Manifold #2. Elementary Definitions매니폴드 #2. 기본 정의: 닫힌집합, 극한점, 폐포, 내부, 경계, 조밀

These are my notes on XylyXylyX’s What is a Manifold? lectures. This post covers lesson 2 of the playlist, Elementary Definitions. Building on the topology from the previous post, this post defines closed sets, limit points, closure, interior, boundary, and density in turn. All of them start from open sets, and the thread of the post is that a topology is a way of deciding what “near” means.

Another way to say a topology: neighborhoods

If points just sit there with no structure, the word “near” means nothing. A topology gives us nearness, and we can do the same job with neighborhoods instead of open sets. For a point $p$ of a set $X$, a subset $N$ is a neighborhood of $p$ under four rules.

1. $p \in N$.
2. If $N$ is a neighborhood of $p$ and $N \subseteq M$, then $M$ is also a neighborhood of $p$. Being a neighborhood is passed up to larger sets.
3. The intersection of two neighborhoods of $p$ is also a neighborhood of $p$.
4. If $N$ is a neighborhood of $p$, there is a neighborhood $W \subseteq N$ of $p$ such that $N$ is a neighborhood of every point $r$ of $W$.

The fourth rule guarantees that every neighborhood contains a smaller one that is a neighborhood of all its own points, which is what makes open neighborhoods exist. (This explanation is added to help understanding and was not covered in the lecture.) Whatever we can build from open sets can also be built from neighborhoods, and there are other approaches based on logic or on computation and measurement. The open set version is the one used most widely in textbooks, so this series goes with open sets.

Closed sets

In a topological space $X$, a closed set is the complement of an open set. For an open set $U$, the set $X \setminus U$ is closed, and the complement of a closed set is open again.

On the plane, the complement of an open ball is the boundary circle together with everything outside it, and this is a closed set. A single point is also closed in the usual topology. The plane minus that point can be built by uniting infinitely many open balls that avoid the point. This would not work with intersections, which are allowed only finitely many times.

Moving the rules for open sets over to complements gives the rules for closed sets. Unions stay closed up to finitely many, and intersections stay closed for any number. So we may also define a topology starting from a collection of closed sets.

Q. Is a closed set the opposite of an open set?
A. It is not the opposite. The empty set $\emptyset$ and the whole set $X$ are both open and closed. In the discrete topology every subset is both, and in the trivial topology a single point is neither open nor closed. (This explanation is added to help understanding and was not covered in the lecture.)

Open neighborhoods and limit points

An open neighborhood is an open set that contains a point $p$. An open set is a neighborhood of every point inside it, so when we speak of neighborhoods, open neighborhoods are usually enough.

Take a subset $S$ of $X$. If every open neighborhood of a point $p$ contains at least one point of $S$, we call $p$ a limit point of $S$. Conversely, if some open neighborhood $U_p$ has $U_p \cap S = \emptyset$, then $p$ is not a limit point of $S$. The point is that no matter how small we make the neighborhood, it still meets $S$.

A limit point need not lie in $S$. A point on the boundary of an open disc $S$ does not belong to $S$, but its open neighborhoods, however small, overlap the disc, so it is a limit point. If we pluck one point out of the disc and leave a hole, that hole is still a limit point. Every point of $S$ is a limit point. A set made of scattered points is treated the same way. If the points crowd toward a point $P$ without end, every open neighborhood of $P$ contains a point of $S$, so $P$ is a limit point. In terms of sequences, this is the case where infinitely many terms of some subsequence fall into the neighborhoods of $P$, and then $P$ is also called a cluster point of the sequence. However the definition is phrased, the core is the same: every neighborhood contains a point of $S$.

One note on terms. The lecture’s limit point includes the points of $S$ themselves, which textbooks call an adherent point. The textbook limit point, the accumulation point, needs every neighborhood to contain a point of $S$ other than $p$ itself. The remark that the discrete topology has no limit points is correct in the accumulation point sense. This post follows the lecture and writes limit point, and separates the two only where they differ. (This explanation is added to help understanding and was not covered in the lecture.)

Q. A point on the boundary does not belong to $S$, so why is it a limit point?
A. The condition for a limit point is only that every open neighborhood contains a point of $S$. It does not require the point to lie in $S$. However small a ball we draw around a boundary point, it overlaps the disc inside.

Closure

The closure $\cl{S}$ of $S$ is the set of $S$ together with all its limit points. The closure of an open disc is the closed disc that includes the boundary. If a region includes its boundary but is missing a few interior points, the missing points are limit points, so the closure fills them in. On the other hand, if the region has a wide hole, the point in the middle of the hole has a small open neighborhood that does not meet $S$, so it is not a limit point, and the closure does not fill that hole. Closure is the operation that closes up $S$ by adding its limit points.

The closure is the smallest closed set that contains $S$, and saying $S$ is closed is the same as saying $S = \cl{S}$. (This explanation is added to help understanding and was not covered in the lecture.)

Interior

A point $p$ belongs to the interior $\Int S$ of $S$ when there is an open set $U_p \subseteq S$ that contains $p$ and lies entirely inside $S$. A point near the boundary is not in the interior, because however we choose an open neighborhood, it sticks out of $S$. Even if $S$ contains its boundary, a point on the boundary is merely an element of $S$ and not an element of the interior.

Three facts about the interior are worth knowing.

• The interior $\Int S$ is always an open set.
• The interior is the union of all open sets contained in $S$.
• $S$ is open exactly when $S = \Int S$.

The first follows from the second, because a union of open sets is open. (This link is added to help understanding and was not covered in the lecture.) The third turns the intuition that “an open set must not include its boundary” into an equation. The interior can also be written using the closure of the complement.

Exterior and boundary

The exterior is the complement of the closure, $\Ext S = X \setminus \cl{S}$. For an open disc the closure is the closed disc, so the exterior is everything beyond the boundary circle, and the boundary circle itself is not in the exterior. The boundary is too close to count as outside. We have to get past the boundary to be in the exterior.

So the boundary is the collection of points that are in neither the interior nor the exterior, which is the same as the closure minus the interior.

We may also say a boundary point is one that can be approached both from inside $S$ and from outside. The lecture takes the definition above as the simplest route. Every point belongs to exactly one of the interior, the boundary, and the exterior.

Dense sets

A subset $S$ of $X$ is dense in $X$ when every point of $X$ either belongs to $S$ or is a limit point of $S$. That is, $\cl{S} = X$. This property depends not only on $S$ but also on the topology of $X$.

A single blob of a region on the plane is not dense. A point away from the region neither belongs to it nor is a limit point. On the other hand, the set $\mathbb{Q}^2$ of all points with rational coordinates is dense in the usual topology. Every open neighborhood of every point contains a point with rational coordinates, so every point of the plane is a limit point.

Give the same set the discrete topology and things change. If a point $P$ is not in $\mathbb{Q}^2$, then $\{P\}$ itself is an open neighborhood that does not meet $\mathbb{Q}^2$, so $P$ is not a limit point. In the discrete topology $\mathbb{Q}^2$ is not dense, and the only dense subset is $X$ itself. When the topology changes, the limit points change, and so does density. This is what it concretely means that a topology decides nearness.

Try it by hand

Pick a topology on a set of four points and press points to build a subset $S$. The interior, closure, exterior, boundary, and density of $S$ are computed at once. In the discrete topology the only dense set is the whole set, and in the chain topology the single point $\{a\}$ is already dense.

Q. What does it mean when the interior equals $S$ itself?
A. It means every point of $S$ has an open neighborhood inside $S$, so $S$ is an open set. In the toy above, press points until the interior equals $S$, and the line asking whether $S$ is open switches to “yes”.

In the next post we look at the separation properties, which say how well a topology keeps two points apart, using several topologies on an interval as examples.