A lacking understanding of the basic axiomatic construction of $\emptyset$

Question

I am truly embarrassed to admit my lacking knowledge of the Set Theory. Something always boggled my mind:

Suppose that we have an empty set $\emptyset$. By extension, we can write it as $$\emptyset = \{\}$$ which denotes the fact, which clearly states that $\emptyset$ is NOT a mathematical "nothing", rather it is a set whose power is $0$ (please, correct me if I'm wrong). We write $$\mathrm{card}(\mathcal{A}) = 0 \Longleftrightarrow \mathcal{A} = \emptyset.$$ As previously stated, an empty set is a set, which contains "nothing", but is not "nothing" by itself. Here I bump into a nasty problem. I immediately saw an iterative extension, which is (because of my lacking knowledge) very disturbing: $$\emptyset = \{\} \neq \{\emptyset\} \neq \{\{\emptyset\}\} \neq \{\cdots \{ \emptyset \} \cdots\}.$$

Let me explain the notation: Suppose that we have a set $\mathcal{A}$ which includes an empty set $\emptyset$. By standard deduction we conclude that $\mathrm{card}(\mathcal{A}) = 1$. But if an empty set is something that includes "nothing" shouldn't be any number of iterations of empty sets be considered "empty" only at the last "iteration"? Therefore, we can have a custom $\mathrm{card}(\mathcal{A})$ which really consists only of "nesting" one empty set.

I kindly ask you, can you present me the physiology of an empty set so that all the ambiguities (I am of course guilty of them) will be cleansed? I apologise if my question is too broad, but I can't put it another way.

EDIT: It would be helpful if you defined a mathematical "nothing" (is there such an area of mathematics that deals with the idea of "nothing"; sorry if I'm getting too philosophical).

Answer 1

Let me see if I can explain how one proves that each of the sets in that chain is distinct from the next.

First, you're quite correct that $\emptyset\neq \{\emptyset\},$ by Extensionality, since $\emptyset\notin\emptyset.$

Now, let's say we've just shown that one of the sets in the iteration isn't equal to the next. That is, we've just shown that $A\neq B,$ where $B=\{A\}.$ The next set in the iteration will be $\{B\}.$ As we have just shown that $A\neq B,$ then by Extensionality, we have $\{A\}\neq\{B\},$ meaning $B\neq\{B\}.$ Iterating this argument takes us as far along the chain as we care to go.

In fact, we can go further and say that none of the sets in the chain are equal, but this takes quite a bit more work. However, Extensionality immediately shows that $\emptyset$ is distinct from all other sets in the chain, since the rest have (exactly) one element.

Answer 2

$P(A)$ is the power set of $A$ or the set of all subsets of $A.$

The empty set is a subset of every set. And every set is a subset of itself.

$P(\emptyset) = \{\emptyset\} \ne \emptyset.$

For any set $A$ the cardinality of the power set.

$|P(A)|=2^{|A|}$

and

$|P(\emptyset)|=1$

Answer 3

The set $\{\emptyset\}$ does not contain nothing; it contains the empty set $\emptyset$. As you correctly observe, "$\emptyset$ is NOT a mathematical "nothing,"" so $\{\emptyset\}$ has one element. Note that in particular this means $\emptyset\not=\{\emptyset\}$ (as again you correctly observe); among other things, this means $\{\emptyset, \{\emptyset\}\}\not=\{\emptyset\}$.

A common analogy here is to think of sets as boxes. $\emptyset$ is an empty box; $\{\emptyset\}$ is a box containing another box, and that other box is empty. Clearly these are different objects. I find this analogy misleading in general (for instance, a box containing two empty boxes is different from a box containing one empty box!), but in this case I think it can be illuminating.

Answer 4

An empty set is not nothing ... it is a set that contains nothing, which is different. Think of it as an empty bag .. which is a bag, not nothing.

So: $\emptyset\not = nothing$

And therefore: $\{ \emptyset \} \not = \{ nothing \} = \{ \} = \emptyset$

so $\{ \emptyset \} \not = \emptyset$