Set, Empty Set

Aug 2010
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Set: A named collection of objects.

Empty Set ( {}, or Ø ): Universal name (container) for a set with no objects. The empty set acquires a specific name when objects are added to it.

Example:
Let {} plus a,b,c be A={a,b,c}.
If a,b,c removed, A={}.

Named: identified, designated.
 
Aug 2010
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Some more thoughts on OP.

A set is defined by its contents, described or listed.

Named in broadest sense: denote, designate, identify, refer to, etc.

A common way of naming a set is to use {}. “{a,b,c}” is the name of {a,b,c}.

A is a subset of B if A only contains elements of B doesn’t work for {}, except in the negative sense, and doesn’t prove existence.

A is a subset of B if you can get A by removing none or more elements of B establishes existence of {}, gives that all {} are identical, and shows the empty set is a subset of every set.
 

Plato

MHF Helper
Aug 2006
22,508
8,664
Some more thoughts on OP.

A set is defined by its contents, described or listed.

Named in broadest sense: denote, designate, identify, refer to, etc.

A common way of naming a set is to use {}. “{a,b,c}” is the name of {a,b,c}.

A is a subset of B if A only contains elements of B doesn’t work for {}, except in the negative sense, and doesn’t prove existence.

A is a subset of B if you can get A by removing none or more elements of B establishes existence of {}, gives that all {} are identical, and shows the empty set is a subset of every set.
\(\displaystyle A\subseteq B\text{ if and only if }(\forall x)[ x\in A\Rightarrow x\in B]\)
 
Aug 2010
961
101
\(\displaystyle A\subseteq B\text{ if and only if }(\forall x)[ x\in A\Rightarrow x\in B]\)
The empty set has no members. I know, the negative proof, because A is not false it is true. But you have to prove that exhausts the options, ie,
You haven't shown the empty set exists. Std Def: Set is a collection of objects. No objects, no set.

EDIT I should have been more specific. There is no problem with your definiton for a non-empty set. It's the empty set that is a problem.

EDIT Actually, there is a problem with your standard definition of subset. {a,b} is a subset of {a,b,c}. Are a,b a subset of {a,b,c}? Why not? They are a collection of objects so they satisfy standard definition of a set. a,b are simultaneously members of {a,b,c} and a subset of {a,b,c}. This thread addresses this.
 
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Aug 2010
961
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I misconstrued “definition.”

A set is a collection of objects does not mean a collection of objects is a set. You have to specify the collection of objects as a set as in “consider the set a,b,c,” which is not the same as “consider the elements a,b,c.”

That leaves the empty set, which is not a collection of objects.
Defining a subset by removing objects from an existing set gives the existence of the empty set, and that the empty set is a subset of all sets. The standard definition of a subset given by Plato above then follows as a theorem.

In other words, A is a subset of B if you can get A by removing elements from B. Beautiful, except that the empty subset still does not satisfy the definition of a set as a collection of objects. Back to the first two posts.
 
Aug 2010
961
101
I misconstrued “definition.”

A set is a collection of objects does not mean a collection of objects is a set. You have to specify the collection of objects as a set as in “consider the set a,b,c,” which is not the same as “consider the elements a,b,c.”

That leaves the empty set, which is not a collection of objects.
Defining a subset by removing objects from an existing set gives the existence of the empty set, and that the empty set is a subset of all sets. The standard definition of a subset given by Plato above then follows as a theorem.

In other words, A is a subset of B if you can get A by removing elements from B. Beautiful, except that the empty subset still does not satisfy the definition of a set as a collection of objects. Back to the first two posts.
"Beautiful, except that the empty subset still does not satisfy the definition of a set as a collection of objects." No problem, it is a definition and exists by construction. So strike last sentence: "back to the first two posts."