0
$\begingroup$

I'm trying to understand Bourbaki's definition of universal quantification. The definition is on Page 36 in Theory of Sets as follows:

$(\exists x R) \equiv (\tau_{x} \mid x) R$

$(\forall x R) \equiv \lnot ((\exists x) \lnot R)$

For example:

$R = \in x y$

The resulting formula is: $\lnot (\lnot \in (\tau \space \lnot \in \square y) y)$

The resulting tree is: Bourbaki Tree Universal Quantification

It's my understanding the $\square$ is a distinguished object that satisfies the evaluated truth of $\tau$. Therefore, if there is a distinguished object $\square$ that satisfies $\tau$, i.e. a object that does not satisfy $\in x y$, the universal quantification is false.

Does the quantification function $\tau$ try every object in the domain of discourse? If so, and all the objects in the domain of discourse do not satisfy $\tau$ (i.e. universal quantification should be true), what does $\tau$ evaluate to?

It seems that $\tau$ must evaluate to an object that satisfies $\in x y$ for the formula to be true, however I'm unsure how this occurs as every object in the domain of discourse must be tested first for universal quantification to be true.

Any guidance here appreciated. Thanks

$\endgroup$
7
  • 1
    $\begingroup$ Why are you reading this in Bourbaki? $\endgroup$ – Asaf Karagila May 14 '19 at 16:07
  • $\begingroup$ @Asaf Karagila Of course there is an alternate definition given by Church: $\forall x$ ___ is True if the value of ____ is True for all values of x. $\forall x$ ___ is False if the value of ____ is False for any value of x. Also defined $\lnot ((\exists x) \lnot R)$. $\endgroup$ – user644059 May 14 '19 at 16:24
  • $\begingroup$ You haven't answered my question. Also, I believe that is due to Tarski, not Church. $\endgroup$ – Asaf Karagila May 14 '19 at 16:26
  • $\begingroup$ @Asaf Karagila I'm interested in the Bourbaki view. The definition is in Church's Introduction to Mathematical Logic, however could be from someone else. $\endgroup$ – user644059 May 14 '19 at 16:27
  • 2
    $\begingroup$ @Nick Except for historical curiosity, there's almost no reason to read Bourbaki's Theory of Sets. A term of length 4,523,659,424,929 gives a pretty clear critique that the formal system presented is poorly made resulting in large amounts of needless complexity. The discussions referenced and had here point to other issues. Theory of Sets is not particularly necessary for the later volumes if those are your real goal. $\endgroup$ – Derek Elkins left SE May 14 '19 at 19:19
1
$\begingroup$

The concept here is that by some means, $\tau_x(R)$ picks - once for all time - an otherwise completely arbitrary value of $x$ such that $R$ is true. If there is no value of $x$ for which $R$ is true, then $\tau_x(R)$ simply picks an arbitrary object.

Think of it as some munificient deity has compiled a list of all possible relations involving $x$, and for each assigned at random a value that makes it true, provided that such a value exists. Otherwise, it assigned a value completely at random. Once this list and assignments are established, $\tau_x(R)$ will always be the value assigned to $R$ in the list. Since the values assigned are random, when $R(\tau_x(R))$ is true, the only things that are knowable about $\tau_x(R)$ are the theorems that can be proved from $R(\tau_x(R))$. And when $R(\tau_x(R))$ is false, the only things knowable about $\tau_x(R)$ are things than can be proven for all values.

Of course, these are just the intuitions behind the operator. From the formalist view of Bourbaki in this book, it is really just strings of symbols being manipulated in accordance with certain rules.

But say what you will about their shortcomings (and I agree), I will still hold a fond spot for formalism in general and Bourbaki's in particular, for first demonstrating to me that the question of "what is the true mathematics" is meaningless. As long as the axioms you've chosen are consistent, your theory is just as "true" as any other. That is mathematical freedom.

$\endgroup$
13
  • $\begingroup$ I get the first part about the random value that makes it true. I'm confused on the statement that false "it assigned a value completely at random". The statement in the book says if an object doesn't satisfy $\tau_x (R)$ then it "represents an object about which nothing can be said". So something other than empty set? Also, I'm still confused on how the tree above evaluates to true, which would show $\forall x$ is a true relation. Thanks $\endgroup$ – user644059 May 15 '19 at 0:52
  • 1
    $\begingroup$ @Nick $\tau_x(\bot)$ is basically like adding a new constant to the set theory, call it $c$. This new constant has no axioms associated with it. You don't know anything about it. This includes negative information. You don't know if $c=\varnothing$, but you also don't know if $c\neq\varnothing$. For every predicate, $P$, that isn't either true for every set or false for every set, $P(c)$ is undecidable. That's what it means by "an object about which nothing can be said". $\endgroup$ – Derek Elkins left SE May 15 '19 at 1:12
  • 1
    $\begingroup$ Exactly as Derek says. When $R$ is always false, $\tau_x(R)$ is just a random object. It does not need to be some new thing. "An object about which nothing can be said" just means we are unable to prove anything about this object from $\lnot R(\tau_x(R))$. It can have properties, but we have no way of knowing what those properties are. $\endgroup$ – Paul Sinclair May 15 '19 at 1:25
  • $\begingroup$ As for your tree, it is the tree of the expression $(\forall x)(x \in y)$. Why would you expect this to evaluate as true? You've not even defined $y$ yet, so how would you expect to prove anything about it. Later on, Bourbaki will prove $(\forall y)(\exists x)(\lnot(x \in y))$. $\endgroup$ – Paul Sinclair May 15 '19 at 1:38
  • $\begingroup$ @Paul Sinclair For the tree, $y$ can be defined to be a set that all $x$ is a member of. $\endgroup$ – user644059 May 15 '19 at 20:35
0
$\begingroup$

See page 20 :

Let us consider the assertion $B$ as expressing a property of the object $X$; then, if there exists an object which has the property in question, $\tau_X(B)$ represents a distinguished object which has this property; if not, $\tau_X(B)$ represents an object about which nothing can be said.

The source is the so-called Hilbert's Epsilon Calculus :

The intended interpretation is that $ε x A$ denotes some $x$ satisfying $A$, if there is one.

Quantifiers can be defined as follows:

$$∃x A(x) \equiv A(ε x A)$$

$$∀x A(x) \equiv A(ε x (¬ A)).$$

Compare with page 36 :

$(\tau_x(R) \mid x)R$ is denoted by "there exists $x$ such that $R$".

The intuition is (quite) simple : if there are some objects $X$ such that $A$ holds of them, the "choice operator" $\tau_X$ will pick up one of them and obviously $A$ will hold of it.


Consider now the example with the universal quantifier : $\forall x R(x)$ [where $R(x)$ is $\in (x,y)$, i.e. $(x \in y)$].

If $\forall x R(x)$ holds, this means that $\lnot R(x)$ holds of no object.

According to the above specification, $\tau_x (\lnot R)$ will pick up an object whatever, and $\lnot R(x)$ does not hold of that object, i.e. $\lnot [((\tau_x (\lnot R)) \mid x) (\lnot R)]$.

In plain language : if $\lnot R$ does not hold of an object whatever, this means that $R$ holds of every object.

$\endgroup$
1
  • $\begingroup$ Given this, in the tree above where an object does not satisfy $\tau$, $\tau$ evaluates to "an object about which nothing can be said". I'm confused however, because I would suspect that $\forall x$ evaluates to true if every $x$ satisfies this quantification. $\endgroup$ – user644059 May 14 '19 at 17:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy