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A set $X$ with a subset $\tau\subset \mathcal{P}(X)$ is called a topological space if:

  1. $X\in\tau$ and $\emptyset\in \tau$.
  2. Let $L$ be any set. If $\{A_\lambda\}_{\lambda\in L}=\mathcal{A}\subset\tau$ then $\bigcup_{\lambda\in L} A_\lambda\in\tau$.
  3. Let $M$ be finite set. If $\{A_\lambda\}_{\lambda\in M}=\mathcal{A}\subset\tau$ then $\bigcap_{\lambda\in M} A_\lambda\in\tau$.

Let $\emptyset=\mathcal{A}=\{A_\lambda\}_{\lambda\in N}$, i.e $N=\emptyset$. Then by 2:

$$\bigcap_{\lambda\in N} A_\lambda=\{x\in X; \forall \lambda\in N\text{ we have }x\in A_\lambda\}=X\in\tau,$$

since $N$ is empty. And by 3:

$$\bigcup_{\lambda\in N} A_\lambda=\{x\in X; \exists \lambda\in N\text{ such that }x\in A_\lambda\}=\emptyset\in\tau,$$

since $N$ is empty. Then 2 and 3 implies 1.

Many books define a topology with 1,2 and 3. But I think that 1 is not necessary because I was prove that 2,3 $\Rightarrow$ 1.

Am I right?

Andrés E. Caicedo
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Gaston Burrull
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3 Answers3

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The problem is that as you formulated that $\bigcap\varnothing$ is the set of all elements $x\in X$ that for every $A\in\varnothing$ we have $x\in A$, this is satisfied by all the elements of $X$.

Note that $\bigcup\varnothing$ is well-defined in ZF since the axiom of union says it is a set, and we can prove that this set is indeed $\varnothing$. However $\bigcap\varnothing$ is not well defined, because as I remark above, it can result with the collection of "everything", which in set theory is not a set at all, and in this case - not even empty.

Asaf Karagila
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  • Thanks for your approach and clarification for axiomatic set theory. – Gaston Burrull May 31 '12 at 07:46
  • But I think that $\bigcap \mathcal A_\lambda =X$ since every $A_\lambda$ is a subset of $X$. I don't full understand. – Gaston Burrull May 31 '12 at 22:16
  • @Gastón: Take $X=\mathbb N$ and $A_n={n}$ what $\bigcap{A_n\mid n\in\mathbb N}$? In this case it is the empty set. One would expect that $\bigcap\varnothing=\varnothing$, but in fact it is not well-defined and requires an additional "axiom" which defines exactly what is the meaning of $\bigcap A$ (much like there is an axiom telling us what is the meaning of $\bigcup A$). – Asaf Karagila May 31 '12 at 22:19
  • Is really true that $\bigcap\mathcal{A}$ need an aditional axiom like the $\bigcap{\emptyset}$ monster? I think that both dont must be the same sets. – Gaston Burrull May 31 '12 at 22:22
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    @Gastón: Axioms simply describe in full (or so we hope) the properties we want from symbols and objects. E.g. $\bigcup A={u\mid\exists v\in A: u\in v}$. On the other hand, $\bigcap A={u\mid\forall v\in A:u\in v}$ or $\bigcap A={u\in\bigcup A\mid\forall v\in A: u\in v}$? The two forms are equivalent only when $A\neq\varnothing$. This means that you need to choose one specific form and stick with it. If you define the first form then $\bigcap\varnothing$ is the whole universe, which in set theory is a problem. In topology a bit less. – Asaf Karagila May 31 '12 at 22:28
  • In both cases, once you chose the form you wish to work with, I can find you a topology that either you cannot produce the empty set as an open set; or you cannot produce the whole space as an open set. – Asaf Karagila May 31 '12 at 22:29
  • Must you mean that never 2,3 implies 1?. – Gaston Burrull May 31 '12 at 22:37
  • If we assume the first form I must be correct saying that 2,3 implies 1 in any axiomatic set theory because is a vacuously logic argument. – Gaston Burrull May 31 '12 at 22:41
  • @Gastón: No. I mean that sometimes 2,3 do not imply 1. Therefore 1 is not provable from 2,3. Let $X=\mathbb N$. Consider the topology in which a set $A$ is open if and only if $4\in A$. This is closed under arbitrary unions as well finite intersections. If you chosen that $\bigcap\varnothing=\mathbb N$ then you cannot generate the empty set. You need to explicitly specify $4\in A$ or $A=\varnothing$. – Asaf Karagila May 31 '12 at 22:42
  • does exist an axiomatic set theory such that always we can be sure that 2,3 implies 1? – Gaston Burrull May 31 '12 at 22:45
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    @Gastón: I don't know. I'm sure that if you try hard enough you can come up with something. What would that be good for? I don't know. It is much easier to add this seemingly-superfluous axiom, and say that for "the intersection of two open sets is open". – Asaf Karagila May 31 '12 at 22:52
  • A last question. Why assuming "first form" we cant deduce that 2,3 implies 1? I think if we assuming first definition of intersection we are done. Since in this case $\bigcap A= X$ and $\bigcup A=\emptyset$. – Gaston Burrull May 31 '12 at 23:00
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Yes, you’re right. However, it’s more convenient to include (1). First, the fact that $\varnothing$ and $X$ are both open sets is important enough to be worth emphasizing. Secondly, many people are a bit uncomfortable with the union or intersection of an empty collection, just as many are a bit uncomfortable with the sum or product of the empty collection of real numbers. If you include (1) as part of the definition, you don’t have to deal with this awkwardness. This is especially important when you’re teaching elementary topology: most students at that stage definitely have trouble with the union and intersection of an empty collection.

Correction: I accidentally inverted one negation mentally when I thought about this the first time. You’re right about $X$, but not about $\varnothing$. Let $S=\{x\in X:\forall A\in\varnothing(x\in A)\}$; then $S$ is actually $X$, not $\varnothing$. To see this informally, ask yourself how there could be an $x\in X\setminus S$: there would have to be an $x\in X$ such that $\exists A\in\varnothing(x\notin A)$. But there isn’t any $A\in\varnothing$, so there is no such $x$, and $S=X$. (And I see now that Asaf has already pointed this out in his answer.)

Brian M. Scott
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  • Thanks Brian I was probably thinking that I make a mistake but I just use formal logic and vacuously for determining some truth value. – Gaston Burrull May 31 '12 at 07:45
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    Suppose I have two different spaces $X$ and $Y$. Then $\bigcap \emptyset$ can't be both $X$ and $Y$, since $\bigcap \emptyset$ does not have any sort of extra parameter in it for the space. So in general there is no reason to assume that $\bigcap \emptyset = X$. This is independent of any set-theoretical difficulties with the empty intersection; assuming the empty intersection is defined, it can only be one thing. – Carl Mummert May 31 '12 at 11:28
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    @Carl: Perfectly true, but ${x\in X:\forall A\in\varnothing(x\in A)}$ is not quite the same animal as $\bigcap\varnothing$, and Gastón was interested in the former. – Brian M. Scott May 31 '12 at 11:30
  • @Brian M. Scott: $\bigcap \mathcal{A}$, for example stated in bullet 3 at the top of the question, depends only on $\mathcal{A}$, not on some other parameter $X$. The computation of $\bigcap \mathcal{A}$ lower down isn't correct for that reason. – Carl Mummert May 31 '12 at 11:35
  • @Carl: Alternatively, the computation is correct, and the definition isn’t. I was responding to the computation; that’s precisely why I defined $S$ as I did. I probably should have responded to both and pointed out the inconsistency, but between these comments and Asaf’s answer that seems a little superfluous now. – Brian M. Scott May 31 '12 at 11:38
  • @BrianM.Scott Why you make a correction? I was write $S=\bigcap \mathcal{A}=X$. I am right in both. – Gaston Burrull May 31 '12 at 22:12
  • @CarlMummert $\bigcap \mathcal{A}=\bigcap A_{\lambda}$ does depend of $X$ since every $A_\lambda$ must be a subset of $X$. – Gaston Burrull May 31 '12 at 22:15
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    @Gastón Burrell: the underlying set theory defines $\bigcap B$ when $B$ is any set (or leaves it undefined). For any space $X$, looking at $\bigcap \emptyset$, every $Z \in \emptyset$ is a subset $X$, but $\bigcap \emptyset$ can be equal to at most one space, not to all of them. – Carl Mummert Jun 01 '12 at 02:33
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I don't know if there's still interest in this thread, but I'm going to answer anyway.

As Asaf points out, $\bigcap \emptyset$ may not exist, and it certainly doesn't equal $X$. The problem occurs because subsets don't remember the larger set they were cut from. However, this is a quirk of how subsets are defined. And, the problem can be avoided by redefining the meaning of "subset." This will allow us to rewrite the definition of a topological space.

Firstly, lets agree that by "function", we will mean an ordered triple $(f,X,Y)$. Then we can define that a subset of $X$ is a function $A : X \rightarrow 2,$ where $2 = \{0,1\}.$

Now for a bit of notation. As shorthand for $A(x)=1$, lets write $x \propto A,$ which can be read "$x$ is an element of $A$."

Lets also write $A \diamond X$ to mean that $A$ is a subset of $X$. Note that the subset relation is no longer transitive, so we also need a containment relation. So if $A,B \diamond X$, lets write $A \subseteq B$ in order to mean that for all $x \in X$ it holds that if $x \propto A$, then $x \propto B$.

Also, lets write $2^X$ for the powerset of $X$. Formally, we define $$2^X := \{A : X \rightarrow 2 | A \mbox{ is a function}\}.$$

Thus $A \diamond X$ if and only if $A \in 2^X$.

Finally, lets write $A = \{x \in X | P(x)\}$ in order to mean that $A \diamond X$, and that $x \propto A$ iff $P(x)$.

Given these conventions, we can define the intersection of $\mathcal{A} \diamond 2^X$ as follows.

$$\bigcap \mathcal{A} = \{x \in X|\forall A \propto \mathcal{A} : x \propto A\}$$

Unions can be defined similarly.

Finally, the payoff: letting $\bot$ denote the least subset of $2^X$, and letting $\top$ denote the greatest subset of $X$, we see that

$$\bigcap \bot = \top.$$

We're now in a position to rewrite the definition of a topological space.

A set $X$ with a subset $\tau \diamond 2^X$ is called a topological space if:

  1. For all $\mathcal{A} \subseteq \tau$ it holds that $\bigcup \mathcal{A} \propto \tau$.

  2. For all finite $\mathcal{B} \subseteq \tau$ it holds that $\bigcap \mathcal{B} \propto \tau$.

goblin GONE
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