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I’ve just started second year at the University of Bristol, of which one of my modules is Metric Spaces.

I looked up what it means for a set to be well defined and also saw that $X = \{\emptyset\}$ is considered non-empty.

Continuing with that though process I wondered if the set defined below,

Let $X = \{\emptyset, \{\emptyset, \{\emptyset, ... \}\}\}$

is well defined.

My questions are is $X$ countable, could you define a metric on it and does it have any interesting properties or applications.

EDIT: Within ZFC set theory, $X$ violates the axiom of foundation. However, I still am interested in knowing if the set above has ... replaced with strictly finite repetitions, is defined as a set.

And if so, can you not define the set such that it has n repetitions. Then, the set would be defined $\forall n\in\mathbb N$, exept in the case of infinity

Thanks

djones
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    Can you define $X$ in precise terms? Does $X={\emptyset, X}$ ? Then the axiom of separation prevents $X$ from being a set – Jakobian Oct 18 '18 at 15:43
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    It looks like $X$ has only two elements, one of which is empty, and one of which is of infinite cardinality. – saulspatz Oct 18 '18 at 15:44
  • @saulspatz I half agree with you, the other set is again a set with two elements one of which is empty and the other has again two elements... Basically this set contains itself infinitely many times. – Yanko Oct 18 '18 at 15:46
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    I think the set $\omega$ is closer to what the OP has in mind. – eyeballfrog Oct 18 '18 at 15:47
  • @ Yanko Yes, you're right. I think I got lost in the braces. – saulspatz Oct 18 '18 at 15:47
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    I'm sorry, a mistake, it's axiom of foundation, not axiom of separation – Jakobian Oct 18 '18 at 15:51
  • 'Well definedness' is somewhat slippery notion and depends on contexts. So for your question to receive good response, it would better to include more contexts or examples. – seoneo Oct 18 '18 at 15:51
  • What’s the $\omega$ set? – djones Oct 18 '18 at 15:53
  • I don't think the OP is thinking that $X$ contains itself. I thing the OP is confused about the old set with sets as elements confusion. That set s/he describes is perfectly well defined. Is countable. And is actually in many context actually the set of natural numbers where $0 = \emptyset; 1 = {\emptyset} = {0}; 2 = {0, 1} = {\emptyset, {\emptyset}},$ etc. – fleablood Oct 18 '18 at 16:01
  • Looks like you are discussing the Von Neumann Ordinals. https://en.wikipedia.org/wiki/Natural_number#Von_Neumann_ordinals – Doug M Oct 18 '18 at 17:32
  • Such set does not exist!. This is because it contains itself. See This question for reference – Yanko Oct 18 '18 at 15:48
  • Perhaps it is almost worth noting that this is almost a technicality. The set doesn't exist because of the axiom of foundation, and there is no real "mathematical" reason to make that axiom part of ZFC. All of ordinary mathematics can be done inside ZFC without the axiom of foundation -- to simplify, the axiom of foundation exists because set theorists find sets like the one in the question pathological. – Mees de Vries Oct 18 '18 at 15:50
  • Thanks, is there a list of all the axioms (i’ve not covered any specific definitions of a set yet and there seem rather a few)? – djones Oct 18 '18 at 15:51
  • This Wikipedia page has a list of such https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory – Yanko Oct 18 '18 at 15:52
  • @JDunn I think you might be interested in the Axiom of infinity. – Yanko Oct 18 '18 at 15:54
  • "This is because it contains itself." Why do you say that? The OP is reusing a the same variable, $X$ is two context but they describe two different objects. $X = {\emptyset, {\emptyset}, {\emptyset,{\emptyset}},.....}$ is perfectly well defined. In fact, in ZFC context, $X = \mathbb N$. – fleablood Oct 18 '18 at 16:05
  • Oh, wait... no it isn't $\mathbb N$. The set the OP describes has only two elements. ANd the second element is ... $X$. Oh.. I see. The set is $X = {\emptyset, X}$ defined recursively. Yes, that is poorly/illegally defined.... – fleablood Oct 18 '18 at 16:08
  • @MeesdeVries: You've got it wrong. Foundation makes the universe smaller. Developing mathematics can be done in ZFC-Foundation, but in fact it can be done in ZFC (with Foundation). It's not that set theorists find $X={X}$ to be perverse somehow, it's just that you have no means to build that set, other than postulating its existence. And it's not very useful, in general, so we don't. – Asaf Karagila Oct 18 '18 at 18:03
  • @asafkaragila, the first few sentences seem to be in no way in conflict with what I said; I know that that is true, and it's what I was saying in my comment. The last sentence of your reply sounds to me like one possible interpretation of the word "pathological" (or "perverse") – Mees de Vries Oct 18 '18 at 18:14
  • @MeesdeVries: Well, it was mostly because you seemed to blame set theorists for that odd and technical axiom which is useless to anyone else... – Asaf Karagila Oct 18 '18 at 18:15
  • Who else came up with it? – Mees de Vries Oct 18 '18 at 18:19

1 Answers1

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If $X = \{\emptyset, \color{blue}{\{\emptyset,\{\emptyset\{..... \}\}\}}\}$ then $X$ has two elements. Element number 1: is: "$\emptyset$, and element number 2 is $ \color{blue}{\{\emptyset,\{\emptyset\{..... \}\}\}}$. So this would not just be countable. It would be finite. And it would not just be finite. It would have $2$ elements; Count them $2$ only.

But....

$ \color{blue}{\{\emptyset,\{\emptyset\{..... \}\}\}} = X$.

So your set is $X = \{\emptyset, X\}$.

This violates the current axioms of set theorem (ZFC) that do not allow sets to have themselves as elements (else we would get Russell's paradox of "the set of all sets that do not have themselves as elements"[1]).

However it would be well defined to define a set as:

$X = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\},....\}$.

We can more formally define this as

$X_0 = \emptyset$

$X_1 = \{X_0\}$.

$X_2 = \{X_0, X_1\}$.

....

$X_{k+1} = \{X_0, X_1, ...... X_k\}$

....

$X = \{X_0, X_1, ...........\}$

That is perfectly acceptable and well defined. Indeed in a set construction model, this is the very definition of the natural numbers. That is.

$0 = \emptyset$

$1 = \{ 0\}$

$2 = \{0, 1\}$

$k = \{0, 1, 2, ...., k-1\}$

$\mathbb N = \{0,1,2,3,......\}$.

....

Addendum:

The above construction is:

$0 \mapsto X_0 = \emptyset$

$k+1 \mapsto X_{k+1} = \{X_0, ....., X_k\} = \{X_0, ...., X_{k-1}\} \cup \{X_k\} = X_k \cup \{X_k\}$.

We go from $X_k \to X_{k+1}$ by adding one element set each step.

$|X_k| = k+1$ and we do this "infinitely" we get:

$\mathbb N \mapsto X = \{X_0, X_1, .....\}$.

And $|X| = |\mathbb N| = $ countably infinite.

From your comments it seems you are thinking of a different construction where we don't bootstrap a single larger set each step but that we consider every possible subset.

$Y_{\alpha} = X_0 = \emptyset$

$Y_0 = P(Y_{\alpha}) = X_1 = \{\emptyset, \{\emptyset\}\}$.

$Y_1 = P(Y_0) = \{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}$.

Notice $Y_1\ne X_2$ because $Y_1$ allows us to have have the element $\{\{\emptyset\}\}$ which is not allowable in our construction of $X_1$.

Notice each step of bootstrapping $X_{k+1} = X_k \cup \{X_k\}$ we are adding one element each step. With each step of bootstrapping $Y_{k+1} = P(Y_k)$ we are doubling the number of elements.

Thus $|Y_k| = 2^k$ (except for $\alpha$ but that was just notation for a starting point). As if $Y = \cup_{i = 0}^{\infty} Y_i = P(P(P(......(\emptyset).....)))$ then $|Y| = 2^{|\mathbb N|}$ which is uncountable.

Also if we have a natural construction of $\mathbb N\to X$ via $k \mapsto X_k$, we have a natural consturction of $(0, 2]\to Y$ via:

$1: \emptyset$

$\frac 12: \{\emptyset\}$

$ 1 + \frac 12: \{\emptyset, \{\emptyset\}\}$

....

$\frac 1 {2^k}: \underbrace{\{....\{}_{ktimes}\emptyset\}....\}$.

...

$\frac 1{2^k} + \frac 1{2^j} + ....: \underbrace{\{....\{}_{ktimes}\emptyset\}....\}\cup \underbrace{\{....\{}_{jtimes}\emptyset\}....\}\cup ...$

[1] Is the set of all sets that do not have themselves as elements, an element of itself or not?

fleablood
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  • That’s perfect, was trying to find a way to edit $X$ to be defined non-recursively. Additionally, does the way you’ve defined it makes it its own power set and as such, is there a similar property for the Naturals. Thanks – djones Oct 18 '18 at 16:30
  • In ZFC that is the natural numbers. Which... tends to be a bit abstract for students seeing it the first time. – fleablood Oct 18 '18 at 16:36
  • Also, cantor’s theorem says the power set of a countable set is uncountable, so is the set you define above uncountable. If so, how is it analogous to the naturals? – djones Oct 18 '18 at 16:37
  • The above has a very strict linear progressive structure. You can't ever leave out the empty set. If you have one set, you can't leave out a "lower" set. And you can't have arbitrary subsets in it. So the recursively defined set about follows the inductive axioms/definitions of the naturals. Now if we consider the set of all *possible* subsets of X, i.e. P(X) we will get an uncountable set. In this case $X = \mathbb N$ and $P(\mathbb N)\cong \mathbb R$ – fleablood Oct 18 '18 at 16:50
  • See, in the above the number $7$ is actually the set ${0,1,2,3,4,5,6}$. And $4 = {0,1,2,3}$ is an element of $7$ and it is also a subset of $7$. But the set ${0,1, 3,5}$ although it is a subset of $7$ it is not itself an element of $7$. So the potential subsets (which can be any collection of elements) far outnumber the elements (which must follow a specific structure). – fleablood Oct 18 '18 at 16:55
  • ${0, 1,3,} = $ ... bear with me ...$={\emptyset,{\emptyset,{\emptyset}},{\emptyset,{\emptyset,{\emptyset,{\emptyset}}}}}$ which is not an element of $X$. It is not an element of $X$ because we "skipped" the element $2 = ,{\emptyset,{\emptyset,{\emptyset}}}$ which needs to be included if we are also going to included the "higher" $3={\emptyset,{\emptyset,{\emptyset,{\emptyset}}}}$ – fleablood Oct 18 '18 at 17:04