# The Geometric Viewpoint

geometric and topological excursions by and for undergraduates

## The Long Line

Topology can be best described as the study of certain “spaces” and the properties they have. Now it is important to figure out what spaces are essentially the “same” and which are different. We define two spaces to be the “same” if we can transform one into the other continuously, and the transformation we preformed can be undone continuously as well. Now you may be asking, “what exactly do you mean by a ‘space’?” A topological space is defined as a set $X$ with an associated set $\mathscr{T}$ consisting of subsets of $X$ which satisfies certain properties. The elements of $\mathscr{T}$ are declared to be the open sets of our topological space $(X,\mathscr{T})$.

Now a major aspect of topology is to define properties that different spaces could have and what these properties should “say” about a space. For example, one may want to determine when a space is connected together or when it can be broken up into different pieces. Or one may wish to determine when any two points in a space can be connected together by a path through the space. Now at first glance these two definitions seem to be describing the same property but in fact they aren’t. The first definition describes when a space is “connected” and the other when a space is “path-connected.” It turns out that “path-connected” is a stronger claim about a space than just “connected.” In other words, there are space which are connected but not path-connected, for example The Topologist’s Sine Curve . However every path-connected space is also connected. One may begin to wonder what other properties in topology share this type of connection. In other words, which properties imply other properties and which do not. Now to show that one property implies another, one must start from the most general assumptions and come up with some mathematical proof. However to show that some property does not imply another, one must simply come up with a counter example. It is the later strategy that this blog is concerned with. I will be discussing a particular topological space, “the long line,” that can be used as a counter example to certain properties of a space, namely different levels of “compactness.”

Before we begin discussing the long line it will be useful to have an overview of an order topology. Now on the real line $\mathbb{R},$ the order topology ends up coinciding with the usual definition of open set, namely the union of open intervals. The key property of the real line that allows us to have this definition of open intervals is that it is totally ordered. This simply means that given any two points $a,b \in \mathbb{R}$, either $a\leq b$ or $b\leq a$. It turns out that we can define an order topology on any set which is totally ordered by some relation $\leq.$ Given a totally ordered set $(X,\leq)$ we can define an order topology on $X$ by first letting $\mathscr{T}''$ be the set consisting of all sets of the form $\{y : y < a\}$ and $\{y : b for any $a,b \in X$. We then create another set $\mathscr{T}'$ of all the finite intersection of elements of $\mathscr{T}''$ and then let our topology $\mathscr{T}$ be all the sets which can be expressed as the arbitrary union and finite intersection of elements of $\mathscr{T}'$. The reader should think about this construction in terms of the real line and note that we end up producing all the open intervals and unions of open intervals of the real line. Refer to the image below to see how the construction comes together.

Example of the creation of an open interval with elements of T”.

We are almost ready to construct the long line, but first we must make one more detour, into the world of set theory. The principle object we will use in our construction of the long line is the ordinal. Ordinals are just a very special type of well ordered set. Now a well ordered set is very similar to a totally ordered set with the additional property that every non-empty subset has a minimum element. An amazing fact from set theory is that every set can be well ordered. It turns out that the ordinals can be thought of as the standard well ordered sets. In fact, every well ordered set can be put into a bijective correspondence which preserves order with a unique ordinal. Now you may be wondering what’s so special about these ordinals. An ordinal  $X$ can be defined as a well ordered set with the property that each element $a \in X$ is exactly the set of all elements in $X$ which precede $a$. In other words, $X$ is an ordinal if for every element $a \in X$, $a = \{x \in X : x. The you may be wondering if such a thing even exists. Refer to the images for a glimpse of the finite and the first few infinite ordinals. You may be interested in a more thorough description of ordinals, which can be found here.

The first few finite ordinals.

The first few infinite ordinals.

Now it turns out that inclusion will always be the well order on an ordinal. It also turns out that there are a lot of ordinals. In fact the collection of all ordinals is not even a set! Naively we can think that there are just too many ordinals for them to be a set. Now as seen above there are ordinals with an infinite number of elements. In fact the first infinite ordinal can be used to make sense of the natural numbers. We often denote the first infinite ordinal as $w$. Amazingly there are different sized infinities, and $w$ can be thought of as the “smallest” infinity. Sets which can be put in a bijective correspondence with it are deemed “countable.” The next size of infinity is known as “uncountable” and we will let $\Omega$ stand for the first uncountable ordinal.

After the long build up we are finally ready to define the long line. The long line can be thought of as taking uncountably many copies of the interval $\lbrack 0,1)$ and “stacking” them end to end. For comparisons sake, we can think of the positive real line as countably many copies of the same interval “stacked” end to end. The long line must be very long indeed! While this definition may provide a good image, it leaves little to work with as far as properties go. Here is a more precise definition, the long line $L$ is the cartesian product $\Omega \times \lbrack 0,1)$ where the elements are ordered lexicographically. In other words, given two elements $(\alpha, a)$ and $(\beta,b)$ with $(\alpha, a)\leq (\beta,b)$, either $\alpha \leq \beta$ or in the case of equality, $a \leq b$. It is easy to see that this is a total order and so we can construct the order topology on $L.$ While this definition may seem strange, it is actually very easy to visualize. Take the set of all non-negative real numbers as an example. We can think of this set as the cartesian product $\mathbb{N} \times \lbrack 0,1)$ (keeping in mind that the natural number $\mathbb{N}$ can be fully described by the ordinal $w$). First note that we can think of elements $(n,d)$ of our set as telling us first the integer part of the number and then the decimal part. Now I ask the reader to consider how they would compare the size of two different positive real numbers. First you would compare the integer parts, and if they were equal you would then move on to the decimal parts. That’s comparing lexicographically! So the long line $L$ is just a much “longer” version of that example. The image below provides a description of our analogy to the long line, unfortunately it is very difficult to create a visual for an uncountable well ordered set.

The real line as a cartesian product.

Now you may wonder, “how much longer is the long line?” Perhaps the best way to compare the “length” of these lines is by looking at sequences. It is common knowledge that any strictly increasing sequence of real numbers does not converge. This is easily seen and accepted and it may be easy to conclude the same fact about the long line as well, however that would be a mistake.

Lemma 1: Every increasing sequence converges in the Long Line.

Proof: Suppose $(x_n)$ is an increasing sequence in the long line. Now consider the first element of each term of our sequence (remember elements of the long line are doubles, the first being the ordinal, and the second being the decimal part). Let $(\alpha_n)$ be the sequence of first elements. Therefore $(\alpha_n)$ is an increasing sequence of ordinal numbers. Now I will present it as fact that every increasing sequence of ordinal numbers has a limit point. Also $\Omega$ can never be the limit point of a sequence of countable ordinals. Therefore $(\alpha_n)$ must converge to a countable ordinal and therefore an ordinal that is represented in the long line. Now, if $(\alpha_n)$ never reaches a point where it remains constant, then we never have to consider the decimal part of the sequence since the limit point of the sequence of ordinals together with 0 will be the limit point of our sequence. So suppose that eventually $(\alpha_n)$ becomes a constant sequence. Let $\alpha'$ be the eventual constant term. Now we will consider the sequence of decimal parts of all terms after the sequence becomes constant in terms of the ordinals. Let $(d_m)$ be this sequence. Since $\lbrack 0,1)$ is bounded and $(d_m)$ must be increasing, it is easy to conclude that it converges, possibly to 1 in which case we take the point $(\alpha'+1,0)$ as our limit point for the original sequence. Therefore we can conclude that every increasing sequence converges in the long line. $\square$

Now amazingly with this one fact we glean even more information about the long line. For instance the long line is sequentially compact. First a quick definition, a topological space is sequentially compact if every sequence in the space has a convergent sub-sequence. Before I prove this I will prove a quick lemma about sequences in a totally ordered set.

Lemma 2: Every sequence in a totally ordered set has a monotone sub-sequence.

Proof: Suppose $(X,\leq)$ is a totally ordered set. Now let $(x_n)$ be a sequence in $X$. Let $x_p$ be a peak of the sequence if for all $n\geq p : x_n \leq x_p$. Now clearly there are two cases to consider, either $(x_n)$ has infinitely many such peaks or it has finitely many. First suppose that $(x_n)$ has infinitely many peaks, $\{x_{n_1}, x_{n_2},\ldots \}$. Therefore we can take $(x_{n_p})$ as our sub-sequence and clearly by definition this sequence must be decreasing. Now suppose our sequence has only finitely many peaks. Therefore there is a last peak $x_{n_{0-1}}$. Now consider the term $x_{n_0}$. Since this term is not a peak there must exist another term $x_{n_1}$ which is greater than $x_{n_0}$. We can then find a term $x_{n_2} \geq x_{n_1}$ since $x_{n_1}$ was not a peak. We can continue this process, creating a sub-sequence $(x_{n_m})$ which is increasing. $\square$

Now we are ready to prove that the Long Line $L$ is sequentially compact.

Proof: Suppose $(x_n)$ is a sequence in $L.$ By the second lemma we know we can find a monotone sub-sequence $(x_{n_m})$. Now first suppose that $(x_{n_m})$ is increasing. Then by the first lemma $(x_{n_m})$ converges. Now suppose that $(x_{n_m})$ is decreasing. Now since the long line is bounded below we know that $(x_{n_m})$ must converge. So $(x_n)$ has a convergent sub-sequence. $\square$

So the long line is sequentially compact. Of course the next question to ask is if the long line is compact. Interestingly it is not. Now the topological definition is a little different then the one given in most calculus classes. A topological space $X$ is compact if every open cover $\mathscr{U}$ of $X$ has a finite sub-cover. Now a cover is just a collection of open sets where the union of the collection is equal to the entire space.

Lemma 3: The Long Line $L$ is not compact.

Proof: Consider the collection of open sets $\mathscr{U'} = \{ (\alpha, \alpha+1) : \alpha \in \Omega \}$. We can then add to this collection sets of the form $((\alpha,\frac{2}{3}),(\alpha+1, \frac{1}{3}))$ giving us a cover $\mathscr{U}$. To see that this is a cover just note that the only points $\mathscr{U'}$ only misses the points with no decimal part, and note that the added collection of sets catch all the ordinals with no decimal part. Finally note that if any set of $\mathscr{U}$ is removed, we will no longer have a cover of $L$ and since $\mathscr{U}$ is clearly infinite we can conclude that $L$ is not compact. $\square$

This then leads us to one more interesting aspect of the long line; the long line is not metrizable. Now before I can explain what a metrizable space is, I will give a brief description of a metric space. A metric space is a set together with a “distance” function that determines how far away two points are. Open sets are then the union of open balls, which is just the set of all points strictly less than some radius from a given point. For instance, the usual distance function, $d$, on $\mathbb{R}$ is just $d(a,b) = |a-b|$. Open balls then correspond to sets of the form $\{x\in \mathbb{R} : |a-x| < \epsilon\}$. Now a metrizable space is a topological space where one can create a distance function on the set that then creates precisely the same open sets that were in the original topology. Now metric spaces, and therefore metrizable spaces, usually have “nicer” properties than general topological spaces, especially when it comes to equivalent features. The property that interests us here is than in a metric space, and so a metrizable space, the concept of sequentially compact and compact coincide. In other words, if a metric space is compact, then it is sequentially compact, and likewise in reverse. So immediately we can see that the long line cannot be metrizable since it is sequentially compact but not compact. So it would be impossible to create a “distance” function, which made sense, on the long line which lead to the construction of all the open sets we have.

Now you may be wondering what’s the point of creating the long line. You may ask yourself why anyone should care. Aside from just being able to work with a weird topological space, the creation and examination of the long line has multiple benefits. First off, it gives us a concrete counter example to properties that seem so similar. If all you ever work with are metrizable spaces, you won’t ever be able to really see how sequentially compact and compact are different. The long line is just one example of the importance of searching for counter examples. One may think that it is possible for spaces to have some properties and not have others, but until a concrete counter example is created, it’s all just conjecture.

For more information on the long line refer to Counter Examples in Topology, Steen and Seebach.

Editor’s note: The author of this post Josh Hews was a student in the Fall 2014 Topology course at Colby College taught by Scott Taylor. Submissions to the blog of essays by and for undergraduates on subjects pertaining to geometry and topology are welcome. For more information see the “Submit and Essay” tab above.

This entry was posted in Uncategorized. Bookmark the permalink.