ that IS an ordered pair".

It takes getting used to. Here's how I understand it.

A relation is a set of ordered pairs with such and so property.

The empty set is a set of ordered pairs. How many? Zero. The empty set is a set containing zero ordered pairs. The empty set is also a set of purple zebras.

That's the first conceptual hurdle. If the empty set is a set, then it's a set containing zero things. Zero ordered pairs, zero purple zebras.

Now that the empty set is a set of ordered pairs, how could it fail to be a relation? Well, the ordered pairs it contains must fail to satisfy the condition that makes a set a relation. But you could never find ordered pairs in the empty set that fail to satisfy a condition.

So, the empty set is a relation. Because it's a set of ordered pairs (an empty set of ordered pairs), and all the ordered pairs (of which there aren't any) satisfy the given condition.

I don't claim this is fully satisfactory from an intuitive standpoint. But I do claim that if you run enough empty set arguments through your mind, after a while they'll stop bothering you.

As the great John von Neumann said: In mathematics you don't understand things. You just get used to them.

 

[mbs]

You have to look carefully at the definition of a relation. It should be something like...

$R\text{ is a relation iff... }$

$\forall x : x\in R \Rightarrow x\text{ is an ordered pair}$.

If the predicate "$x\in R$" is false for every $x$, the statement "$x\in R \Rightarrow x\text{ is an ordered pair}$" must be true for every $x$. This is because in general if statement "$A$" is false, then statement "$A\Rightarrow B$" is true, regardless of whether "$B$" is true or false.

There is nothing in the definition of a relation that says it cannot be empty. In math the basic idea is to take a definition exactly at face value. With some definitions there's a tendency to assume more based a common understanding from real life or the way we normally use language with implied meanings. In math and logic there are no hidden assumptions. You only assume exactly what you're told.

 

[StatOnTheSide]

Thanks Steve. I also found a link to a previous discussion on the same topic in Physics Forums (I cannot point out to the url as the website would not allow me to but the title is "Empty relation") where HallsOfIvy and Hurkyl also made good comments.

I feel most comfortable with the following explanation. "A relation is basically a subset of the set of ordered pairs XxY. Since ∅$\subset$XxY, it is a relation."

About functions: "a, function from (or on) X to (or into) Y is a relation f such that dom of f = X and such that for each x in X there is a unique element y in Y with (x,y) $\in$f . This is a 'for each x, there is a y' which is a conditional statement unlike the existantial statement like 'there EXISTS x in X such that y in Y satisfies...' and hence, as it is not existantial and as it is a conditional statement, ∅ will be a function too."

I do believe that it is a matter of how ∅ fits into the defition of relations and functions. If at all the defitions were existantial in nature rather than being conditional, GOD knows what would have happened to the fate of ∅.

Also, @ Steve, nice one on the ordering of exciting functions/relations :).

 

[920118]

Any n-place relation R is just a subset of the cartesian product of some domain U, i.e., R$\subseteq$Ux...xU. One of the first things you should've learned is that the empty set is a subset of every set, so it's trivial that there is an empty relation. I hope that helps.

Heheheheh... Sorry, missed that people wrote :(

 

[StatOnTheSide]

920118: I know right? Things can be expressed so well in formal language. Sometimes, Halmos tries words and I just hate it.
There are so many places where the words are confusing but when you put is formally, it starts to make sense.