Filed under: Math, Topology | Tags: Math, order, pretty pictures, set theory, topology

Today I’m going to talk about two more or less unrelated ideas: a partial order on the set of topologies over a space, and topologies that “come automatically” with a totally ordered set. Hopefully my post will be a good deal shorter than last time.

Look at the set of all topologies on a set . As these are just sets of subsets of , the subset relation forms a partial order on them. Say , or is **coarser** than , when ; more intuitively, when every open set under is also open under . Likewise, is **finer** than . To remember which is which, imagine the open sets forming a sifter out of . Coarser topologies will have “bigger holes” and so it will sift the flour more coarsely. Adding open sets to a topology is just like adding more wires to the sifter, and making its sifting finer. (I got the terminology from Munkres, but other authors use “smaller” or “weaker” instead of “coarser.” I think some even use “stronger” instead of “coarser.” Watch out!)

The coarseness relation on any set has a natural minimum and maximum. The coarsest is the topology with the fewest possible open sets, i. e. two: and . We call this the **indiscrete** or **trivial topology**, or . The finest topology is the topology with the most possible open sets, i. e. every subset of . This is called the **discrete topology** or . Neither topology is very interesting, but it’s good to know what they look like. Notice that is generated by the (sub)basis , and is generated by the (sub)basis (the set consisting of the points of )!

It’s easy to talk about the coarseness ordering in terms of bases. If every basis element of is a union of basis elements of , then . This is equivalent to saying that for all , there is a such that $x\in U^\prime\subset U$. This language is often preferable and we’ll see why shortly.

We talked about topologies on last time: you should now be able to show that the cofinite topology is coarser than the Euclidean topology, which is in turn coarser than the Sorgenfrey topology. Both relations are strict; for example, there is no Euclidean open set around the point that is also a subset of . Since coarseness is anti-symmetric, we can use it to prove that the topologies generated by balls and cubes in are the same: any ball contains a cube around each point and vice versa, so each topology is coarser than the other, so they are the same.

The second idea is our first example of a topology “generated” from another structure. As I said, this is actually quite a common phenomenon. Suppose is totally ordered by a relation . Then the **order topology** on generated by is the topology generated by the subbasis of “rays” and . (Infinity might not really make sense in — say, if is finite — but the sets defined this way do. The infinity symbols are just placeholders to say that the rays go on as far as possible.) A basis for the same topology consists of those rays and the sets .

The Euclidean topology on is the same as the order topology. The same order gives order topologies for every subset of , and we call these **induced order topologies**. For example, the induced order topology on is generated by the subbasis of rays and . In particular, all sets are open… so this is just the discrete topology! Similarly, gets the discrete topology. The rationals do *not* get the discrete topology: open sets there are generated by the basis . Point sets like $\{0\}$ are not open because there is no greatest rational less than , or least rational greater than it.

Define the set to be , with the usual order on , and with for all . Then the order topology here is *almost* discrete: any is open, but the open sets containing are just the rays . The set is not open. This is an example of an **ordinal**, which we’ll talk about soon. The ordering of the ordinals gives them all “almost discrete” topologies like this, and the ways in which they’re not discrete are interesting.

Regrettably, there isn’t an order on for giving the Euclidean topology as its order topology. There are some weird order topologies we can create, though. Define the dictionary order on by if for for some , and under the usual order on . An example for is shown at right. Since this is a total order, there’s an interval between any two points, and it consists of all of the vertical lines between them, as well as the downwards ray from the first point and the upwards ray from the second one. (What does a closed set look like?) In the Euclidean topology, the two rays form *some* boundary for this open set, but the set does not have boundary where the rays end: it is neither open nor closed. Likewise, Euclidean balls are neither open nor closed in the dictionary order topology. So the two topologies are **incomparable**: neither is coarser.

Soon, some more natural topological ideas, like closure and boundary. I’ll also set up a blogroll, a background, and a header. My computer is brokie brokie and I need to take it in to be repaired, so the background and header might take a while… :-(

**4 Comments so far**

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[…] this using the universal property above! And if has two topologies, and , with under the coarseness ordering, then the subspace topology of on some subset of is also at least as coarse as the subspace […]

Pingback by Subspaces, and more continuity « Gracious Living and the Two Meat MealNovember 7, 2010 @ 06:19[…] order topologies, one thing to do is look for separations into two rays, which means that if some closed ray or […]

Pingback by Connectedness « Gracious Living and the Two Meat MealNovember 9, 2010 @ 09:30what are conditions such that a topology will be induced by an total order?

Comment by AnonymousMay 17, 2011 @ 20:26I’m not quite sure what you mean. Any total order induces a topology with subbasis the open rays. If you’re asking whether, given a topology on a set, you can tell whether it can be induced by a total order on that set, then that’s a more interesting question. A good reference is here, specifically the top of the second page, or Google “orderability problem” or “linearly ordered topological space” It looks like there’s no easy answer, but there are some pretty cool ones: for example, a connected and locally connected space is orderable iff for any three proper connected subsets, two of them don’t cover the space. (Think about it!)

Since a subspace of a linearly ordered topological space isn’t necessarily orderable (for example, {-1} union (0,1) in the real line), but has a lot of the same nice properties, another interesting question is how to tell which spaces can be expressed as such subspaces. This is the “suborderability problem.”

Comment by thetwomeatmealJune 14, 2011 @ 20:55