prove without using contradiction.that the empty set is the subset of every set

You should be able to do this from the definitions of empty set and subset. Show your work and we can help you find the answer.

You should be able to do this from the definitions of empty set and subset. Show your work and we can help you find the answer.

 

What is the definition of the empty set??

It is the set having no elements.

It is the set having no elements.

 

That does not help in the above proof

First of all, you very specifically asked "What is the definition of the empty set," triclino. Mr. Skeptic answered.

 

Secondly, that definition does help. The set {D} is not a subset of the set {A,B,C}. Nor are {A, E} and {A,B,C,F}. Why is that?

Like I said, you need both definitions.

Perhaps i should have mentioned in my 1st post,that i have in my mind a proof.

 

But i am interested to know if there are any other proofs.

 

 

Apart of those that Wikipedia offers

Don't beat around the bush then! Post your proof.

 

You are the one with the obsession over trivial proofs, triclino. Nobody else really cares.

O.k this a trivial proof and i am waiting for another trivial proof :

 

Suppose x does not belong to the set ,A .But x does not belong to the empty set.

 

Hence if x does not belong to A ,then x does not belong to the empty set .

 

Thus : [math]\emptyset\subseteq A[/math]

O.k this a trivial proof and i am waiting for another trivial proof :

 

Suppose x does not belong to the set ,A .But x does not belong to the empty set.

 

Hence if x does not belong to A ,then x does not belong to the empty set .

 

Thus : [math]\emptyset\subseteq A[/math]

 

That is almost correct, but you are missing one line. The definition of subset is:

 

A is a subset of B means that for all x in A, x is in B.

 

You haven't come up with that precise statement for the empty set and A, so you still need one more line. You are quite close, though.

=Uncool-