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

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Homework Statement


Let S be a set such that for each set A, we have S\subseteqA. Show that S is an empty set.

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

The Attempt at a Solution


Solution. Consider any set A and a set S such that S\subseteqA. Choose any x\inS. Then since S\subseteqA we also have x\inA. From proposition 10.6 we know that if x\subseteqempty set, then x\subseteqA. 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?
 
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DeadOriginal said:
I feel like this proof is horrible and doesn't flow at all. Can someone give me some pointers?

Clearly \emptyset \subseteq S and by hypothesis S \subseteq \emptyset. Can you put this together?
 
jgens said:
Clearly \emptyset \subseteq S and by hypothesis S \subseteq \emptyset. Can you put this together?

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.
 
Nevermind. Beat to the punch.
 
DeadOriginal said:
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.

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?
 
jgens said:
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?

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?
 
DeadOriginal said:
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?

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

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.
 
DeadOriginal said:
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.

That is clearer, thank you. Do you accept that \emptyset is a set?
 
  • #10
Yes I do.
 
  • #11
DeadOriginal said:
Yes I do.

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
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
DeadOriginal said:
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. 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?

Yeah. The only thing I would change is that I would say S is the empty set. The empty set is provably unique.
 
  • #14
jgens said:
Yeah. The only thing I would change is that I would say S is the empty set. The empty set is provably unique.

Done.

Thanks so much for your help! I need to stare at this till it makes more sense to me.
 
  • #15
DeadOriginal said:
Thanks so much for your help! I need to stare at this till it makes more sense to me.

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
DeadOriginal said:
Done.

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

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.
 
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