8

I'm reading the first book by Bourbaki (set theory) and he introduces this logical symbol $\tau$ to later define the quantifiers with it. It is such that if $A$ is an assembly possibly contianing $x$ (term, variable?), then $\tau_xA$ does not contain it.

Is there a reference to this symbol $\tau$? Is it not useful/necessary? Did it die out of fashion?

How to read it?

At the end of the chapter on logic, they make use of it by... I don't know treating it like $\tau_X(R)$ would represent the solution of an equation $R$, this also confuses me.

Community
  • 1
Nikolaj-K
  • 12,249
  • 4
    Just out of sheer interest, what reason do you have for studying Bourbaki's set theory? – Asaf Karagila Aug 01 '12 at 09:15
  • @AsafKaragila: I want a concise introduction which states all the things it uses. I'd be thankful for a more modern reference!! Now I'm using Bourbaki and Jech and wikipedia. I try to write the things down while reading it to get a good overview, and it might be a little pedantic but when e.g. Jech starts writing "$a\ne b$" without saying that this symbol "$\ne$" is supposed to express the negation of "$a=b$" I get uneasy. As a remark, I'm really a physicist and I never took a logic/set-theory lecture. – Nikolaj-K Aug 01 '12 at 09:27
  • 1
    I think $\tau$ is supposed to be some kind of choice operator, but I haven't looked at the definition. (I don't read French.) As you have surmised, its use is unfashionable nowadays. – Zhen Lin Aug 01 '12 at 09:27
  • 1
    @Nick: You remind me one of my teachers which is extremely pedantic. When I took the advanced logic course we spent two weeks proving that there is a unique way to interpret a sentence, even without parenthesis everywhere. He even had two different symbols for equality, one in the language and one in the meta-language. On the other hand, when I was 12 or so I already knew the symbol $\neq$, so I suppose using it without specifying what its meaning is not the worst thing that can happen in a book. – Asaf Karagila Aug 01 '12 at 09:33
  • @AsafKaragila: I know the symbol too of course, it's obvious. But I care about the definition for implementation reasons. A priori "$\ne$", "$\int$" and "$\Gamma$" could all mean the same. Suddently writing "$a\ne b$" is the same as "my_cat_loves_fish(a,b)" for an AI, say. I'd like a book which spares me the work to come up with all these things "$(a\ne b)\ :>\ \text{not}(a=b)$" myself. PS: I think you told me about your teacher in my set builder notation bracket question. – Nikolaj-K Aug 01 '12 at 09:37
  • Found it here. – Nikolaj-K Aug 23 '14 at 13:37
  • See in SEP : The Epsilon Calculus. – Mauro ALLEGRANZA Sep 11 '14 at 12:17

2 Answers2

5

Adrian Mathias offers the following explanation here:

Bourbaki use the Hilbert operator but write it as $\tau$ rather than $\varepsilon$, which latter is visually too close to the sign $\in$ for the membership relation. Bourbaki use the word assemblage, or in their English translation, assembly, to mean a finite sequence of signs or leters, the signs being $\tau$, $\square$, $\lor$, $\lnot$, $=$, $\in$ and $\bullet$.

The substitution of the assembly $A$ for each occurrence of the letter $x$ in the assembly $B$ is denoted by $(A|x) B$.

Bourbaki use the word relation to mean what in English-speaking countries is usually called a well-formed formula.

The rules of formation for $\tau$-terms are these:

Let $R$ be an assembly and $x$ a letter; then the assembly $\tau_x(R)$ is obtained in three steps:

  1. form $\tau R$, of length one more than that of $R$;
  2. link that first occurrence of $\tau$ to all occurrences of $x$ in $R$
  3. replace all those occurrences of $x$ by an occurrence of $\square$.

In the result $x$ does not occur. The point of that is that there are no bound variables; as variables become bound (by an occurrence of $\tau$), they are replaced by $\square$, and those occurrences of $\square$ are linked to the occurrence of $\tau$ that binds them.

The intended meaning is that $\tau_x(R)$ is some $x$ of which $R$ is true.

Community
  • 1
Zhen Lin
  • 90,111
  • Thanks! Do you happen to be able a more modern reference, basically for a foundational treatment of first order logic? – Nikolaj-K Aug 01 '12 at 09:58
  • 1
    I liked Forster's Logic, induction and sets, but you may find it a bit too informal. – Zhen Lin Aug 01 '12 at 10:11
  • Hmm... don't know whether it's only me that the link to planetmath seems to be dead. –  Aug 10 '16 at 15:56
3

The Hilbert operator was introduced by Hilbert in 1923. Hilbert was convinced of the axiom of choice as an indispensable principle but refrained to introduce it in its logical system. Instead he postulated another axiom, similar in contents to the axiom of choice, introducing new kind of terms: the $\tau$-terms (in fact Hilbert used the letter $\epsilon$ instead of $\tau$ but Bourbaki preferred the latter to avoid any confusion with the set-theoretical $\in$ symbol).

Hilbert's operator offers a systematic mechanism of attaching to a first-order predicate $R[x]$ a $\tau$-term by the procedure you described above, i.e. $\tau_x(R)$. Formally this is a string where all the occurrences of $x$, if any, are muted and tightly bound to the initial $\tau$ symbol by an upper link (the largest link indicates the scope of the binding).

The intended intuitive meaning is according to Bourbaki the following:

"If there exists at least one object for which the predicate $R[x]$ is true, then $\tau_x(R)$ represents some kind of an "ideal" (Hilbert's terminology) "preferred" (Bourbaki's terminology) object of the theory for which the predicate $R[x]$ is certainly true. Otherwise, strictly nothing can be said about it".

According to this naïve interpretation $(\exists x)R$ can be considered as an alias for the relation $R[\tau_x(R)]$ where the $x$ variable is in fact muted and bound in all of its occurrences in $R$. Similarly $(\forall x)R$ can be considered as an alias for the string $\neg(\exists x)(\neg R[x])$, which becomes now expandable into an explicit formula based on Hilbert's $\tau$-term.

Two logically equivalent 1st-order predicates in variable $x$ should lead to the same ideal $\tau$-term. The axiom schema about $\tau$-terms and equality ensures that: $(R\iff S)\implies \tau_x(R)=\tau_x(S)$.

The net result of that schema is that when a predicate is one-to-one in $x$:

$$R[x] \wedge R[x'] \implies x=x'$$

then $R[x]\implies x=\tau_x(R)$. The $\tau$-term attached to predicate $R$ is the "obligated" logical solution of $R[x]$, if such a solution does exist. This was the axiom schema postulated by Hilbert.

Conversely if such a solution does really exists, i.e. if $(\exists x) R[x]$ literally $R[\tau_x(R)]$ is true, then $R[x]$ logical equation is equivalent to $x=\tau_x(R)$: the converse implication in Hilbert's axiom schema is also true.

$\tau$-term $\tau_x(R)$ can thus be regarded as the result of some particular choice (an ideal one) among all possible logical solutions of predicate $R[x]$ and can thus be constructed from a choice function which usage is set-theoretically justified by the axiom of choice.

Pete
  • 790
  • Thank you! If $(\exists x)R$ is alias for $R[\tau_x(R)]$, then what if $(\exists x)R$ is actually false? You imply $\tau_x(R)$ still makes sense then. – Nikolaj-K Dec 29 '15 at 14:03
  • $\tau_x(R)$ always make sense as a perfeclty determined object of the theory. $(\exists x)R$ is false means simply and litteraly $\tau_x(R)$ is not a solution of the predicate $R$. One can say nothing about it ($R$ is not a property of this object). This is perfectly consistent. If the object thought as to be the prefered one to solve the predicate $R$ when a solution exists, is itself not a solution how could any other object of the theory be a solution too ? – Pete Dec 29 '15 at 16:36
  • In case $(\nexists x)R$ you can think that $\tau_x(R)$ is the same as $\tau_x(\neg R)$; you can even add an axiom to that effect. It is not strictly necessary, but I find that it helps my intuition. – Vladimir Reshetnikov Sep 16 '21 at 18:49