MHB ZFC and the Axiom of Foundation

[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 Foundation which reads as shown below:https://www.physicsforums.com/attachments/5072Can someone explain this Axiom and give some simple examples ...

I am perplexed by my own example ... as follows ...

Consider the set $$a = \{ 1, 2, 3 \}$$

[Note that $$1, 2, 3$$ are sets - previous post by Deveno ... ]

Now $$1 \cup a = 1 $$

and $$2 \cup a = 2$$

and $$3 \cup a = 3$$

... ? ... what is the member of a which is disjoint from a ...

Can someone clarify this issue and explain how the Axiom works ...

Hope someone can help ..

Peter

 

[Deveno]

Let's look at a very simple set:

$A = \{a\}$.

Since $a \in A$, the axiom applies, because $A$ is non-empty. So we have an element of $A$ which is disjoint from $A$. Well, the only element to be found is $a$. So:

$a \cap A = \emptyset$. Let's write this a bit differently:

$a \cap \{a\} = \emptyset$.

I claim this means $a \not\in a$.

For suppose we had $a \in a$. Since $a \in A$, this would mean $a$ and $A$ have a common element, $a$. But this violates:

$a \cap A = \emptyset$.

The purpose of this, is to establish that there is no such set as:

$S = \{S,\{S,\{S,\dots,\}\}\}$

in other words "the buck stops somewhere" (this principle is called "foundation" or "well-founded-ness").

It establishes that $a$ and $\{a\}$ are always *distinct* sets. This distinction is often glossed over in analysis, where the real number $x$, and the singleton subset $\{x\}$ are often conflated (by such vague language such as "consider the point $x$").

Interestingly enough, if $\phi(x)$ is the predicate; $x \not\in x$, we see that $\phi(x)$ is true for any non-empty set $x$.

So if the collection of all sets was a set, it would be a Russell set, whose very existence negates itself. This contradiction shows the collection of all sets is not a set.

As for your example, you are confusing $1$ with $\{1\}$. For example:

$\{1\} \cap a = \{1\}$, but:

$1 \cap a = \emptyset$ (since $0 \not\in a$).

 

[Math Amateur]

Let's look at a very simple set:

$A = \{a\}$.

Since $a \in A$, the axiom applies, because $A$ is non-empty. So we have an element of $A$ which is disjoint from $A$. Well, the only element to be found is $a$. So:

$a \cap A = \emptyset$. Let's write this a bit differently:

$a \cap \{a\} = \emptyset$.

I claim this means $a \not\in a$.

For suppose we had $a \in a$. Since $a \in A$, this would mean $a$ and $A$ have a common element, $a$. But this violates:

$a \cap A = \emptyset$.

The purpose of this, is to establish that there is no such set as:

$S = \{S,\{S,\{S,\dots,\}\}\}$

in other words "the buck stops somewhere" (this principle is called "foundation" or "well-founded-ness").

It establishes that $a$ and $\{a\}$ are always *distinct* sets. This distinction is often glossed over in analysis, where the real number $x$, and the singleton subset $\{x\}$ are often conflated (by such vague language such as "consider the point $x$").

Interestingly enough, if $\phi(x)$ is the predicate; $x \not\in x$, we see that $\phi(x)$ is true for any non-empty set $x$.

So if the collection of all sets was a set, it would be a Russell set, whose very existence negates itself. This contradiction shows the collection of all sets is not a set.

As for your example, you are confusing $1$ with $\{1\}$. For example:

$\{1\} \cap a = \{1\}$, but:

$1 \cap a = \emptyset$ (since $0 \not\in a$).

Hi Deveno ... thanks for the help ...

Still reflecting on your post ...

But just one quick question ... ...

... you write ...

" ... As for your example, you are confusing $1$ with $\{1\}$. For example:

$\{1\} \cap a = \{1\}$, but:

$1 \cap a = \emptyset$ (since $0 \not\in a$). ..."

I do not follow your reasoning in arguing that $1 \cap a = \emptyset$ because $0 \not\in a$ ... can you explain this further ... how exactly does the fact that $0 \not\in a$ imply that $1 \cap a = \emptyset$ ...

Hope you can clarify and explain this issue ...

Peter

 

[Deveno]

As a set, $1 = \{0\}$.