Understanding ZFC and the Axiom of Infinity: Simple Explanation and Examples

[Math Amateur]

I am reading Micheal Searcoid's book: Elements of Abstract Analysis ( Springer Undergraduate Mathematics Series) ...

I am currently focussed on Searcoid's treatment of ZFC in Chapter 1: Sets ...

I am struggling to attain a full understanding of the Axiom of Infinity which reads as shown below:View attachment 5055I am at a loss to see how this Axiom as stated leads to the definition of infinite sets ... indeed I am not even sure how it leads to the definition of finite sets! ... can someone please explain the mechanics of this in simple terms ... hopefully including a simple example(s) ...

It would also be good to know, in particular how all the natural numbers "can be gathered together in a set" ... ...

Help will be much appreciated ...

Peter

 

[Deveno]

Let's call our set $I$ (for *infinite*).

Since $\emptyset \in I$, $I$ contains a set with no elements (that is, we can form sets of zero cardinality).

Since $\emptyset \cup \{\emptyset\} = \{\emptyset\} \in I$, we know $I$ contains a set with just a single member (a *singleton* subset), so we can form sets of cardinality 1 (at least one such set. Later, when you learn how to define functions as sets, we can create other singleton sets by mapping from our one-element set as the domain of a function).

Similarly, we have a two-element set in $\{\emptyset\} \cup \{\{\emptyset\}\}$ which has the two distinct elements $\emptyset$ and $\{\emptyset\}$.

One can use induction to show that for any finite natural number $n$, that $n \cup \{n\}$ has $n+1$ elements (if you do this right, it's "true by definition").

Now there are many subtle ways to characterize "finite" (some of them quite involved), but as I indicated in another post, we can regard $I$ as a "superset" of the natural numbers, and one way to characterize finite is that there exists a bijection (again, this is a special kind of set) to some $n$ for $n \in \Bbb N$.

Now if $I$ were finite, there would be some bijection $f:I \to \{0,1,2,\dots,n-1\} = n$

Since $\Bbb N \subseteq I$, we have the inclusion injection $\iota$:

$k \mapsto k$

and then $f \circ \iota$ would be an injection of $\Bbb N$ onto $n$!. The impossiblity of this is sometimes called "the pigeonhole principle", we are forced to map the first $n$ elements of $\Bbb N$ to the different $n$ elements of $n$, and then $n+1$ has "nowhere to go", as all the image slots have been filled. And we cannot "double up", because $f \circ \iota$ must be injective!

(Note: these claims about injective and surjective functions and compositions actually entail some logical formalism to back them up, but they can be codified as well-formed formulae that can be quantified over).

Note that positing the existence of $I$ doesn't make $I$ BE the natural numbers, $I$ might be $\Bbb Z[x]$, or the reals, or the Euclidean plane. It just contains a COPY of the natural numbers, it might have "other stuff in it too".

 

[Math Amateur]

Let's call our set $I$ (for *infinite*).

Since $\emptyset \in I$, $I$ contains a set with no elements (that is, we can form sets of zero cardinality).

Since $\emptyset \cup \{\emptyset\} = \{\emptyset\} \in I$, we know $I$ contains a set with just a single member (a *singleton* subset), so we can form sets of cardinality 1 (at least one such set. Later, when you learn how to define functions as sets, we can create other singleton sets by mapping from our one-element set as the domain of a function).

Similarly, we have a two-element set in $\{\emptyset\} \cup \{\{\emptyset\}\}$ which has the two distinct elements $\emptyset$ and $\{\emptyset\}$.

One can use induction to show that for any finite natural number $n$, that $n \cup \{n\}$ has $n+1$ elements (if you do this right, it's "true by definition").

Now there are many subtle ways to characterize "finite" (some of them quite involved), but as I indicated in another post, we can regard $I$ as a "superset" of the natural numbers, and one way to characterize finite is that there exists a bijection (again, this is a special kind of set) to some $n$ for $n \in \Bbb N$.

Now if $I$ were finite, there would be some bijection $f:I \to \{0,1,2,\dots,n-1\} = n$

Since $\Bbb N \subseteq I$, we have the inclusion injection $\iota$:

$k \mapsto k$

and then $f \circ \iota$ would be an injection of $\Bbb N$ onto $n$!. The impossiblity of this is sometimes called "the pigeonhole principle", we are forced to map the first $n$ elements of $\Bbb N$ to the different $n$ elements of $n$, and then $n+1$ has "nowhere to go", as all the image slots have been filled. And we cannot "double up", because $f \circ \iota$ must be injective!

(Note: these claims about injective and surjective functions and compositions actually entail some logical formalism to back them up, but they can be codified as well-formed formulae that can be quantified over).

Note that positing the existence of $I$ doesn't make $I$ BE the natural numbers, $I$ might be $\Bbb Z[x]$, or the reals, or the Euclidean plane. It just contains a COPY of the natural numbers, it might have "other stuff in it too".

Thanks for your help Deveno ... still thinking over this post ...

Sorry to be slow in responding ... away from Tasmania ... traveling in state of Victoria ...

Peter