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I'm currently reading the notes (rather a book) of an MIT preliminary math course for discrete mathematics.

In section on page 39, some ZFC axioms are written and roughly explained.
For example, the "Union" axiom states the following:

$$\forall z. \exists u \forall x. (\exists y. x \in y \space \land y \in z) \iff x \in u.$$

And the "Pairing" axiom states the following:

$$\forall x, y. \exists u. \forall z. [z \in u \iff (z = x \space \lor z = y)]$$

So the "Union" axiom contains something like $\exists u \forall x$, whereas the "Pairing" axiom has a period between both quantifiers: $\exists u. \forall z.$

So what is the point (pun intended) of the period? As far as I can tell, a period indicates that the statement before it relates to the one after it, that is, $\forall x. \exists y$ means "For all $x$, there exists a $y$." OTOH, $\forall x \exists y$ simply means "For all $x$ and for some $y$."

Is that correct? Or what is the meaning of inserting a period?
Besides, why is there no $.$ at the end of the "Pairing" axiom?

cadaniluk
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3 Answers3

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I think the period is used as a substitute for parentheses, to indicate the scope of a quantifier. It seems to mean that the scope of the preceding quantifier extends as far as possible, i.e., to the end of the formula unless it is blocked by an actual parenthesis. Here is the axiom of unions in more usual notation; square brackets have been added to replace the periods:

$$\forall z\ [\ \exists u\ \forall x\ [\ (\exists y\ [\ x\in y\land y\in z\ ]\ )\iff x\in u\ ]\ ]$$

Edit. In a comment, Mauro ALLEGRANZA provided a reference to The Notation in Principia Mathematica.

bof
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    (just as an aside) The use of a dot to eliminate some parentheses goes back to Church and even Russell & Whitehead. The convention was that the dot indicated that what follows, extending as far to the right as possible, should be parenthesized. (That's Church's convention; R&W's was iirc weirder, and involved multi-dots.) The dot was heavier than a period but lighter than a \bullet. This applied to both quantifiers and connectives. E.g. $p \to . q \to r$ means $p \to (q \to r)$. – BrianO Jun 06 '16 at 12:12
  • So the period is not placed arbitrarily and it does have a purpose? Do you have any credible source? Because as of now, it's your word against the others'. Not that I don't believe you, just for assurance. – cadaniluk Jun 06 '16 at 12:34
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    @Downvoter - see The Notation in Principia Mathematica. – Mauro ALLEGRANZA Jun 06 '16 at 13:13
  • @MauroALLEGRANZA Thank you for the reference! I have edited it into my answer. – bof Jun 06 '16 at 19:50
  • I honestly do not see any difference between the statement you wrote and the one without the square brackets. It is also not clear how you can deduce with a single period where the closing parenthesis should be, and the fact that you were able to know where to put them proves they weren't needed. And the notation in PM is completely different...! I challenge you to find a statement that looks like $: p \vee p . \supset . p$ in OP's book. – Najib Idrissi Jun 06 '16 at 19:53
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    @NajibIdrissi Without the square brackets the formula I wrote is either ambiguous or (applying the usual rules of precedence) just wrong as a statement of the axiom of union: the $\iff$ would be read as the main connective, and $x$ and $u$ would be free variables. – bof Jun 06 '16 at 20:58
  • @NajibIdrissi The rule is that the closing parenthesis goes as far to the right as possible. This was explained in my answer, and also in Brian)'s comment and in the article linked by by Mauro ALLEGRANZA. – bof Jun 06 '16 at 21:01
  • @NajibIdrissi As noted in the linked article, the single-dot notation is not the PM notation but a variant thereof. – bof Jun 06 '16 at 21:05
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It has been traditional in mathematical logic to take the quantifiers to have higher operator precedence than the propositional connectives: so $\exists y\,x \in y \land y \in z$ would mean $(\exists y\,x \in y) \land y \in z$. Russell and Whitehead had a convention of using dots to modify precedence and a few authors followed this convention, but it is rather complicated and unintuitive (compared with brackets, which are familiar from algebra) and it fell into disuse.

Nowadays, many authors (particularly computer scientists, like the authors of the notes in question) adopt the convention that quantifiers have lower operator precedence than the propositional connectives and use a dot as a clue to readers familiar with the older conventions. So they write $\exists y.x \in y \land y \in z$ rather than $\exists y(x \in y \land y \in z)$. This agrees with the traditional notation for $\lambda$-calculus, where $\lambda x. x + y$ is not read as $(\lambda x.x) + y$, and is (in my opinion) much more convenient in practice.

In the examples you cite, there is a dot missing after $\exists u$ in the union axiom. This must be a typo. The authors are also using commas to abbreviate nested quantification: $\forall x, y.\cdots$ abbreviates $\forall x. \forall y. \cdots$

Rob Arthan
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The period does not mean anything. You can read the formula as if it were not there, or you can read it as if there was one after each quantifier. It does not change the meaning. It seems that the author has the habit of putting a period at the end of each string of quantifier, but sometimes puts one in the middle, presumably an unimportant typo: it's hard to stay 100% consistent all the time when you're writing a 500+ page book...

However, regarding your last paragraph, you cannot rearrange things as you want. $\forall x \exists y$ (or $\forall x. \exists y$ if you prefer) means that for every $x$, there exists some $y$ such that [...]. There's not other interpretation possible. The $y$ depends on the $x$, and might change if $x$ changes.

On the other hand, the sentence "For all $x$ and for some $y$" is, as most sentences written in a natural language (e.g. English), ambiguous. Either it could mean that the $y$ depends on $x$, in which case you should read it as $\forall x \exists y$; or it means that there's some $y$ such that for all $x$ the next part of the sentence is true, meaning $\exists y \forall x$. As explained in the notes you're reading, these two things are not equivalent.

A simple example of this is: "For all human $x$, there exists some human $y$ such that $y$ is the mother of $x$" is true, as everyone has a mother. But "There exists some human $y$ such that for all human $x$, $y$ is the mother is $x$" is obviously false, as there isn't a single person who's the mother of every human. The first sentence is $\forall x \exists y [\dots]$, the second one is $\exists y \forall x [\dots]$". Because of this ambiguity it's easier to just read the mathematical formula and not try to translate it in English...

Najib Idrissi
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  • In the book I am referring to in my answer it sounds like that meaning is deduced from it's pronunciation in English. I also asked myself why one should do that because, as you said, English is ambiguous! So the meaning of the "composite" quantifiers $\exists y \forall x$ is predefined? – cadaniluk Jun 06 '16 at 19:48
  • @Downvoter There is no such thing as a "composite quantifier". The statement is of the form $\exists y P(y)$, and $P(y) = \forall x Q(x,y)$. This means there's some $y$ such that $P(y)$ is true; in other words there's some $y$ such that for all $x$, $Q(x,y)$ is true. – Najib Idrissi Jun 06 '16 at 19:55