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The question arises from the following statements:

    * " $\varnothing$ is a subset of every set. This fact (that $\varnothing \subseteq A$ for any A) is "vacuously true" (...) " (Enderton - Elements of Set Theory)

Okay, so I understand that the empty set is included in every other set. So far so good.

    * "$\varnothing \in \{\varnothing\}$" (Enderton)

From here, it's possible to see that while $\varnothing$ doesn't contain elements at all, it is the element of the set containing the empty set.

    * "For any two sets there exists a set that contains both of them and nothing else" (Halmos - Naive Set Theory)

Okay, here is where I get into trouble. Since $\varnothing \in \{\varnothing\}$ , it follows that the empty set can be considered a member of another set. Namely, $(\forall A)(\varnothing \in A) $.

Or that's what I assume. If it's true, some weird things happen.

Putting it shortly, since the empty set will be an element of every set, the axiom of pairing won't be true: For every sets $a$ and $b$, a set $A$ will also contain the empty set. So (following Halmos' definition) "for any two sets there exists a set that contains both of them and nothing else" won't be true because a third element $\varnothing $ will always be present.

The same reasoning can be applied to the idea of a singleton: I think it's not possible for a set of a single element to exist if there's always an empty set right there, next to the so-called "unique element". Could be wrong about this one, too.

Any further explanation of the problems that arise will be, in my opinion, wasted time if the premise that caused them is false, so I'll leave it there until someone proves me that the statement is right. Else, the question is answered.

TL;DR: Is $(\forall A)(\varnothing \in A) $ true?

Thanks!

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  • $\begingroup$ No, it is not true. How did you come to the conclusion? $\endgroup$ – Tobias Kildetoft Mar 20 '14 at 8:57
  • $\begingroup$ @TobiasKildetoft The empty set is the element of the set containing the empty set, and this one belongs to {{{}}} and so on. It's accepted that the empty set if a subset of every other set, but just like {} belongs to {{}}, why won't it belong to the other sets if it's included as well? $\endgroup$ – Kevin Languasco Mar 20 '14 at 9:05
  • $\begingroup$ You should distinguish two different cases the first when we consider $\emptyset$ as a subset and in this case $\emptyset\subset A,$ for all $A$ is true and the second case when we consider $\emptyset$ as an element and in this case $\emptyset\in A$ for all $A$ isn't true. $\endgroup$ – user63181 Mar 20 '14 at 9:10
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    $\begingroup$ Element$\ne$subset. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 20 '14 at 9:16
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    $\begingroup$ I would like to point out that "For any two sets there is a set that contains both of them and nothing else" isn't about union. If you have two sets $A$ and $B$, then the set spoken of here isn't $A\cup B$, it's $\{A, B\}$. $\endgroup$ – Arthur Mar 20 '14 at 10:34
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The empty set is always a subset, but not always an element in a given set.

Say the set $A$ consists of the elements $a_1, a_2, \dots, a_n$. It could be infinite too.

Then you have another set which contains the empty set $B =\{ \emptyset \}$.

It is clear that $\emptyset \not \in A$ and $\emptyset \in B$.

The statement For any two sets there exists a set that contains both of them and nothing else doesn't mean that $A$ also includes the empty set, it means there is some set $C$ that contains everything in $A$ and $B$, for example $$C = A \cup B = \{a_1, \dots, a_n, \emptyset\}$$ for which it is true that $\emptyset \in C$. It says nothing about whether the empty set is a member of $A$ or not.

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  • $\begingroup$ So, is the only set that contains $\emptyset$ as an element $\{ \emptyset \}$ ? In this case, B $\endgroup$ – Kevin Languasco Mar 20 '14 at 9:20
  • $\begingroup$ Not the only one, obviously $C$ too. $\endgroup$ – naslundx Mar 20 '14 at 9:23
  • $\begingroup$ Oh of course, you're right. So any set can contain $\{ \emptyset \} $, but not $\emptyset $ directly (only by inclusion as a subset), unless this set is a collection that includes $\{ \emptyset \} $ and, sometimes, other sets. Is this right? $\endgroup$ – Kevin Languasco Mar 20 '14 at 9:34
  • $\begingroup$ Any set has $\emptyset$ as a subset, simply by picking out exactly 0 of the members. But $\emptyset$, or $\{\emptyset\}$, or $\{\{\emptyset\}\}$, ... may or may not be a member of a set. As you said, it depends on whether it is included or not. $\endgroup$ – naslundx Mar 20 '14 at 9:37
  • $\begingroup$ Well that answers my question. It's so clear now, I don't know why the empty set confused me when talking about belonging and inclusion. Thank you! $\endgroup$ – Kevin Languasco Mar 20 '14 at 10:23
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You are confusing the concept of membership with the concept of subset. They are not the same.

The empty set $\varnothing$ is a subset of every set, but it is not necessarily a member of a given set. If $B = \{ \varnothing \}$ (the singleton consisting of the empty set), then both $\varnothing \subseteq B$ and $\varnothing \in B$.

Suppose $A = \{1, 2, 3\}$. Then $$A \cup B = A \cup \{ \varnothing \} = \{ \varnothing, 1, 2, 3 \} \ne A.$$ Your misunderstanding is that you believe that for any set $A$, $A \cup \{\varnothing\} = A$ since $\varnothing \subseteq A$. But as I have pointed out, the property of subset is not the same as membership: for example, if $A = \{1, 2, 3\}$ and $B = \{1, 2 \}$, then $B \subset A$ but $B \not\in A$, because in order for the latter to be true, $A$ would need to contain an element which is the set $B$; e.g., if $C = \{1, 2, 3, \{1,2\}\}$, then $B \in C$.

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  • $\begingroup$ Thank you! I did know the difference between belonging and inclusion, but for some reason the empty set gave me trouble. Thanks for clarifying, again $\endgroup$ – Kevin Languasco Mar 20 '14 at 10:21

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