Is the cardinality of $\varnothing$ undefined?

Question

It is intuitive that the cardinality of the empty set is $0$.

However we are asked to demonstrate this using given definitions/axioms in Tao Analysis I 4th ed ex 3.6.2.

My question arises as I think the given definitions are insufficient to cover the case of the empty set.

The following two definitions are given to us, in addition to the usual basic set theory axioms.

Def 3.6.1 (Equal cardinality) We say that two sets $X$ and $Y$ have equal cardinality iff there exists a bijection $f : X \to Y$ from $X$ to $Y$.

Def 3.6.5 Let $n$ be a natural number. A set $X$ is said to have cardinality $n$, iff it has equal cardinality with $\{i ∈ N : 1 ≤ i ≤ n\}$. We also say that $X$ has $n$ elements iff it has cardinality $n$.

My reasoning is as follows.

The empty set has no bijection with any other set, because it has no elements for a function to map to the codomain. Therefore the first Def 3.6.1 doesn't apply.
Therefore the cardinality of an empty set is not defined.

I would value your guidance on where I have gone wrong.

Answer 1

First, I'll address 3.6.1. Setting $X=\varnothing$, it says that the empty set has the same cardinality as $Y$ if there exists a bijection $\varnothing \rightarrow Y$.

Now, your concern is that the empty set is not in bijection with any other set. This is true, but the empty set is in bijection with itself. Therefore, it is valid to say that $\varnothing$ and $\varnothing$ have equal cardinality.

(Aside: Note that even if the condition were always false, the definition would still "apply" to the empty set, in the sense that you could state facts about whether it and other sets had equal cardinality. It's just that those assertions would always be false.)

Now, I'll address 3.6.5. If we set $n=0$ in the definition, it says that $X$ has cardinality $0$ if it has equal cardinality with the set $\{i \in \Bbb{N} : 1 \leq i \leq 0\}$. Well, what is this latter set? Since there is no $i$ such that $1\leq i \leq 0$, we must have $\{i \in \Bbb{N} : 1 \leq i \leq 0\} = \varnothing$.

Therefore, $X$ has cardinality $0$ if it has equal cardinality with $\varnothing$. Since the empty set has equal cardinality with itself (according to 3.6.1, as we discussed above), this statement is true of the empty set. So, the empty set has cardinality $0$.

EDIT: I just read the comments more carefully and noticed that you're not familiar with the bijection $\varnothing \rightarrow \varnothing$, so I'll go over that.

Recall that, formally, a function $A \rightarrow B$ is a subset $R \subseteq A \times B$ such that, for any $a \in A$, there exists a unique $b \in B$ such that $(a,b) \in R$. Well, if $A$ and $B$ are both the empty set, a function would be a subset of $\varnothing \times \varnothing$ for which this condition holds. Note that $\varnothing$ itself is a subset of this cross product, and it satisfies the condition vacuously: we have to check that it holds "for all $a$ in the domain...", but there are no such $a$ in the domain, so the statement is considered true.

Therefore, $\varnothing$ is a function from $\varnothing$ to $\varnothing$, which we call the empty map. You can also check that it is injective and surjective, because these conditions are vacuously true (just like the function condition). So, it is a bijection from the empty set to itself.

Answer 2

The initial ordinal, $0$, is the set of all of its predecessors. Since $0$ has no predecessors, the initial ordinal is $\varnothing$. The empty map from $\varnothing$ to $\varnothing$ is the function you are looking for.

Answer 3

Here is a possible resolution: Let A be a set with cardinality $n$.

$A\cup \emptyset=A$

$A\cap\emptyset=\emptyset$

According to theorem:

$|A\cup B|=|A|+|B|-|A\cap B|$ $|A\cup B|=|A|+|B|$

when $A\cap B=\emptyset$

(For exercise try to prove it).

Hence:

$|A\cup \emptyset|=|A|+|\emptyset|=|A|$

What do you deduce from that?

Answer 4

OP Doubt is valid , though easily resolved.

(1) Why Current Definition is enough :

$3.6.1$ talks about two sets having the Same Cardinality , without assigning a Natural number to that Cardinality.
Hence $X=Y=\phi$ will have the same Cardinality , which is as yet unknown.
OP is right that there is no Bijection between $\phi$ & every other set , hence $X=\phi$ & $Y \ne \phi$ will not have the Same Cardinality.

We can still claim that $X=\phi$ & $Y=\phi$ have the Same Cardinality.
[[ We have one trivial vacuous Bijection involving $\phi$ , Elaborate Discussion about this at the end ]]

$3.6.5$ assigns Cardinality $n$ when we have $1 \le i \le n$
Eg , we assign $n=1$ to $\{A\}$ & we assign $n=4$ to $\{A,B,C,D\}$
When we have $n=0$ , then $1 \le i \le n$ indicates $1 \le i \le 0$ , which is $1 \le 0$ , which is obviously the empty list. Hence $X=\phi$ works here.

Tao could have been a bit more explicit about this Case , though there is no major concern here.

(2) When we "Demand" that Current Definition is not enough :

Alternatively , let us say the Definitions are really not enough.
It is easy to add $3.6.0$ which "hardcodes" the Empty Set with Cardinality $0$.

(3) Why Current Definition is useful :

We want to make theorems about $\#(A \cup B)$ & $\#(A \cap B)$ (& more !) with $\#A$ & $\#B$ : Then those theorems should work even when $\phi$ occurs.

Hence we can not go with OP claim/thinking that "Cardinality of Empty Set is not defined" : All calculations should involve Natural numbers , with no Exceptions , within the finite Context.
Assigning $0$ to Empty Set makes us add & subtract $0$ where necessary.
Considering it undefined will make things more complicated later.

(4) What is the trivial vacuous Bijection involving $\phi$ :

There is no Bijection between $X=\{A,B,C\}$ & $Y=\{1,2,3,4\}$ , because we will consume all elements of $X$ while $Y$ will have 1 element left over.
There is no Bijection between $X=\{A,B,C,D\}$ & $Y=\{1,2,3\}$ , because $X$ will have 1 element with no matching element in $Y$.

When we have $X=\{A,B,C,D\}$ & $Y=\{1,2,3,4\}$ , then we can map $A$ to either of the 4 elements.
Then we can map $B$ to either of the 3 elements.
Then we can map $C$ to either of the 2 elements.
$D$ will map to the last element.
Hence we have $4!$ Bijections.
In general we have $n!$ Bijections.

Here is one way we can get the Bijection & reduce the number of elements in $X$ & $Y$ , until we get the Null Set :
$\#X=\#Y=4$ : $\{(A,3),(C,1),(B,2),(D,4)\}$ : $4!=24$ Possible Bijections
$\#X=\#Y=3$ : $\{(A,3),(C,1),(B,2)\}$ : $3!=6$ Possible Bijections
$\#X=\#Y=2$ : $\{(A,1),(B,2)\}$ : $2!=2$ Possible Bijections
$\#X=\#Y=1$ : $\{(A,1)\}$ : $1!=1$ Possible Bijection
$\#X=\#Y=0$ : $\{\}$ : $0!=1$ Possible Bijection

We can see that Empty Set has Exactly 1 trivial vacuous Bijection.
We have used all elements of $X$ , while no element of $Y$ is left over !
Hence $\phi$ has a Proper Bijection !