# Alexandrov Spaces or Topology on a finite point set - is there a distinction between open and closed sets?

A topology can be defined as a family of sets that is closed under finite intersections and unions. We then call these open sets.

But, if the point-set is finite, why should we call these open sets, and not closed sets?

Consider:

The union of any number of open sets is open, and the intersection of a finite number of open sets is open.

The union of a finite number of closed sets is closed, and the intersection of any number of closed sets is closed.

But if our family is finite, there is no distinction because we don't have infinite sets to union or intersect. If it is an Alexandrov space, then infinite unions of open sets are always open, so this also removes that distinction.

So, given just a family of sets which is finite, closed under unions and intersections, is there a reason to declare these as "open" rather than "closed" other than convention (and then declaring complements of members of the family the "closed" and "open" respectively)?

I cannot talk about limit points, or any such things, so it seems like in these cases I am just calling them open, and their complements closed, as a matter of convention. Without extra information about the topology in this case, can open sets and closed sets be viewed as "opposites, but qualitatively equivalent"?

• As long as your topology contains all complements, then you don't distinguish between closed and open, i.e. all sets are "clopen". However, it seems you didn't specify that all closed sets are included in the topology. – jdods Mar 10 '15 at 17:44
• I didn't specify that for a reason. I am speaking of any finite family closed under unions and intersections whatsoever. – Jonathan Hebert Mar 10 '15 at 17:45
• Then there is a distinction between closed and open, yes? E.g. see the example given by @rnrstopstraffic below. – jdods Mar 10 '15 at 17:46
• If I give you a finite family of sets which are closed under intersections and unions, such as $\tau=\{\emptyset , \{1,2\}, \{1,2,3,4\}\}$, we call this a topology and declare the members open sets, because of this fact. But, closed sets also have this property that finite intersections and finite unions result in another closed set. Since the family is finite, we cannot talk of infinite unions. What's the distinction other than convention? – Jonathan Hebert Mar 10 '15 at 17:51
• Open sets are members of the topology, close sets may or may not be. That is the distinction! You have to specify a topology. The terms open and closed are only meaningful given a specific topology. – jdods Mar 10 '15 at 17:55

You have made a correct, contentful observation here: given any topology $\tau$ on a finite set $X$, you can take $\tau' = \{ X \setminus U \mid U \in \tau\}$ and that is another topology on $X$, not necessarily homeomorphic to the first, although of course $\tau'' = \tau$.

As you say, this does not generalize to all topological spaces but does generalize to all Alexandrov spaces, which by definition are the topological spaces in which arbitrary intersections of open subsets remain open, hence equivalently arbitrary unions of closed subsets remain closed. We find in fact that for any topological space $(X,\tau)$, $\tau'$ is also a topology on $X$ iff $\tau$ is an Alexandrov topology. Thus your procedure gives an involution on the class of Alexandrov spaces.

This can be better understood via the equivalence between Alexandrov spaces and preordered sets. This equivalence is discussed in the above wikipedia article; for a more succinct version, see 5.16) on this problem set (from a general topology course I am currently teaching).

It is not hard to check that on the preordered set side, this involution takes $(X,\prec)$ to $(X,\prec'$) where $x \prec' y \iff y \prec x$. This operation might be called the "dual preordering".

Using this equivalence we can see for instance that if $\# X \leq 2$ we have $(X,\tau') \cong (X,\tau)$: any preordered set with at most two elements is self-dual. However, there is a partially ordered set with 3 elements which is not self-dual: take two points which are incomparable to each other and each less than a third point. The dual has one point which is less than each of two other points which are incomparable to each other. (We can see the difference between a triangle and an upside down triangle!) If you want to work with total orders, then any finite total order is self-dual, but for instance the order dual of the positive integers $\mathbb{Z}^+$ is the negative integers $\mathbb{Z}^-$, and these are not order-isomorphic. Notice that maximal elements on the preorder side correspond to closed points on the Alexandrov space side, so the Alexandrov space associated to $\mathbb{Z}^+$ has no closed points whereas the Alexandrov space associated to $\mathbb{Z}^-$ has one.

I do not know of any application of this involution on Alexandrov spaces, but it is a nice observation and every nice observation is at least potentially useful...

• Very informative, thank you! – Jonathan Hebert Mar 19 '15 at 14:56

You are correct. You could take any topology $\tau$ on a finite space $X$, and then $\tau'=\{X\setminus U:U\in\tau\}$ would also be a topology on the space. However, this topology may not be equivalent to the first.

• Intuitively, I think they should be equivalent, and vary only superficially. So what properties might vary between the two topologies? And, does this also apply for Alexandrov spaces? – Jonathan Hebert Mar 10 '15 at 19:53

After some discussion with the OP, I think we've gotten to the root of the matter. That is, a finite topology $T=\{U_1,\ldots, U_n\}$ is a finite collection of open sets. Then we can discus the "complementary topology" $T^c=\{U_1^c,\ldots, U_n^c\}$ which is a collection of closed sets (closed relative the the topology $T$). However, we could easily have started off with $T^c$ as our topology and gotten its complementary topology ${T^c}^c=T$. Hence, in some sense, distinguishing between open and closed sets is arbitrary since a finite topology generates a collection of closed set which are also a topology, and that topology will generate the original topology as its collection of closed sets.

As the question has now been edited: Yes, there is a distinction between open and closed sets in that open sets are members of a particular topology. However, the collection of closed sets for a particular given finite topology is also a topology.

You get to decide which sets are open as long as your topology obeys the rules of topology: (1) the empty set and its complement are both open and closed; (2) finite intersections are open; (3) arbitrary unions are open.

It could be that all sets are both open and closed in a particular topology, e.g. the trivial topology $\{\emptyset, S\}$.

You will still have limit points, it's just that likely a convergent sequence is eventually constant.

For example consider the topology $\mathcal{T}=\{\emptyset,\{0\},\{1\},\{0,1\}\}$ (the set of all subsets of $\{0,1\}$). All sets are both open and closed.

Consider what happens when we remove a set to form a new topology: $\mathcal{T}_{0}=\{\emptyset,\{1\},\{0,1\}\}$. Now the set $\{0\}$ is closed but not open. The collection of closed sets is $\{\emptyset,\{0\},\{1\},\{0,1\}\}$. These sets are closed with respect to the given topology.

To answer your question: The distinction between open and closed is not a matter of convention, per se, nor are they "opposites" or "qualitatively different". The distinction between open and closed is a matter of guaranteed membership in the topology of the former. You are correct that for a finite topology, the collection of closed sets will also be a topology.

In the topology $\mathcal{T}$ above, all sets are closed and open, therefore they are neither opposite nor qualitatively different. In the topology $\mathcal{T}_0$, there is one set which is closed but not open. It is kind of the "opposite" of an open set (in that it is a complement of an open set), but it could be argued as being qualitatively identical since it is just a discrete set, a singleton.

A counterexample would be to consider the topology $\tau$ on $\{1,2,3,4\}$ given by $\tau=\{\emptyset , \{1,2\}, \{1,2,3,4\}\}$ Notice that this is a topology, but this topology does not contain $\{3,4\}$ which is closed as it is the complement of an open set.

• I think you misunderstood the question. – Jonathan Hebert Mar 10 '15 at 17:31