Math Is Fun Forum

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Can I prove it this way?

Prove that the empty set is a subset of every set.

If a subset A has all the elements that of the universal Set U , then the complement of subset A is a empty set , then the empty set is in U , then it's a subset of U .

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Numbers are the essence of the Universe

Ricky

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Registered: 2005-12-04

Posts: 3,791

Re: Can I prove it this way?

If a subset A...

What is A a subset of?

then the complement of subset A is a empty set , then the empty set is in U

How do you reach this conclusion?

Typically, this is proved by the following.

For all x in A, if x is in B, then A is a subset of B.

This holds vacuously true for the null set.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Can I prove it this way?

A is a subset of U , then the complement of A must be a subset of U , since , A has all the elements that are in U , then the complement of A must be an empty , and is in U.

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Numbers are the essence of the Universe

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Can I prove it this way?

What I need to prove is  , Empty set is the subset of every set.

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Numbers are the essence of the Universe

Ricky

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Registered: 2005-12-04

Posts: 3,791

Re: Can I prove it this way?

I still don't understand your proof, but that's what I did.

For all x in A, if x is in B, then A is a subset of B.

This holds vacuously true for the null set.

Perhaps it is easier to see with an equivalent definition subset.

For all x not in B, x is not in A.

Let E be the empty set, A be some arbitrary set.  Then for all elements x (let alone only those not in A), x is not in E.  So E is a subset of A.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Can I prove it this way?

I undertand you prove ,
  My prove was,  like U={1,2,3,4} , Let A be a subset of U , and C be the complement of subset A, then C is also a subset of U,
If A={1,2,3} , then C={4}, if A has all the elements, then C is an empty set. 
If U is any random set, then there must be a subset C which is empty.

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Numbers are the essence of the Universe

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Can I prove it this way?

I understand you up to here:

if A has all the elements, then C is an empty set.
If U is any random set, then there must be a subset C which is empty.

Why must C be the subset?

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Can I prove it this way?

Because C is a complement of A , and A is a subset of U

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Numbers are the essence of the Universe

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Can I prove it this way?

Because C is a complement of A , and A is a subset of U

You sound like you're assuming the conclusion.  Any element in C must also be in U.  But the special case is where there are no elements in C.  This is what you're trying to prove.  By saying the same logic applies when C has no elements is the equivalent of saying that the null set is the subset of U, nothing new is shown.  You must show how this logic applies.

Also, don't call U the universal set, this can be misleading.  Let U be any arbitrary set.

Edit: Anyways, definition pushing is typically the way to go in most proofs, if you can.  Definitions are typically set up so that certain facts become trivial.   This is one of them.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Can I prove it this way?

well the way i understand this, a set A is a subset of B if for all elements in A, there is a matching element in B, the empty set has no elements, so by default any set includes the empty set? same way that any number + 0 is still the same number.

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JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Can I prove it this way?

The way to prove it – as Ricky has pointed out – is by a vacuous proof.

A statement of the form “if P then Q” is said to be vacuously true iff P is false. (It does not matter whether Q is true or not. As long as P is false, the whole of “if P then Q” is true.)

Now, the definition of “A is a subset of B” is: “For all x, if x is in A, then x is in B”. If you let A be the empty set and B be any set, you can clearly see that the statement “x is in A” is false for all x. Therefore the statement “if x is in A, then x is in B” is automatically true if A is the empty set. Hence A (the empty set) is a subset of B (any set).

(Note: Whether x is actually in B or not is immaterial. We are not interested in the truth value of “x is in B”; we are only interested in the truth value of “if x is in A, then x is in B”. And this is automatically (vacuously) true if “x is in A” is false.)

Last edited by JaneFairfax (2007-03-22 04:57:32)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Can I prove it this way?

I think Stanley is just trying to understand whether his proof works or not.  Not how to prove it.

Hopefully what I said made sense.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Can I prove it this way?

Yah , I understand .

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Numbers are the essence of the Universe