1
$\begingroup$

Given that a topology on a set X is defined as a subset of the powerset of X that satisfy certain conditions and the powerset of X itself is a topology over X, I conclude, that the definition is impredicative. Is this line of reasoning correct?

$\endgroup$
  • 2
    $\begingroup$ No. The fact that $\mathcal{P}(X)$ is a topology of $X$ is not needed to define topologies. Rather, it is a consequence of how topologies are defined. $\endgroup$ – Jackozee Hakkiuz Jun 18 '18 at 21:03
  • $\begingroup$ Why is it impredicative? $\endgroup$ – herb steinberg Jun 18 '18 at 21:04
  • $\begingroup$ All topologies are subsets of the powerset, this is an impredicative property for all topologies : " A property (more accurately, a linguistic expression for the relevant property) is called non-predicative if it contains a bound variable such that the object to be defined falls within its range of applicability. A property is called predicative if it contains no such bound variables." The quotation is from encyclopediaofmath.org/index.php/Non-predicative_definition - although a similar definition can be found in Carnap's Logical Syntax of language. @JackozeeHakkiuz $\endgroup$ – val 72 Jun 18 '18 at 21:13
  • 1
    $\begingroup$ @val72: Even though a topology happens to be a subset of the power set, you don't need to mention that fact in a definition. You can simply say that every element of the topology has to be a subset of $X$. $\endgroup$ – Henning Makholm Jun 18 '18 at 21:20
  • 1
    $\begingroup$ On the other hand, you'll probably need an "impredicative" quantification in order to express that the topology must be closed under arbitrary unions. But it's not clear to me that the concept of "impredicative" is even very meaningful when you're defining a property (presumably to be applied later to something you have already made by then) rather than a particular object. $\endgroup$ – Henning Makholm Jun 18 '18 at 21:26
4
$\begingroup$

The notion of impredicativity is context-dependent and not always clear, but in this case, I don't think you can argue that because the powerset $\Bbb{P}(X)$ of a set $X$ gives a topology on $X$ (the indiscrete topology), then the definition of a topology as a subset ${\cal T}$ of $\Bbb{P}(X)$ satisfying certain conditions is impredicative. That would be analogous to arguing that because the number $2$ is even, then the definition of an even number as a number divisible by $2$ is impredicative.

To relate this to your quotation from the Encyclopedia of Mathematics, note that $2$ in the definition of even-ness and $\Bbb{P}$ in the definition of topology are not bound variables: $2$ is a constant symbol and $\Bbb{P}$ is a function symbol. The "object to be defined" in the definition of a topology is the class of all topologies. The definition does not involve quantification over this class.

$\endgroup$
  • 1
    $\begingroup$ You used the exact same analogy I had in mind. $\endgroup$ – Derek Elkins Jun 18 '18 at 21:34
  • $\begingroup$ @DerekElkins: interesting! home is where the heart is, I guess. $\endgroup$ – Rob Arthan Jun 18 '18 at 21:38
  • $\begingroup$ In my definition, any topology defined in terms of the definition is the bound variable. The power set of X is one value the variable can take and the range of the variable is the power set. $\endgroup$ – val 72 Jun 18 '18 at 21:44
  • $\begingroup$ See the last two sentences in my updated answer. $\endgroup$ – Rob Arthan Jun 18 '18 at 21:48
  • $\begingroup$ @RobArthan: I think you are right.$ (\forall x)(x\in\mathcal{T}\leftrightarrow x\subseteq\wp(X)\land\ldots)$ where $\mathcal{T}$ is the class of all topologies and the definiendum $\mathcal{T}$ is not a value of x $\endgroup$ – val 72 Jun 18 '18 at 22:29
1
$\begingroup$

Following up on Rob Arthan's answer I attempt a more detailed one:

To begin with, Carnap wrote in the Logical Syntax of Language: “A thing is called impredicative (in the material mode of speech) when it is defined (or can only be defined) with the help of a totality to which it itself belongs. This means (translated into the formal mode of speech) that a defined symbol $\mathfrak{a}_1$ is called impredicative when an unrestricted operator with a variable to whose range of values $\mathfrak{a}_1$ belongs, occurs in its chain of definitions” (p.163)

The term “operator” above, signifies a quantifier. In the case of sets the aforementioned definition should be understood as follows: a set $X$ defined as follows,$(\forall x)(x\in X\leftrightarrow\ A\left(x\right))$ for some predicate $A$, is called impredicable if for the formulation of $A$ are used quantifiers that bound variables with range of values which includes the definiendum $X$.

Let us now try to define the set of topologies $\mathcal{T}$ on a set $X$:

$(\forall x)[x\in\mathcal{T}\leftrightarrow x\subseteq\mathbb{P}(X)\land\ \emptyset\in x\land\ Χ∈x∧(∀y)(∀z)((y∈x∧z∈x)→y∩z∈x)∧(∀w)(∃φ)(∀t)((t∈w↔(t∈x∧(∃i)(i∈\mathbb{N}∧φ(i,t)))→∪w∈x)]$

where $\mathbb{P}(X)$ the power set of $X$, $\mathbb{N}$ the set of natural mumbers and $\varphi(s,t)$ any function that defines a countable family of subsets of $X$.

It can be easily verified that the definition is not impredicative since $\mathcal{T}$ is not the value of any bound variable in the definiens to the right of the first occurrence of the equivalence.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.