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[itex]\subseteq[/itex]A. Show that S is an empty set.

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

The Attempt at a Solution


Solution. Consider any set A and a set S such that S[itex]\subseteq[/itex]A. Choose any x[itex]\in[/itex]S. Then since S[itex]\subseteq[/itex]A we also have x[itex]\in[/itex]A. From proposition 10.6 we know that if x[itex]\subseteq[/itex]empty set, then x[itex]\subseteq[/itex]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?
 
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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 [itex]\emptyset \subseteq S[/itex] and by hypothesis [itex]S \subseteq \emptyset[/itex]. Can you put this together?
 
jgens said:
Clearly [itex]\emptyset \subseteq S[/itex] and by hypothesis [itex]S \subseteq \emptyset[/itex]. 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 [itex]\emptyset \subseteq S[/itex] and [itex]S \subseteq \emptyset[/itex], then we can say that [itex]S=\emptyset[/itex] but I don't understand what you mean by, by hypothesis [itex]S \subseteq \emptyset[/itex].
 
DeadOriginal said:
I am trying to see how this works but I don't get it.

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

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 [itex]S\subseteq\emptyset[/itex] we are claiming that [itex]\emptyset[/itex] 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 [itex]S\subseteq\emptyset[/itex] we are claiming that [itex]\emptyset[/itex] 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 [itex]S\subseteq\emptyset[/itex].
 
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 [itex]S\subseteq\emptyset[/itex].

That is clearer, thank you. Do you accept that [itex]\emptyset[/itex] is a set?
 
DeadOriginal said:
Yes I do.

Since [itex]S[/itex] is a subset of every set and the empty set is a set, this mean [itex]S \subseteq \emptyset[/itex] (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?
 
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?
 
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.
 
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.
 
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 [itex]S \subseteq \emptyset[/itex] feel free to ask the other members. I just don't know how else to explain it :/
 
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.