Manifold #17. Metric Spaces매니폴드 #17. 거리 공간

These are my notes on XylyXylyX’s What is a Manifold? lectures. This post covers lesson 17 of the playlist, Metric Spaces. After making a Möbius strip as a quotient space in the last post, this time we step back and look at the metric space, which was born before topology.

Why go backwards?

Topology is usually learned starting from metric spaces. A distance gives open balls, open balls give open sets, and open sets give a topology. This series did it the other way and defined a topology first. Looking back, the topology is the more fundamental idea, and the modern approach starts from it. In history, though, the distance came first.

When we want to say that two elements of a set are close or far, topology says it abstractly with open sets and separation axioms. A distance says it literally, by attaching one real number to the two elements. A metric space carries the idea that two points in the plane have a distance over to an arbitrary set.

The conditions on a distance function

A distance function (metric) on a set $X$ is a function $d : X \times X \to \R$ that takes two elements and gives a real number, and for all $x, y, z \in X$ it satisfies four conditions.

1. $d(x, y) \ge 0$. There is no negative distance.
2. $d(x, y) = 0$ implies $x = y$, and the converse holds. The distance is 0 only for the same point.
3. $d(x, y) = d(y, x)$. The distance there equals the distance back.
4. $d(x, z) \le d(x, y) + d(y, z)$. This is called the triangle inequality. One side of a triangle cannot be longer than the sum of the other two.

All of them are the distance of plane geometry carried over word for word. (Condition 1 follows from the other three, because $0 = d(x, x) \le d(x, y) + d(y, x) = 2\, d(x, y)$. This explanation is added for understanding and was not covered in the lecture.)

We cannot give a distance function to just any set. Which topological spaces can have a distance function is the subject of the metrization theorems, which the lecture names without proof. The other direction always works. A set with a distance function always gets a topology.

Here are three examples.

• Euclidean distance: on $\R^n$, $d(x, y) = \sqrt{\sum_i (x_i - y_i)^2}$. It comes from plane geometry, so it satisfies the four conditions.
• Hub distance: pick one reference point (the hub) in the plane and call it $h$. If two points lie on one line through $h$, use the usual distance. Otherwise set $d(x, y) = d(x, h) + d(h, y)$. It is like a rail network where every trip must pass through a big city, and we can check the four conditions one by one.
• Discrete distance: $d(x, x) = 0$, and for $x \ne y$ always $d(x, y) = 1$. Every point other than itself is exactly 1 away.

The topology a distance makes

With a distance function we can define the open ball of center $x$ and radius $r$.

When the first post used open balls, it assumed without saying that a distance existed, because we cannot measure a radius $r$ without a distance function. The topology that has these open balls as a base is the metric topology, and on the plane it equals the open ball topology, that is, the usual topology. By convention, when a set and a distance function are given, the topology is read as the metric topology. A space with a distance can be given another topology, but then it has to be stated separately.

Change the distance in the toy below. Even for the same center and radius, the shape of the ball depends on the distance.

(The taxicab distance $|x_1 - y_1| + |x_2 - y_2|$ and the maximum distance $\max(|x_1 - y_1|, |x_2 - y_2|)$ are distances added to the toy to compare shapes, and were not covered in the lecture.)

The Euclidean ball is a disc, the ball of the maximum distance is a square, and the ball of the taxicab distance is a diamond. For these three, balls with the same center contain each other, as the dotted circles in the toy show. Around any point, a ball of one distance contains a small ball of another distance, so they make no difference in deciding which sets are open, and the three give the same topology. (This containment and the ratios of the radii are a computation added for understanding and were not covered in the lecture.)

The hub distance and the discrete distance are different. In the discrete distance, a ball of radius 1 or less is the single center point, and for a point away from the hub, a hub-distance ball of small radius is only a segment on a line. It contains no ordinary disc, so these two give a topology different from the usual one. The lecture only says that open balls can be made from such distances but they are much weirder. (The verdict that these two distances differ from the usual topology is added for understanding and was not covered in the lecture.)

Q. If the distance function differs, does the topology differ too?
A. Not necessarily. What fixes a topology is which open sets arise, and even when the values of the distance differ, balls that contain each other give the same topology. The Euclidean, taxicab, and maximum distances are examples.

The plane has a topology even without a distance

The real line $\R$ gets its usual topology without any distance. Take the open intervals $(a, b)$ as a base, and all we need to specify such an interval is the two endpoints $a$ and $b$. An open ball given by a center and a radius is only one way.

The plane $\R^2 = \R \times \R$ comes with the product topology automatically. A pair of open intervals $(a, b)$ and $(c, d)$ of the two lines becomes an open box $(a, b) \times (c, d)$, and these boxes are a base. The lecture calls this the box topology. No distance appeared anywhere.

Compared with the open ball topology, the two topologies are the same. An open ball can be written as a union of open boxes, and an open box can be written as a union of open balls. Picture making the boxes smaller and smaller to fill the inside of a ball. If one base can make the other, the topologies the two bases make are the same.

The same topology does not automatically bring a distance. The space $\R^2$ with only a topology and the space with a distance function $d$ added on top have different structure. The latter is more elaborate, and when we only write the metric space $(\R^2, d)$ we read it as carrying the metric topology. The $\R^n$ that a coordinate map of a manifold goes to is the former. It has only the usual topology and no distance. So the values in $\R^n$ are just coordinates, and the distance between points is told by a separately chosen metric.

Q. If the topology from open balls and the topology from open boxes are the same, doesn’t the plane already have a distance?
A. The topology is the same, but a distance is structure that has to be given separately. We have to pick and add one distance to get the radius that measures an open ball. The boxes are made without a distance.

The two definitions of continuity agree

The definition of continuity we saw earlier speaks in terms of topology. A function $f : X \to Y$ between topological spaces is continuous if for every open set $U$ of $Y$ the preimage $f^{-1}(U)$ is open in $X$. The definition used before topology was sorted out speaks in terms of distances. A function between metric spaces $(X, d)$ and $(Y, \rho)$ is continuous as follows.

For every point $x \in X$ and every $\varepsilon > 0$ there is some $\delta > 0$ such that $d(x, w) < \delta$ implies $\rho(f(x), f(w)) < \varepsilon$.

This is called the epsilon-delta definition. Read with open balls it says $f(B(x, \delta)) \subset B(f(x), \varepsilon)$. The conclusion of this section is that when both spaces carry the metric topology the two definitions agree, and we show both directions.

To go from the topological definition to epsilon-delta, fix a point $x$ and an $\varepsilon$. The ball $B(f(x), \varepsilon)$ is open, so its preimage is open and contains $x$. An open set contains a small open ball around each of its points, so some $B(x, \delta)$ lies inside the preimage. That is, $d(x, w) < \delta$ implies $f(w) \in B(f(x), \varepsilon)$.

To go from epsilon-delta to the topological definition, take an open set $U$ of $Y$ and a point $x$ in the preimage. Since $f(x) \in U$ and $U$ is open, some $B(f(x), \varepsilon) \subset U$. Build $B(x, \delta)$ with the $\delta$ that epsilon-delta gives. Every point of this ball goes into $B(f(x), \varepsilon)$, hence into $U$. So $B(x, \delta) \subset f^{-1}(U)$, and since every point of the preimage has a ball around it inside, the preimage is open.

Q. In the epsilon-delta definition, may $\delta$ change when $\varepsilon$ or the point $x$ changes?
A. Yes. The order is $\varepsilon$ first and $\delta$ after. It is enough to find one suitable $\delta$ for each point and each $\varepsilon$.

This link matters because writing about manifolds keeps bringing up continuous functions and homeomorphisms. A homeomorphism is a continuous one-to-one correspondence whose inverse is also continuous.

Closed sets and limits of sequences

With a distance, closed sets can be described by limits too. A sequence of points $x_1, x_2, \dots$ converges to a point $L$ if for any $\varepsilon > 0$ there is some $N$ such that $n > N$ implies $d(x_n, L) < \varepsilon$. Then $L$ is called the limit. This does not mean reaching 0. It means squeezing into an arbitrarily small size, and important facts of infinitesimal mathematics are proved by this squeezing.

A closed set is a set whose complement is open. In a metric space we can say it this way too.

A set $F$ is closed exactly when the limit of every convergent sequence of points in $F$ lies in $F$.

If the boundary lies outside $F$, a sequence in $F$ can approach the boundary and its limit lands outside $F$. So such a set is not closed. A closed set contains all of its boundary points. We ignore sequences that do not converge. Some books handle odd sequences, such as one whose odd terms keep jumping away, by speaking of convergent subsequences, and other books look only at convergent sequences.

We show the equivalence in two directions.

Suppose the complement is open, and a sequence in $F$ converges to $L$. If $L$ were outside $F$, some $B(L, \delta)$ would lie entirely in the open complement. Then every term of the sequence in $F$ would be at least $\delta$ away from $L$, so it could not converge to $L$. This is a contradiction, so $L \in F$.

Conversely, suppose $F$ contains all its limits but the complement is not open. Then some point $e$ of the complement has the property that every ball $B(e, \delta)$ around it meets $F$. For each $n$, pick a point $x_n$ of $F$ inside $B(e, 1/n)$. This sequence lies in $F$ and converges to $e$. Since $F$ contains all its limits, $e \in F$, which contradicts $e$ being a point of the complement. So the complement is open.

Telling it apart from the metric of relativity

The lecture ends with one request. The distance function $d$ here is the metric used when dealing with topological spaces, and it differs from the metric tensor $g_{\mu\nu}$ placed on the spacetime manifold in general relativity. The metric tensor takes two vectors and gives a value, and that value can be 0 or less, and it can be 0 even when the two vectors are not equal. So conditions 1 and 2 above collapse and only symmetry remains. Such a metric is called a pseudo-Riemannian metric.

The two should not be mixed up when studying. The distance between points on a manifold is fixed by the metric tensor, and the $\R^n$ that holds the coordinates has no distance. The lecture says there is a connection, but a very awkward one.

In the next post we go one step beyond topology to the first step of algebraic topology, homotopy.