Manifold #5. Compactness, Connectedness, and Topological Properties매니폴드 #5. 컴팩트함, 연결성, 위상적 성질

These are my notes on XylyXylyX’s What is a Manifold? lectures. This post covers lesson 5 of the playlist, Compactness, Connectedness, and Topological Properties. In the previous post we reached homeomorphisms. This time we look at the most important properties a homeomorphism preserves, compactness and connectedness.

Homeomorphisms and topological properties

A homeomorphism only makes sense between topological spaces. For two spaces $(X, \mathcal{T}_X)$ and $(Y, \mathcal{T}_Y)$, a function $f : X \to Y$ is called a homeomorphism if it is a bijection and both $f$ and its inverse $f^{-1}$ are continuous. Since both are continuous, $f$ sends open sets of $X$ to open sets of $Y$ and $f^{-1}$ does the reverse, so the open sets of the two topologies are paired one to one. One topology is deformed into the other, and no open set disappears and no two merge into one.

Whether two spaces are homeomorphic depends on whether at least one such function exists. There may be a huge number of candidates, and finding one is enough. From this definition comes the topological property: a property that two homeomorphic spaces must share. Countability and the separation properties from the last post are examples, and so are compactness and connectedness in this one. It is used in two directions. If we find one homeomorphism, every topological property of one space carries over to the other. In the opposite direction, if we find a property that only one space has, for example $X$ is second countable and $Y$ is not, then no matter how long we search there is no homeomorphism. This is the usual way to show that two spaces are not homeomorphic.

A donut and a coffee cup are the famous example. Each has one hole, and we can match every point of the donut with a point of the cup and deform one into the other so that open sets match up. Topology has no distance and no shape, so a homeomorphism is just a loose deformation of form. Taking a square to a parallelogram by an affine transformation belongs to geometry. Topology is geometry stripped of all measurement, which is why it lies under every geometric structure.

Q. If there is a continuous function from the donut to the coffee cup, are they not homeomorphic already?
A. The inverse must be continuous too. With a continuous function alone, distinct points can collapse into one or open sets can vanish, and then we cannot come back. A homeomorphism has to be a bijection that is continuous in both directions.

Covers and subcovers

To define compactness we need a cover. A cover means what the name says. For a subset $A$ of a topological space $X$, a collection of open sets $\{ U_i \}$ whose union contains $A$ is called an open cover of $A$.

If a part of the open cover still covers $A$, that part is called a subcover. A subcover made of finitely many sets is a finite subcover. Now we define compactness this way. A set $A$ is compact if every open cover of $A$ has a finite subcover. However huge the cover, finitely many of its sets are enough, and this lets us treat something infinite as if it were finite. So to show a set is not compact, a single cover is enough. If we build even one open cover with no finite subcover, the set is not compact.

An open rectangle is not compact

In the plane $\R^2$ with the usual topology, take a rectangle $X$ without its boundary. $X$ is itself an open set, so one set covers it, and two overlapping rectangles cover it, as do three, or $n$. A set in the cover may contain points outside $X$. It only has to contain $X$ as a subset.

But we can also build a cover that uses infinitely many sets. Cover half of $X$ with a first rectangle, place the next one with a little overlap to cover half of what is left, and repeat forever. The union of these rectangles, which approach the boundary without end, covers all of $X$. Any point near the boundary is eventually caught by some rectangle, and the boundary itself is not in $X$, so it need not be covered. Yet if we choose only finitely many from this cover, some points outside the rectangle nearest to the boundary are always left uncovered. This cover has no finite subcover, so the open rectangle is not compact.

Not every infinite cover behaves like this. A cover that tiles the whole plane with open rectangles forever also covers $X$, but just a few of those that meet $X$ already cover it. This cover is no evidence for non-compactness.

A closed rectangle is compact

Change to a closed rectangle that includes its boundary, and the method above is blocked. Interior points are still covered, but the boundary points are never caught by the rectangles that keep halving. To cover a boundary point, the cover must contain an open set around that point, and an open set does not sit on the boundary but holds a thin band on the inside of that point. The band overlaps the rectangles above at some stage, and from then on it ends with finitely many sets. Having the boundary or not is all that separates compactness in the usual topology of the plane. (This explanation is a rough sketch, and the rigorous proof is the Heine-Borel theorem, which says a closed and bounded set in $\R^n$ is compact. It was not covered in the lecture.)

With the toy below you can do the same thing on an interval. For the open interval $(0, 1)$, whichever finitely many sets you choose from the cover, the part near 0 is left. For the closed interval $[0, 1]$, the open set $W$ containing 0 guards the end, and finitely many sets cover it.

Q. The open rectangle has a finite size in the plane, so why is it not compact?
A. Compactness is decided by covers, so it has nothing to do with the size we see. If a cover exists that approaches the boundary without end, finitely many sets can never finish the job.

Compactness has several weaker forms as well. Whichever one we use, a homeomorphism preserves it as it is.

Connectedness

There are many definitions of connectedness, and we look at two. First we define being broken apart. If $X$ is the union of two open sets $G$ and $H$ that do not overlap, then $X$ is disconnected.

If even one such split exists, the space is disconnected, and if none exists in any way, it is connected. (The two sets must be nonempty. The split $G = \emptyset$, $H = X$ exists in every space, so it must be excluded. This condition was not covered in the lecture.)

Let us follow why connectedness is a topological property. Suppose there is a homeomorphism $f : X \to Y$ and $X$ is broken into $G$ and $H$. Since $f$ sends open sets to open sets, $f(G)$ and $f(H)$ are open sets in $Y$, and since it is a bijection their union is $Y$ and they do not overlap. Then $Y$ is disconnected too. What the function $f$ actually is does not matter here. Only the existence of a homeomorphism is used. Studying another function $g$ that is continuous but not a homeomorphism does not change the fact that the two spaces are homeomorphic.

Path connectedness

A concept close to connectedness but different is path connectedness. To speak of joining two points $x$ and $y$ with a line in a general set, we use a continuous function from the interval $[0, 1]$ into $X$.

If such an $f$ exists, the two points are joined by a path. $f$ need not be a homeomorphism, and continuity guarantees the line-like shape. If any two points of a space can be joined this way, the space is path connected. A disconnected set cannot be path connected, since we cannot build a continuous function that joins its two pieces.

A path connected space is connected. The converse does not hold. The legendary counterexample is $\sin(1/x)$. As $x$ goes to 0, $1/x$ grows without bound and the sine oscillates forever. The set is this graph with a segment over 0 attached.

This set is connected but not path connected, because there is no continuous function that goes from the oscillating graph over to the segment on the $y$ axis. (The lecture says briefly that the graph is connected because it crosses 0. Writing the exact set as above was added for understanding.)

Q. Why is a path connected space connected?
A. Suppose a path connected space were split into $G$ and $H$. We could take a path $f$ from a point of $G$ to a point of $H$. Then $f^{-1}(G)$ and $f^{-1}(H)$ would split the interval $[0, 1]$ in two, but the interval cannot be broken, which is a contradiction. (This explanation is added for understanding and was not covered in the lecture. The lecture only says it can be proved.)

Holes and simple connectedness

Two path connected spaces can still differ. Say $X$ has no hole and $Y$ has its middle missing. In both, any two points can be joined by a path. But in $Y$, a path that passes beside the hole, a path that goes around once, and a path that goes around ten times are fundamentally different, because no continuous shrinking or stretching turns one into another.

The field that studies functions that deform paths continuously is homotopy. In $X$, any path between two given points can be deformed into any other, and a path connected space in which all paths deform into each other this way is called simply connected. Loosely, it has no holes. Simple connectedness is also a topological property. Homology is a different story and we do not cover it here.

Summary

If we find even a single homeomorphism, the two spaces are homeomorphic, and all topological properties such as the separation properties, countability, compactness, and connectedness are the same. Homeomorphisms matter because they only deform a topology by matching open sets in both directions, so no measurement such as distance or shape takes part. In the next post we finally define a topological manifold with these tools.