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
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?