Manifold #3. Separation매니폴드 #3. 분리: T0, T1, T2와 하우스도르프

These are my notes on XylyXylyX’s What is a Manifold? lectures. This post covers lesson 3 of the playlist, Separation. In the previous post we learned to talk about nearness with open sets. This time we give the same line five different topologies and compare how well each one keeps two distinct points apart, using the separation axioms $T_0$, $T_1$, and $T_2$.

The same line, different topologies

Even when the set of points is the same, a different topology gives a different space. This post shows that with five topologies on a line. Each topology is given by a base, the basic sets from which the open sets are built. The topology consists of all unions of base sets, closed under overlaps (finite intersections) as well. The lecture has a passage that sounds as if arbitrary intersections were also allowed, but a topology only allows finite intersections.

1. The usual topology. On the whole real line $\mathbb{R}$, the open intervals $(a, b)$ form a base. This is the one-dimensional version of the open balls from the previous post.
2. The nested interval topology. On the interval $(0, 1)$, the open sets are the empty set, the whole interval, and the sets of the form $(0, 1-\tfrac{1}{n})$ for $n = 2, 3, 4, \dots$, so $(0, \tfrac12), (0, \tfrac23), (0, \tfrac34), \dots$ go in one after another. The open sets are stacked one inside another.
3. The closed interval topology. On the interval $[-1, 1]$, the base consists of sets $[-1, a)$ with $a > 0$ and sets $(b, 1]$ with $b < 0$. The intersection of the two kinds is $(b, a)$, so open intervals containing 0 become open sets, and the empty set and the whole set are included. The name means the endpoints $\pm 1$ are included. In real analysis these sets are called half-open intervals, but they are elements of the topology, so they are open sets. By definition, an open set is a set that belongs to the topology. An interval that does not contain 0 is not open.
4. The cofinite topology. On $[0, 1]$, an open set is a set with finitely many points removed, together with the empty set. Removing a few points leaves an open set, and so does removing a million, as long as the number is finite. The union of two such sets leaves only some of the removed points, and the intersection gathers all removed points but still finitely many, so the rules of a topology hold.
5. The lower limit topology. On $\mathbb{R}$, the half-open intervals $[a, b)$ form a base, which includes the left end and omits the right end. It has more open sets than the usual topology and is a fairly famous one.

Checking the rules for unions and intersections shows that all five are topologies.

Three levels of separating two points

The separation axioms make the question “can open sets tell two distinct points $p, q$ apart?” more and more demanding.

• $T_0$: at least one of the two points has an open set that contains it but not the other point. This is also called a Kolmogorov space. It is the weakest condition.
• $T_1$: we can separate from either side. That is, there is an open set containing $p$ but not $q$, and another containing $q$ but not $p$. The lecture phrases it as the two points each sitting in their own open set. It is sometimes called a Fréchet space, though the lecturer says the name is not certain.
• $T_2$: the two points each sit in an open set, and those two open sets are disjoint. This is called a Hausdorff space.

$T_2$ implies $T_1$, and $T_1$ implies $T_0$. The condition must hold for every pair of distinct points for the space to be $T_i$. If even one pair fails, the space does not reach that level.

Q. What does $T_1$ need beyond $T_0$?
A. $T_0$ may separate in only one direction, but $T_1$ must separate in both. For the same reason, a single point is a closed set in a $T_1$ space. (This equivalence is added to help understanding and was not covered in the lecture.)

Checking each topology

The nested interval topology is not even $T_0$. The right ends of the open sets $(0, 1-\tfrac1n)$ are only $\tfrac12, \tfrac23, \tfrac34, \dots$, so if two points lie between consecutive values, no open set separates them. For example, two points between $\tfrac12$ and $\tfrac23$ are either both inside an open set or both outside it, and $T_0$ breaks at this pair.

The closed interval topology is $T_0$ but not $T_1$. The key is that only open sets containing 0 are available. Two points on opposite sides of 0 can be separated by trimming just one end. Two points $0 < p < q$ on the same side can be separated by $[-1, q)$, which contains only the nearer point $p$. But there is no open set that contains the farther point $q$ and leaves out $p$, because an interval containing 0 that reaches the farther point also passes the nearer one. Separation works in only one direction, so the space is $T_0$ and not $T_1$. Also, every nonempty open set contains 0, so no two open sets are disjoint, and it is not $T_2$ either.

The cofinite topology is $T_1$ but not $T_2$. For two points $p, q$, the set with $q$ removed is an open set containing $p$ but not $q$, and the set with $p$ removed does the opposite. So it is $T_1$. But two cofinite open sets share infinitely many points, so no pair of disjoint open sets exists. The two open sets above are almost the same set, differing by one point. It is surprisingly hard to find a space that is $T_1$ but not $T_2$, and the cofinite topology is a simple example.

The usual topology is $T_2$. Put each point in a small open interval, small enough that the two intervals do not overlap. So it is also $T_0$ and $T_1$.

The lower limit topology is also $T_2$. If $p < q$, take $[p, m)$ and $[q, q+d)$, where $m$ is the midpoint of $p$ and $q$ and $d$ is half of $q-p$. These two sets do not overlap. (This fact is added to help understanding and was not covered in the lecture.)

In summary:

Try it by hand

Pick a topology and set two points with the sliders. The open sets that separate the two points are drawn as bars, and the toy tells you whether a pair of disjoint open sets exists. In the cofinite topology the points separate in both directions wherever you put them, but a disjoint pair never appears. The toy works on [−1, 1], so the nested interval topology is the shifted copy of $(0, 1)$, using open sets of the form [−1, c) with end values $0, \tfrac13, \tfrac12, \tfrac35, \dots$ ($1-\tfrac2n$).

Q. In the closed interval topology, does putting one point at 0 and another elsewhere give $T_1$?
A. It does not. Every nonempty open set contains 0, so there is no open set that contains the other point without containing 0. In the toy, place one point at 0 and only one direction says “yes”.

Why Hausdorff matters

Almost every space used in physics is $T_2$. Texts on manifolds often say “assume a second countable Hausdorff space”, and this phrase bundles together conditions that come from topology. Without the Hausdorff property, a sequence converging to a point can have several limits, and tools such as differentiation fall apart. (This explanation is added to help understanding and was not covered in the lecture.)

There are conditions stronger than $T_2$. They separate closed sets, instead of points, by disjoint open sets. A space where a point and a closed set can be separated is regular, and one where two disjoint closed sets can be separated is normal. The lecturer said they always mix up which name goes with which and chose the order above, which agrees with standard usage. As an example that loses these properties, the slotted disc topology on the plane is introduced. It splits the plane by a line, uses open discs on each side, and builds open sets so that the line behaves like the discrete topology. The lecture skipped the details, so this post does not go into them either.

Subspaces and products

At the end of the lecture two common ways to build new topologies are briefly introduced. One is the subspace topology. For a line $L$ in the plane $\mathbb{R}^2$, the intersections of $L$ with the open sets of the plane are the open sets of $L$, and then $L$ inherits the usual topology of the line. The other is the product topology. For a set $S$ with topology $\mathcal{T}_S$ and a set $T$ with topology $\mathcal{T}_T$, the product set $S \times T$ naturally gets a topology $\mathcal{T}_S \times \mathcal{T}_T$.

In the next post we cover the countability axioms and continuous functions.