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I think this question is not asked here. I apologize in advance if I am wrong.

I have the following two definitions (Joaquín Olivert, Estructuras de álgebra multilineal, 1996):

Class.- A class is an abstract object $C$ which permits to decide if their elements belong to it or not.

Set.- A class $C$ is said to be a set if there exists a class $D$ such that $C\in D$. A class which is not a set is called a proper class.

With this in mind, I'm trying to understand if $A=\{\{\emptyset\}\}$ is a set or not and why. It seems to me the answer should be not but with these definitions, I think $\{ A \}$ is a class and then $A$ is a set. On the other hand, with this reasoning, every class would be a set and that is false (e.g. Russell's set).

Any help please?

Thanks

EDIT.

It's been a long long time, but I think it's never to late to add some context. I had just attended a seminar on set theory when I asked this question. The speaker introduced ordinals, and gave the example of $\{\{\emptyset\}\}$ as something which is not an ordinal. I mistakenly understood set (and hence proper class as a counterexample).

As it has been mentioned on one answer, there is an axiom in one of the theories (I don't remember which one) that says

If $A$ is a set, then $\{A\}$ is also a set.

This answers the question.

Nevertheless, I am grateful to those who provided other comments or answers, it helped me learn a lot about Set Theory. Thank you

Dog_69
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2 Answers2

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In ZF(C) set theory, the Axiom of Pairing states that if $x$ and $y$ are sets, then $\{x,y\}$ is a set too. Applying this with $x=y$ gives that whenever $x$ is a set, $\{x\}$ is a set too (because it has the same elements as $\{x,x\}$ and is therefore equal to it).

Thus,

  • $\varnothing$ is a set because of the Axiom of the Empty Set. (Sometimes this is not considered an axiom, in which case $\varnothing$ is still a set because the Axiom of Infinity states that there exists some set, and then the Axiom of Separation can be used to remove all of its elements).
  • $\{\varnothing\}$ is a set because of Pairing applied to $\varnothing$.
  • $\{\{\varnothing\}\}$ is a set because of Pairing applied to $\{\varnothing\}$.

Your definition of "set" sounds like it doesn't belong to pure ZF(C), but more likely to NBG or MK set theory. However, these theories have similar axioms of Empty Set and Pairing, so the argument (at least at the non-formal level I'm writing at) is the same.

hmakholm left over Monica
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$A=\{\{\varnothing\}\}$ is a set, not a proper class, because we have $A=\{\{\varnothing\}\}\in\{A\}=\{\{\{\varnothing\}\}\}.$ And $\{A\}$ is not a class either, because we have $\{A\}\in\{\{A\}\}.$

Axiomatically, one usually assumes that $\varnothing$ is a set, and for any set, the singleton containing that set is a set (for example via the axiom of pairing), so $\{\varnothing\},$ $\{\{\varnothing\}\}$, etc are all sets, and not proper classes.

In the usual universes of axiomatic set theory, for something to be a proper class, and not a set, it must be very big. For example the class of all sets, the class of all cardinals, or the class of all ordinals. These classes are not contained in any other classes.

ziggurism
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  • Wait... who says that, in this framework, ${{{ \emptyset }}}$ is a class? – Stefan Mesken Jan 09 '18 at 22:07
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    "one usually assumes" [emphasize is mine] My point exactly. There isn't sufficient information to answer OP's question. Who are we to judge what OP or the author considers to be a set/class if he doesn't tell us? Is ${{{\emptyset}}}}$ an abstract object that allows us to decide if their members belong to it? I don't know... sounds like utter nonsense to me. – Stefan Mesken Jan 09 '18 at 22:13
  • @StefanMesken If you need to make some assumptions in order to make the question answerable, I think it is allowable to do so, as long as you state your assumptions. Hopefully my answer has done that. If not, please feel free to let me know how it may be improved. – ziggurism Jan 09 '18 at 22:18
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    @StefanMesken I don't' know if the following helps you but the author says: ''Esentially, the axiomatic we are going to follow is ZF together with notable contributions of Neumann-Bernays-Gödel (...) and some details given by J. L. Kelly (...).". – Dog_69 Jan 09 '18 at 22:23
  • It's not really clear what you have assumed. For example you say "For something to be a proper class, and not a set, it must be very big." That's not true at all. Classes are 'very big' in most common set theories but they certainly don't need to be. Singletons can be classes in the right framework. – Stefan Mesken Jan 09 '18 at 22:23
  • @Dog_69 Yes, that helps a lot and should be added to your question! – Stefan Mesken Jan 09 '18 at 22:23
  • I don't know practically anything about Set Theory, but ziggurism and Henning Makholm's answers seems to me clear, correct and very rigorous (may be except for the intuitive explanation of proper classes as an enormous families). – Dog_69 Jan 09 '18 at 22:27
  • @ziggurism: With your first reasoning, I think you can prooved the Russell's set is in fact a set, since $R\in{R}$ and ${R}\in {{R}}$. I mean, I think the correct reasoning is the second. – Dog_69 Jan 09 '18 at 22:39
  • @Dog_69 my first argument is implicitly assuming that ${{{\varnothing}}}$ is a set, and therefore its element is a set. That assumption is justified more explicitly in my second paragraph. There is no such argument for ${R}R$. – ziggurism Jan 10 '18 at 19:00
  • @ziggurism Okey – Dog_69 Jan 10 '18 at 19:24