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I came upon the following definition for closure,

Given a subset of a topological space $X$, the closure of $A$ is defined as the intersection of all closed sets containing $A$.

How is this definition well-defined? We need to know if the expression, $$\bar{A} := \bigcap_{U \in S} U $$ where $S$ is the collection of closed sets containing $A$, exists, and is unique?

(We do know $S$ is nonempty as $X \in S$. But why does this guarantee existence of intersection? More generally, is the intersection of the elements of any nonempty collection of sets well-defined?) I think I am now a bit confused on the fundamentals of sets.

May someone explain? Thank you so much.

EDIT: Thank you so much for all the replies. I have now read some basic set theory from Enderton's text. Here is my attempt to prove from scratch (Which is pretty much the same as the comments). Please do tell me if any part is incorrect.

The three axioms which I use:

Power Set Axiom (PSA) $$\forall a, \exists B \forall x ( x \in B \Leftrightarrow x \subseteq a)$$

Exstensionality Axiom (EA) $$\forall A \forall B [ \forall x (x \in A \Leftrightarrow x \in B ) \Rightarrow A = B ] $$

Axiom of Separation (AoS) For each formula $f(x)$ not containing $B$, the following is an axiom: $$ \forall t_1 \ldots \forall t_k \forall c \exists B \forall x ( x \in B \Leftrightarrow x \in c, f(x)) $$

Well-defineness: Given a Topological Space $(X, \mathcal{T}_X)$ and $A \subseteq X$. Firstly, $\mathcal{T}_X$ is a well-defined set. This is because, by the PSA, set $\mathcal{P}(X)$ exists and is unique by EA.

So by the AoS, $\mathcal{T}_X : = \{ x \in \mathcal {P}(X) : f(x) \}$ (where $f(x)$ is a formula of $x$ for its openess) exists, and is unique by EA.

Define the collection of closed sets which contains $A$ by, $$ \mathcal{C} := \{ x \in \mathcal{P}(X) : x^c \in \mathcal{T}_X, A \subseteq x \}$$ which exists by AoS and is unique by EA. Also, $\mathcal{C}$ is nonempty ($X$ is in the set) so the set $$ \bigcap \mathcal{C} := \{ x \in X : \forall y \in \mathcal{C}, x \in y \} = \bar{A}$$ again exists by AoS, and is unique by EA, so the closure is well-defined.

Bryan Shih
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  • Notice that $A\subseteq U$ for all $U$ that contains $A$ (by the definition of closure), so if $A$ is non-empty then $\overline{A}$ is non-empty as well. And notice that $A\subseteq \overline{A}$ – Masacroso Jul 04 '16 at 06:03

4 Answers4

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If you're confused about the fundamentals, I'd suggest going over the first chapter in Munkres's book. I'd also recommend the section "Background in set theory" from these notes by Prof. Curtis McMullen, which illuminates the axiomatic foundations of set theory known as the ZFC axioms, the most widely used today.

As you said, the intersection is over a nonempty collection, and therefore is non-trivial. Moreover, it is unique, because there's nothing it could possibly depend on; the subset of closed sets that contain $A$ is a well-defined set of subsets of $X$, so the intersection of all of them is well-defined too (this follows by axiom IV from McMullen's notes, Unions, and taking complements).

amakelov
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If $A\in S$, then $\bigcap S \subseteq A$. Thus by the axiom of separation, $\bigcap S$ exists if $S\ne\emptyset$. Because it can be given as an explicit class builder:

$$\bigcap S=\{x\mid\forall y\in S,x\in y\},$$

it is unique. Thus the intersection of any nonempty collection of sets is a well-defined set. (The intersection of the empty set is also well-defined, but equals the universe, $\bigcap\emptyset=V$, which is not a set.) In topology, though, we want more than that: we want to know that the closure is closed, which follows from one of the axioms of a topology - any arbitrary intersection of a nonempty collection of closed sets is closed.

Mario Carneiro
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    For the intersection of the empty set, in this context I would understand it to be not the universe of set theory, but that of the current discourse, i.e. the space $X$. But that is of course not important for the OP at this point. – Carsten S Jul 04 '16 at 09:42
  • @CarstenS In axiomatic set theory it's a hassle to set this up, because $X$ can't be the universe since it's a set (otherwise we wouldn't even be able to consider subcollections of the powerset of $X$). You can use "relativized intersection", but then that means hidden parameters, which could be a potential cause of confusion for OP's "well-definedness" concerns. I prefer to just use a notation like $X\cap\bigcap S$ to explicitly indicate relativized intersection when relevant. – Mario Carneiro Jul 04 '16 at 12:42
  • You are right, I was thinking about this as a hidden parameter. I find it useful in this context, because I can for example check that something is a topology by checking that it is closed under unions and finite intersections, where the latter condition implies that the whole space is open. But it's a matter of taste. – Carsten S Jul 04 '16 at 12:49
  • @CarstenS In type theory, your approach is the usual one; in that case $X$ is the universe, but there is no guarantee that it's the only one, and the set of all topologies on $X$ lives in a different (higher) universe. Relativized intersection can also be written as a simple class builder: $$X\cap\bigcap S={x\in X\mid\forall y\in S,x\in y},$$ and it is obviously a set, even if $S$ is empty, since it is a subset of $X$. – Mario Carneiro Jul 04 '16 at 12:53
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Sure, let $S = \{ C | C^c \in \tau, A \subset C \}$.

Then $x \in \overline{A}$ iff $x \in C$ for all $C \in S$.

copper.hat
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First $S = \{U : U \mbox{ is closed set containing } $A$\}$.

$U\in S$ contains $A$ so $\cap_{U \in S} U$ contains A , So Closure of $A$ is well defined.

Seongqjini
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