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In ZFC, for example, there is no universal set, so what does it mean to write $\forall x (\cdots)$, i.e., for all sets something is true? Does it avoid the problem by quantifying over all elements but not allowing those elements to form a set?

A related question I have is about quantifying over infinite sets. If $X$ is a finite set with $n$ elements $x_1, x_2, x_3, \ldots, x_n$, Then it is reasonable to define $\forall x\in X :P(x)$ to be the statement $P(x_1)\wedge P(x_2)\wedge\cdots\wedge P(x_n)$ (which can be formalized using a recursive definition). But what about infinite sets? It seems that if $Y$ is a countably infinite set with elements $y_1,y_2,\ldots$, then $\forall y\in Y: P(y)$ would mean $P(y_1)\wedge P(y_2)\wedge\cdots$, which we couldn't finish stating. For uncountable sets, we would miss an element in between every time we count upwards.

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    $\begingroup$ ...which is why universal quantification is not the same as a conjunction, and why lattices are not always complete lattices. Universal quantification is part of the underlying first-order logic of set theory, so there is no need to have a universal set to quantify over. $\endgroup$ – Arturo Magidin Apr 15 '12 at 21:54
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    $\begingroup$ @russell11 For each cardinal $\kappa$ the language $L_{\kappa,\omega}$ allow you to use conjunctions of length $\kappa$ i.e., $\bigwedge_{\xi<\kappa} P(x_\xi)$. $\endgroup$ – azarel Apr 15 '12 at 21:59
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Universal quantification is given by first-order logic. It does not depend on a set theory, such as ZFC. It is asserted by ZFC that every term is a set; at least this is an interpretation of the Axiom of Extensionality, that two terms are equal if they act the same way as sets.

While it is useful motivation to think of the universally quantified statement as an "infinite" conjunction, this is not literally true and cannot be formalized "using a recursive definition". If you want to quantify "universally" over all elements of a set, the natural way to do this is simply $\forall x (x \in X \implies P(x))$.

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