Explaining the empty set: subset or element of set

Greetings, I have a question regarding the difference between an element of a set and a subset, with regards to the empty set. What is the difference between the two statements:

**(a) Is x an element of {x} and (b) Is x a subset of {x} ?**

I understand that the empty set is a subset of every set, but I don't see why it, as a subset, is not considered an element of every set.

Thank you in advance!

Re: Explaining the empty set: subset or element of set

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Originally Posted by

**aprilrocks92** Greetings, I have a question regarding the difference between an element of a set and a subset, with regards to the empty set. What is the difference between the two statements:

**(a) Is x an element of {x} and (b) Is x a subset of {x} ?**

I understand that the empty set is a subset of every set, but I don't see why it, as a subset, is not considered an element of every set.

The statement that "the empty set is an element of every set" is FALSE. It is not.

Let $\displaystyle S=\{x,y,z\}$ then $\displaystyle \emptyset\notin S~.$ It is not listed there.

Let $\displaystyle T=\{x,y,\emptyset\}$ then $\displaystyle \emptyset\in T~.$ It is listed there.

Re: Explaining the empty set: subset or element of set

Rudin says the empty set is a subset of every set because you can't show that it is not a subset. I know, tricky.

OK. Plato is right. A subset of a set is not a member of a set unless specifically included in the set. {a,b} is a subset of {a,b,c} but not a member. {a,b} is a subset of {a,b,{a,b},c} and a member.

Sorry, repititious, just trying to fix it in my mind.

Re: Explaining the empty set: subset or element of set

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Originally Posted by

**aprilrocks92** I understand that the empty set is a subset of every set, but I don't see why it, as a subset, is not considered an element of every set.

The __contents__ of the empty set (nothing) is an element of every set. I don't know if there is a symbol for the contents of the empty set.

Re: Explaining the empty set: subset or element of set

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Originally Posted by

**aprilrocks92** but I don't see why it, as a subset, is not considered an element of every set.

You seem to be under the impression that a "subset" is an "element". That is not true. An "element of a set" is one of the members of the set. A subset is a set of objects that happen to also be in the original set. {x} contains the **element** x. x, here, is not a "subset" of {x} because it is NOT a set!

Re: Explaining the empty set: subset or element of set

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**Hartlw** The empty set is a subset of every set. I know, tricky.

It is not tricky at all:

$\displaystyle (\forall x~\&~\forall A)[\text{ if }x\in\emptyset\text{ then }x\in A\text{ is a true statement.}]$ A false statement implies any statement.

That is the very definition of subset so $\displaystyle \emptyset\subseteq A$.

Re: Explaining the empty set: subset or element of set

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Originally Posted by

**HallsofIvy** (aprilrocks92) seems to be under the impression that a "subset" is an "element". That is not true. An "element of a set" is one of the members of the set.

Actually, I assumed that aprilrocks92 may have been studying transitive sets.

Much like the *enlarged universe* of non-standard analysis.

Let $\displaystyle U=A\cup \mathcal{P}(A)$. Now for any set $\displaystyle A$ we have $\displaystyle \emptyset\in U~\&~\emptyset\subseteq U$.

Re: Explaining the empty set: subset or element of set

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Originally Posted by

**Plato** It is not tricky at all:

$\displaystyle (\forall x~\&~\forall A)[\text{ if }x\in\emptyset\text{ then }x\in A\text{ is a true statement.}]$ A false statement implies any statement.

That is the very definition of subset so $\displaystyle \emptyset\subseteq A$.

You quoted me out of context. That is not what I meant by tricky.

The tricky part is the proof, which is basically: “the assumption that ɸ is not a subset is false, therefore ɸ must be a subset.” The problem with that standard proof is you haven’t shown that ɸ exists in the first place.

Re: Explaining the empty set: subset or element of set

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**Hartlw** the assumption that ɸ is not a subset is false, therefore ɸ must be a subset.” The problem with that standard proof is **you haven’t shown that ɸ exists in the first place.**

You are a person after R L Moore's own heart.

He has been called the greatest math teacher ever by Keith Devlin.

Moore was one of the founding fathers of *Point Set Topology*. His foundational book has no empty point set in it. He was stead fast in his denial it was possible. Look at the list of his students. You see there some of the most important mathematicians of the 20th century. On that list is Mary Ellen Rudin who was married to your offed quoted Walter Rudin. Moore is the only person who has had five of his students to be president of the MAA.

Of course, all of mathematics is the result of the human brain. It is from definitions and/or axioms. Existence is not an issue. The empty set is very much part of modern mathematics.

Re: Explaining the empty set: subset or element of set

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Originally Posted by

**Plato** You are a person after

R L Moore's own heart.

He has been called

the greatest math teacher ever by Keith Devlin.

Moore was one of the founding fathers of

*Point Set Topology*. His foundational book has no empty point set in it. He was stead fast in his denial it was possible. Look at the list of his students. You see there some of the most important mathematicians of the 20th century. On that list is Mary Ellen Rudin who was married to your offed quoted Walter Rudin. Moore is the only person who has had five of his students to be president of the MAA.

Of course, all of mathematics is the result of the human brain. It is from definitions and/or axioms. Existence is not an issue. The empty set is very much part of modern mathematics.

Thanks for that post. I really didn't appreciate it until exercising the subject:

http://mathhelpforum.com/peer-math-r...empty-set.html

Basically,

Remove a,b,c from the collection of objects a,b,c and you are left with nothing.

Remove a,b,c from the set {a,b,c} and you are left with {}.