# Empty set and subset?

1. Dec 10, 2006

### pivoxa15

In my book, it says

"We agree to regard the empty set as a subset of every set. Thus any non-empty set S has just two improper subsets, the empty set and the set S itself; all other subsets of S are proper."

Does this sound right?

I thought the improper subset is the set which is the same as the original set. Why is the empty set also an improper subset?

Later in the book it said the empty set is a subset of every set. So the empty set is a subset of the universal set. Than it said the complement of the universal set is the empty set. From this information it implies the complement of the universal set is a subset of the universal set. Which is a contradiction?

Last edited: Dec 10, 2006
2. Dec 10, 2006

### matt grime

It is called improper because it is useful to do so. You are looking for too much meaning in the adjective.

I don't know how you deduced that contradiction. It is perfectly possible for a set E to be a subset X and X^c. As long as E is the empty set anyway. Now do you see why it is important that we don't include the empty set as a proper subset?

The empty set is different from other sets. In particular the statement

If x is in the empty set then ANYTHING AT ALL

is always true, since 'x in the empty set' is always false.

This is one reason why we might choose to exclude it from the 'proper subsets'.

3. Dec 10, 2006

### HallsofIvy

Textbooks, unfortunately, vary on this. Some texts use the word "proper" subset to mean a set that is strictly contained in another: A is a proper subset of B if there exist a point x in B that is not in A. Using that definition, the empty set is a "proper" subset of any set (except, of course, the empty set). Other textbooks specifically define "proper" subset to mean a nonempty subset having that property.

Textbooks that use the first definition often have to use the phrase "non-trivial proper subset" where texts using the second can just say "proper subset"!

4. Dec 10, 2006

### kesh

my hunch is you're extrapolating x can't be an element of a set and its complement (by definition) to something similar about subsets.

if a set is a subset of any set and its complement then it is the empty set

5. Dec 10, 2006

### verty

Just practicing...

P subset Q and P subset ~Q
-> for all x ((x elem P -> x elem Q) & (x elem P -> x elem ~Q))
-> for all x (x elem P -> (x elem Q & x elem ~Q))
-> for all x (x elem P -> (x elem Q & x notelem Q))
-> for all x (x elem P -> bottom)
-> for all x (x notelem P)
-> P = 0

0 subset U & 0 subset ~U
-> for all x ((x elem 0 -> x elem U) & (x elem 0 -> x elem ~U))
-> for all x ((x elem 0 -> x elem U) & (x elem 0 -> x notelem U))
-> for all x (x elem 0 -> (x elem U & x notelem U))
-> for all x (x elem 0 -> bottom)
-> for all x (x notelem 0)
-> true (by definition of 0)

6. Dec 10, 2006

### pivoxa15

It was late at night and I made a mistake in my opening post about the deduction. I have changed it now.

Here is the edited version.

'Later in the book it said the empty set is a subset of every set. So the empty set is a subset of the universal set. Than it said the complement of the universal set is the empty set. From this information it implies the complement of the universal set is a subset of the universal set. Which is a contradiction?'

Last edited: Dec 10, 2006
7. Dec 10, 2006

### matt grime

Why is that a contradiction? As has been explained by at least two people in this thread, the empty set is different from other sets owing to the fact x in E is false for all x if and only if E is the empty set, thus one can make any deduction one wishes vacuously.

8. Dec 10, 2006

### verty

(U is the universal set)

~U subset U
-> for all x (x notelem U -> x elem U)
-> for all x (~(x notelem U) or (x elem U))
-> for all x (x elem U or x elem U)
-> for all x (x elem U)
-> true (by definition of U)

9. Dec 10, 2006

### pivoxa15

I jumped to the conclusion too quickly without rational thought.

I will try to prove the result that the complement of the universal set U' is a subset of U.

U = universal set
E = empty set

Defn: The complement of a set A, is the set A' such that every element in A' is not in A.

Proof:

(for every x)(if x is in U' then x is not in U) where x is chosen from U because it's the universal set

=> E=U' is the only choice.

Lemma: E is a subset of every set
Proof:
Suppose E is not a subset of some set A.
Then there contains an element x in E that is not part of A.
The consequent is false. So the antecedent must also be false in order for the conditional statement to be true.
This leads to E is a subset of A.
Hence E is a subset of every set.

=> E is a subset of U

=> U' is a subset of U

QED

Hence this statement is not a contradiction but can be proved. Correct proof?

Last edited: Dec 10, 2006
10. Dec 10, 2006

### matt grime

For the 4th time, any reasoning that starts 'for all x in X' is vacuously true when X is the empty set since the precedent is false. If x is in the empty set this implies x is not in the empty set, it implies x is not the square root of minus 1....

11. Dec 10, 2006

### pivoxa15

You have tried to start with what I alleged to be a contradiction and tried to deduce a correct result hence showing it is not a contradiction.

Whereas I have tried to prove my alleged contradiction to show it is not a contradiction. It's two different ways of doing the same thing isn't it.

12. Dec 10, 2006

### pivoxa15

I understand that. I believe I have used this fact correctly in my proof. Or have I made a mistake regarding this fact?

13. Dec 10, 2006

### verty

I'm no mathematician; I don't think I can answer that. I'm trying to figure out if I did the right thing.

14. Dec 10, 2006

### verty

Well I'm learning but evidentally if showing that assuming some statement leads to a contradiction means that statement is false, then showing that no contradiction arises must corroborate it.

I wouldn't say I tried to prove or disprove it but rather that I merely reduced it to a more elementary form and in doing so it became evident that accepting it implies a necessarily true consequent, so by accepting it we can never derive a falsehood.

15. Dec 10, 2006

### matt grime

I doubt you have used it correctly becuase you appear to have several lines of proof. It is not a definition that the empty set is a subset of everything, it is a (vacuously) true deduction (in any reasonable logic). If it weren't a subset of X, say, then there is an e in E (empty set) that is not in X. But that is false - there is no e in E since E is empty. I don't see that there is anything to prove at all. The complement of U is the set of u in U that are not in U - this is vacuously empty, and it is vacuously (and I am using the word deliberately, pun intended) a subset of U as well.

16. Dec 10, 2006

### matt grime

Erm, any deduction from a false premise can be true or false, so I wouldn't go round accepting things from which we can deduce *true* a consequence. You have made several leaps that aren't justified. if (A implies B) is true and B is true then we know nothing abuot A.

17. Dec 10, 2006

### verty

Ok, let me try again.

Let U be the universal set.
Then by definition: for all x (x elem U)

for all x (x elem U)
-> for all x (x notelem U -> x elem U)
-> ~U subset U

So U is the universal set -> ~U subset U.
I now show the reverse implication:

~U subset U
-> for all x (x notelem U -> x elem U)
-> for all x (not (x notelem U) or x elem U)
-> for all x (x elem U or x elem U)
-> for all x (x elem U)

By definition, U is the universal set <-> for all x (x elem U)
Therefore ~U subset U -> U is the universal set.

So ~U subset U <-> U is the universal set.
U is the universal set.
Therefore I have shown that ~U subset U is true.

18. Dec 10, 2006

### verty

Ah, I only needed the first half.

19. Dec 10, 2006

### pivoxa15

I have fixed the proof by using a collory that the empty set is a subset of everything.

But you think my proof is trivial. You are right. But for a beginner like me I want to make things rigorous even these basic things, after all it tricked me in thinking that the (proved) result was a contradiction.