I have a confession to make: I am a bibliophile. Reading, owning, perusing, lending, alphabetizing and buying books are all things that make me happy. High on my list are hardcover graphic novels and quality dictionaries. One of the skills you learn quickly while reading a dictionary (so I hear) is how to look up words. Of course the words in a dictionary are laid out in a very orderly fashion; first the ‘A’s then the ‘B’s, etc.. This order turns out to be a useful example of an interesting linear order.

Example: Consider $\{a,b,c\}\times\{a,b\}$ with the dictionary ordering. We get $aa < ab < ba < bb < ca < cb$.

In general to get a dictionary ordering on $A\times B$ out of two linear orders $A,B$ we do the following:

1. Compare first elements. If they are the different, use the ordering on A.
2. If the first coordinates are different, compare the second coordinates. If the second coordinates are different, use the ordering on $B$. If the second coordinates are the same, the elements you are comparing are the same (as they have the same first and second coordinates).

You can extend this process if you want and compare third, fourth or fifth coordinates if you start with three, four or five linear orders. Of course this is just saying something you already know; I don’t need to tell you how to figure out whether ‘oscillate’ comes before ‘ossifrage‘.

Example: Now my fellow sesquipedalians might be interested in the following linear order: Let $D = \{*, a,b,c, \ldots, z\}$ where $* < a < b < \ldots < z$ and $*$ stands for a blank space. Now consider $D^{189819}$ with the dictionary ordering. This will contain every English word both technical and non-technical. Granted it will also contain silly non-words like: “this*word*asserts*that*it*is*a*silly*word”.

Now, down to business. Recall that any linear order can be imbued with a topology called the linear order topology.

Linear Order Topology. Let $(L, <)$ be a linear order. The linear order topology is given by the subbasic open sets $(-\infty, a) := \{x \in L : x and $(b, \infty) := \{x \in L : b .

Here the use of the symbol $\infty$ is purely for notational convenience. In most cases we will be able to say that the intervals $(a,b)$ form a basis for the linear order topology.

Example: The usual topology on $\mathbb{R}$ is given by the linear order topology as every open ball is an open interval and every open interval is a union of open balls. Technically, for the masochists: $\displaystyle (\forall a,b \in \mathbb{R} \cup \{\infty,-\infty\})(\forall x \in (a,b))(\exists \epsilon > 0)(\exists y \in \mathbb{R})[x \in B_\epsilon (y) \subseteq (a,b)]$

Non-example: The Sorgenfrey line $\R_l$ is not given by an order topology. This can be seen by seeing that the order topology on $\mathbb{R}$ is the usual topology, which is not the Sorgenfrey topology. Example: Looking at $\{0,1\} \times \mathbb{R}$ with the dictionary ordering, we can give it the order topology. What does this look like?  Well it looks like two copies of $\mathbb{R}$ sitting next to each other. Everything in the $\{0\}\times\mathbb{R}$ interval is less than everything in the $\{1\}\times\mathbb{R}$ interval. Or, if you prefer, it is the same as $(3,4)\cup(5,6)$.

Now what I am really concerned with are the following two spaces, each with the dictionary ordering: $[0,1]\times[0,1]$ and $(0,1)\times(0,1)$. On first blush they seem like they should have some similar properties. $(0,1)\times(0,1)$. The first thing we notice is that this space is not separable, Lindelöf or second countable, all for (basically) the same reason: The set $\{(x,\frac{1}{2}): x \in (0,1)\}$ is an uncountable, closed discrete set. However, it is first-countable, for around any point there is the obvious countable local base in the second coordinate. Clearly it is not, compact or limit-point compact. As for separation properties, it is obviously Hausdorff and (non-obviously) normal. One way to see this is that a set $C \subseteq (0,1)\times(0,1)$ is closed iff when I take everything in $C$ with a given fixed first coordinate, it is closed in that coordinate.
Showing metrizabilty of this space is an (easy) exercise in Munkres. I thought about it for a long time. It turns out that this space is really just the product of two metric spaces: $(0,1)$usual and $(0,1)$discrete. Looking at it this way is somewhat illuminating: the different coordinates are actually far apart, and the discrete metric captures that.
So to sum up: $(0,1)\times(0,1)$ T2 YES! T4 YES! Compact NO. Limit Point Compact NO. Lindelöf NO. Separable NO. Second Countable NO. First Countable YES! Metrizable YES! $[0,1]\times[0,1]$: So think about this space for a bit. It looks awfully similar to the last space. Looks like this one is also just the product of two metric spaces. Well let’s see. For the same reasons as the last space, $[0,1]\times[0,1]$ is not separable, not second countable, and is Hausdorff. Is this space first countable? Yes, but it takes a bit of thinking to see why the endpoints have a countable local base. For example, what happens at the point $(\frac{1}{2}, 1)$?

Now the best part is that $[0,1]\times[0,1]$ is compact. Hold on a second before you say: “Clearly, Mike”, because you can’t use Tychonoff’s Theorem here. This is not $[0,1]\times[0,1]$ with the product topology. However, we can prove that it is compact by creeping along! The only technicality is showing that, as a linear order, $[0,1]\times[0,1]$ is complete but this happens mostly by inheriting completeness from $\mathbb{R}$.
So now we know that $[0,1]\times[0,1]$ cannot be metrizable (!) as compact and metrizability give separability, which this space does not have. Also, this space must be limit point compact; doesn’t that seem counterintuitive from our experience with $(0,1)\times(0,1)$! Also, we get normality since the space is compact and Hausdorff.
What went wrong with our argument about this space being the product of two metric spaces? Well, we didn’t actually check that the topologies agree, and in fact they don’t. In the metric topology any individual coordinate is both open and closed. In the order topology though, individual coordinates cannot be open, because open sets that contain an endpoint also contain a lot of other coordinates.

So to summarize: $(0,1)\times(0,1)$ $[0,1]\times[0,1]$ T2 YES! YES! T4 YES! YES! Compact NO. YES! Limit Point Compact NO. YES! Lindelöf NO. YES! Separable NO. NO. Second Countable NO. NO. First Countable YES! YES! Metrizable YES! NO.

I think that these two spaces are fairly simple, possess nice properties, but seem to be relatively unknown.