Manifold #4. Countability and Continuity매니폴드 #4. 가산성과 연속성: 제1·제2가산, 연속 함수, 위상동형

These are my notes on XylyXylyX’s What is a Manifold? lectures. This post covers lesson 4 of the playlist, Countability and Continuity. The previous post asked how well a topology keeps two points apart. This post covers two countability axioms, which ask whether open sets can be counted, then continuous functions defined by topology alone, and the best kind of continuous function, the homeomorphism. It also finishes explaining where the condition that a manifold be second countable and Hausdorff comes from.

Open sets piling up around a point: first countable

A point $p$ of a space $X$ lies in many open sets. Inside such an open set there may be a smaller open set that still contains $p$, and repeating the question gives nested open sets that wrap around $p$. For open balls these are balls with shrinking radius. In some topologies this chain stops after three sets, and in others, such as the usual topology of the plane, it goes on without end. In the discrete topology, $\{p\}$ itself is open, so the chain ends at one.

The question is not whether the chain is infinite but whether it is countable. In the plane the radius can be chosen from all the real numbers, so there are uncountably many balls around $p$. But using only rational radii is enough, because for any open neighborhood given, some ball of rational radius lies inside it.

A space is first countable if for every point $p$ there is a countable collection of open neighborhoods such that every open neighborhood of $p$ contains one of them. This collection is called a local base at $p$. First countability is a property checked at each point. The lecture says this collection is nested, but whenever a countable collection exists it can be rewritten as a nested one, so the condition is the same. (This explanation is added to help understanding and was not covered in the lecture.)

Countable building blocks for all open sets: second countable

The second axiom concerns the whole topology, not a point. Every open set is a union of base elements, and if the base can be chosen to be countable, the space is second countable. In the plane, open balls with rational center and rational radius suffice to build an open set of any shape, and there are only countably many such balls. So the plane with its usual topology is second countable.

Second countable implies first countable. Collect the base elements that contain a point $p$, and they form a countable local base at $p$. The converse fails. At this point the lecture seems to say that first countable definitely implies not second countable, but the transcript is garbled, and the meaning is that a first countable space may fail to be second countable. The lower limit topology just below is the counterexample.

A manifold must be second countable because this condition guarantees several properties that keep the space well behaved when coordinate patches are glued together. For example, the partitions of unity used to patch functions together on a manifold are obtained on top of this condition. (This example is added to help understanding and was not covered in the lecture.)

Examples: which topology is countable in which way

We check the topologies from the previous post one by one.

The usual topology. Open intervals with rational endpoints form a countable base.

The nested interval topology. The open sets are only the sets of the form $(0, 1-\tfrac1n)$, so even the number of open sets is countable. Both axioms hold. Countability is checked on its own, even for a pathological space that fails to separate two points.

The closed interval topology. Among the open intervals $(b, a)$ containing 0 and the sets $[-1, a)$, $(b, 1]$, using only rational end values is enough to build every open set. Both axioms hold.

The cofinite topology. An open set containing a point $p$ contains all points except finitely many, so shrinking such sets in a nested way never closes in on $\{p\}$. The lecture explains it as a nested chain of open sets that can never be built. It is not first countable. Since second countable implies first countable, it is not second countable either. This kind of misbehaving space is what we want to avoid with manifolds. This is for the case of uncountably many points, as in the interval $[0, 1]$, and the story changes if the points are countable. The proof for the uncountable case is as follows. Suppose open sets $U_1, U_2, \dots$ containing $p$ are given as a countable list. Each $U_n$ omits finitely many points, so all of them together omit only countably many. Hence there is a point $q$ other than $p$ that lies in every $U_n$, and the open set containing $p$ and omitting $q$ contains none of the $U_n$. (This proof is added to help understanding and was not covered in the lecture.)

The lower limit topology. Taking the open sets $[x, x+\tfrac1n)$ around a point $x$ gives a countable local base, so it is first countable. It has more open sets than the usual topology, yet it is not second countable. Suppose a countable base exists. For every real number $x$, the set $[x, x+1)$ is open, so some base element $B_x$ satisfies $x \in B_x \subseteq [x, x+1)$. Then the left end of $B_x$ is $x$, and distinct $x$ give distinct $B_x$. So the base would be as large as the real numbers and uncountable. The lecture tells the same story with a picture: a point in a gap between the end values of the base cannot be made the left end of an interval. (The proof above is added to help understanding and was not covered in the lecture.)

Saying the lower limit topology is finer than the usual topology means that every open interval of the usual topology is open in the lower limit topology, and there are more open sets, such as $[a, b)$, that the usual topology lacks. Conversely the usual topology is coarser. The key to checking this is that $(a, b)$ is the union of the sets $[a+\tfrac1n, b)$, so it is open in the lower limit topology. A finer topology does not make countability better.

Q. How do first countable and second countable differ?
A. First countable means that at each point we can choose the open sets around the point countably, and second countable means we can choose a countable base that builds the whole topology. Second countable implies first countable, but the converse fails, as the lower limit topology shows.

Continuity defined by topology alone

In calculus, continuity was stated with $\epsilon$ and $\delta$, using distance. With no distance and only a topology, we can still define continuity. We have two topological spaces $X$, $Y$ and a function $f : X \to Y$. Without topologies, $f$ is just a list pairing points with points, and the word continuous has no meaning.

A function $f$ is continuous when, for every open set $V$ of $Y$, the preimage $f^{-1}(V) = \{ x \in X : f(x) \in V \}$ is an open set of $X$. In this definition, open sets travel backward, from $Y$ to $X$. A preimage is always defined even when $f$ is not one-to-one, so it works without an inverse function. The lecture points this out separately.

Saying the same thing with neighborhoods brings back the feel of epsilon-delta. For every point $x$ and every open neighborhood $V$ of $f(x)$, there is an open neighborhood $U$ of $x$ with $f(U) \subseteq V$. If we read $V$ as a measurement error, it says that however narrow the error range, a narrow enough neighborhood of $x$ keeps the measured value within it. The two definitions are equivalent.

Checking continuity thus becomes checking one by one whether preimages are open. A few quick facts follow. (These facts are added to help understanding and were not covered in the lecture.)

• A constant function is continuous between any topologies, because the preimage of any open set is either empty or everything.
• If the domain has the discrete topology, every function is continuous, because every subset is open.
• The identity function $\mathrm{id} : (X, \mathcal{T}_1) \to (X, \mathcal{T}_2)$ between two topologies on the same set is continuous exactly when $\mathcal{T}_2 \subseteq \mathcal{T}_1$, because the preimage of an open set of the target is that very set.

Identity functions between the same line

The last fact lets us settle $f(x) = x$ between all the lines of the previous post. The table gathers the directions from the usual line to each other line and back.

Going from the usual line to the nested interval, closed interval, and cofinite topologies, every open set of the target is already open in the usual topology, so the map is continuous. Going to the lower limit topology, $[a, b)$ is not open in the usual topology, so continuity fails. The reverse directions are the opposite. A small open interval is not open in the nested, closed interval, or cofinite topology, so the maps from those lines back to the usual line are not continuous, while from the lower limit topology to the usual one the map is continuous, because every usual open set is open in the lower limit topology. The lecture says the map from the usual line to the closed interval line is not continuous either, but if both lines sit on the same point set $[-1, 1]$ and the usual topology is the subspace topology there, that map is continuous, because the open sets $[-1, a)$ of the closed interval topology are already open in the subspace topology. It is possible that the lecture’s usual line was an interval without endpoints, so the point sets differed.

The same reasoning applies between the discrete line and the usual line. The identity from discrete to usual is continuous, and the identity from usual to discrete is not. The lecture says no continuous function goes from the usual line to the discrete topology, but constant functions are continuous, so that statement holds only for the identity function.

Homeomorphisms

If a function $f$ is a bijection, its inverse $f^{-1}$ is also a function. If $f$ is continuous and $f^{-1}$ is continuous, $f$ is called a homeomorphism, and the two spaces are regarded as topologically the same. A homeomorphism matches open sets one to one. So we check at the same time that $f$ sends open sets to open sets.

A homeomorphism matters because it preserves properties defined by topology. The separation properties ($T_0$, $T_1$, $T_2$), compactness, and connectedness are such properties. The lecture only states that they are preserved and does not prove it, and the last two are covered in part 5. A function that is merely a continuous bijection falls short of being a homeomorphism. Continuity of the inverse is a separate requirement.

Try it by hand

Pick a topology on each of two three-point sets $X$ and $Y$, and press to decide where $f$ sends each point. Each open set of $Y$ gets a line showing whether its preimage is open in $X$, and the result above tells you whether $f$ is continuous or a homeomorphism. Starting from the identity function with $X$ discrete and $Y$ the chain topology gives a function that is continuous but not a homeomorphism.

Q. Is a continuous bijection always a homeomorphism?
A. It is not. The identity from the discrete topology to the trivial topology is continuous and a bijection, but its inverse is not continuous. In the toy, set $X$ to discrete and $Y$ to trivial, press the identity function, and the results say it is continuous and also not a homeomorphism.

The conditions manifolds need

At the end of the lecture, the lecturer says the topology of a manifold is not one of these odd topologies but is close to the usual topology. So when books on general relativity and the like say they assume a second countable Hausdorff space for manifolds, it is now clear what these conditions point to. We need such conditions to set up coordinates and compute derivatives. The links between the separation axioms and countability have many counterexamples, and the lecture recommends the book of counterexamples in topology as the best book on the subject.

From the next post on, the story turns to manifolds themselves.