Prove that if S is a subset of A then S is an empty set.

1. May 11, 2012

1. The problem statement, all variables and given/known data
Let S be a set such that for each set A, we have S$\subseteq$A. Show that S is an empty set.

3. Relevant equations
10.6 Proposition. For each set A, we have empty set $\subseteq$ A.

3. The attempt at a solution
Solution. Consider any set A and a set S such that S$\subseteq$A. Choose any x$\in$S. Then since S$\subseteq$A we also have x$\in$A. From proposition 10.6 we know that if x$\subseteq$empty set, then x$\subseteq$A. Now x is arbitrary. Thus S must be an empty set.

I feel like this proof is horrible and doesn't flow at all. Can someone give me some pointers?

2. May 11, 2012

jgens

Clearly $\emptyset \subseteq S$ and by hypothesis $S \subseteq \emptyset$. Can you put this together?

3. May 11, 2012

I am trying to see how this works but I don't get it.

I understand that if we can show that $\emptyset \subseteq S$ and $S \subseteq \emptyset$, then we can say that $S=\emptyset$ but I don't understand what you mean by, by hypothesis $S \subseteq \emptyset$.

4. May 11, 2012

gopher_p

Nevermind. Beat to the punch.

5. May 11, 2012

jgens

In your own words: Let S be a set such that for each set A, we have S⊆A. The empty set is a set, yes?

6. May 11, 2012

Let me apologize for being a bit slow with grasping this. Ever since my proof class moved into sets I've been a little slow...

Anyhow, from the way I am thinking about it, the problem is to show that there exists a set such that it is a subset of A. Then by saying that $S\subseteq\emptyset$ we are claiming that $\emptyset$ is such a set. Is that the correct way of looking at it?

7. May 11, 2012

jgens

I can't make sense out of what you're saying, so could you be a little more clear?

8. May 11, 2012

I can't see how saying Let S be a set such that for each set A, we have S⊆A can lead to $S\subseteq\emptyset$.

9. May 11, 2012

jgens

That is clearer, thank you. Do you accept that $\emptyset$ is a set?

10. May 11, 2012

Yes I do.

11. May 11, 2012

jgens

Since $S$ is a subset of every set and the empty set is a set, this mean $S \subseteq \emptyset$ (i.e. S is a subset of the empty set). Not sure what else to say about that. If you're still confused perhaps another member could chime in and help?

12. May 11, 2012

Ok. Here is an attempt at a better proof.

Consider any set A and a set S such that S is a subset of A. From proposition 10.6 we know that the empty set is a subset of S. Now S is a subset of every set A so it follows that S is a subset of the empty set. Thus we have the empty set is a subset of S and S is a subset of the empty set. Therefore S is an empty set.

Does that work?

13. May 11, 2012

jgens

Yeah. The only thing I would change is that I would say S is the empty set. The empty set is provably unique.

14. May 11, 2012

Done.

Thanks so much for your help! I need to stare at this till it makes more sense to me.

15. May 11, 2012

jgens

Sometimes that is the only approach, haha. If you still have difficulty with the $S \subseteq \emptyset$ feel free to ask the other members. I just don't know how else to explain it :/

16. May 11, 2012

SteveL27

There's only one empty set. That's because a set is completely characterized by its elements. There's only one set with no elements, that's the empty set.