Real Topology: Introductory Definitions and Concepts

This video is a first introduction to the terminology and ideas found in point-set topology as it applies specifically to Euclidean spaces. Several of these ideas (compactness and connectedness in particular) will be studied in much greater detail in later sections.

Video. Real Topology

Some Topics Covered Neighborhoods; Properties of open and closed sets; interior, closure, and boundary; compactness and connectedness

1. Neighborhoods, Open and Closed Sets

Given vectors \(x := (x_1,\ldots,x_n) \in {\mathbb R}^n\), we define the Euclidean norm on \({\mathbb R}^n\) to be the quantity \(||\cdot||\):
\[ ||x|| := \sqrt{ \sum_{j=1}^n |x_i|^2}. \]
The Euclidean distance from vector \(x\) to \(y\) (both in \({\mathbb R}^n\)) is defined to equal \(||x-y||\). A fundamental fact usually proved in linear algebra is that this distance satisfies the triangle inequality:
\[ ||x-z|| \leq ||x-y|| + ||y-z||. \]
For now, we will take this fact for granted.
A neighborhood in \({\mathbb R}^n\), often denoted \(N_r(x)\) is a set with center \(x\) and radius \(r > 0\) defined by
\[ N_r(x) := \left\{ y \in {\mathbb R}^n \ : \ ||x-y|| < r \right\}. \]
A key feature of the definition os that the inequality is strict. In the one-dimensional case, the neighborhood \(N_r(x)\) is simply equal to the interval \((x-r,x+r)\) and the distance between two real numbers is simply the absolute value of the difference. Note that in this case, the triangle inequality is relatively easy to prove.
An open set in \({\mathbb R}^n\) is any set which can be written as a union of any number of neighborhoods (including uncountably many).
A closed set in \({\mathbb R}^n\) is any set which is the complement of an open set.

1.1. Basic Facts about Open Sets

The empty set \(\emptyset\) is open. (It's trivially an empty union of neighborhoods.)
The whole space \({\mathbb R}^n\) is open. (Take the union of all neighborhoods of all centers and all radii.)
Unions of any number of open sets (including uncountably many) are open.
Proof
If each set \(U_\alpha\) in some collection \(\{U_\alpha\}_{\alpha \in \Lambda}\) is open, one can write the union itself as the union of all neighborhoods \(N_r(x)\) which are contained in \(U_\alpha\) for some \(\alpha\).
If \(x\) is a point in an open set \(U\), then there is some radius \(r\) such that \(N_r(x) \subset U\).
Proof
Since \(U\) is open, there is some neighborhood \(N_{r_0}(y)\) such that \(x \in N_{r_0} (y)\). Fixing \(r = r_0 - ||x-y||\) works. First, observe that \(x \in N_{r_0} (y)\) forces \(r > 0\) by definition of the neighborhood, so \(r\) is a valid radius. Then, every \(z \in N_r(x)\) satisfies \(||x-z|| < r_0 - ||x-y||\); then \(||y-z|| \leq ||x-z|| + ||x-y|| < r_0\), which means \(z \in N_{r_0}(y)\).
Any intersection of finitely many open sets is open.
Proof
Every \(x \in U_1 \cap \cdots \cap U_N\) admits positive radii \(r_1,\ldots,r_N\) such that \(N_{r_j}(x) \subset U_j\) for each \(j = 1,\ldots,N\). Then taking \(r = \min\{r_1,\ldots,r_j\}\) shows that \(N_r(x) \subset U_j\) for each \(j\) and consequently \(N_r(x) \subset U_1 \cap \cdots \cap U_N\). We can now write the intersection of \(U_1,\ldots,U_N\) as a union of neighborhoods of some (center-dependent) radius for each \(x\) in that intersection.

1.2. Basic Facts about Closed Sets

The whole space \({\mathbb R}^n\) is closed. 2. The empty set \(\emptyset\) is closed. 3. Intersections of closed sets (including uncountably many) are closed. (Think: DeMorgan's Laws.) 4. If \(x\) is any point in the complement of a closed set, then there is some radius \(r\) such that all of \(N_r(x)\) belongs to the complement as well. 5. Any union of finitely many closed sets is closed.

1.3. Other Basic Facts

Some sets are both open and closed (e.g., \(\emptyset\) and \({\mathbb R}^n\)).
Some sets are neither open nor closed (think \(\mathbb Q\): every open interval contains some rational numbers but also contains some irrational numbers).
For any set \(E\), we may always define its interior \(E^{\rm{o}}\) to be the union of all open sets it contains. The interior is itself an open set. Informally, it represents the unique largest open set contained in \(E\). For example, \(\mathbb{Q}^{\rm{o}} = \emptyset\) because \(\mathbb{Q}\) contains no intervals.
For any set \(E\), we may define its closure \(\overline{E}\) to be the intersection of all closed sets containing \(E\). Informally, it represents the unique smallest closed set that contains \(E\). Also \(\overline{E} = ((E^c)^{\rm{o}})^c\). For example, \(\overline{\mathbb Q} = {\mathbb R}\) because \(\mathbb{Q}^c\) also contains no intervals.
We define the boundary \(\partial E\) to be those points in \(\overline{E} \setminus E^{\rm{o}}\). For example, \(\partial \mathbb{Q} = {\mathbb R}\).

2. Compactness

If \(E \subset {\mathbb R}^n\) is a set, we say that \(\mathcal U\) is an open cover of \(E\) when the elements of \(\mathcal U\) are themselves open sets in \({\mathbb R}^n\) and the union of those open sets contains \(E\).

Example

If \(E = [0,1)\), then the intervals \(\{(-n,1-\frac{1}{n})\}_{n=1}^\infty\) form an open cover of \(E\). That's because each member of the collection is an open set and the union of all sets in the collection contains \(E\).

Given a set \(E\) and an open cover \(\mathcal U\) of \(E\), a subcover \(\mathcal U'\) of \(\mathcal U\) is any subset of \(\mathcal U\) which is also an open cover of \(E\).

We say that a set \(E \subset {\mathbb R}^n\) is compact when every open cover \(\mathcal U\) always has a finite subcover.

Example

The most trivial example of a compact set is a finite set. Suppose \(E = \{x_1,\ldots,x_n\}\) and let \(\mathcal U\) be an open cover of \(E\). Since it's a cover, every point \(x_j \in E\) belongs to some set \(U_j \in \mathcal U\). This means that \(\{U_1,\ldots,U_n\}\) is a subcover of \(\mathcal U\).

Compactness is a far-reaching and important topological generalization of the concept of finiteness. Compact sets aren't always finite sets, but they have many of the same key properties that finite sets have.

3. Connectedness

A nonempty set \(E \subset {\mathbb R}^n\) is called disconnected when there are two open sets \(A\) and \(B\) with the following properties:

Every point \(x \in E\) belongs to either \(A\) or \(B\).
No point \(x \in E\) belongs to both \(A\) and \(B\).
Neither of \(A\) or \(B\) contains the entire set \(E\) (and as a corollary, both \(A\) and \(B\) contain some point in \(E\).)

Example

Any set \(E \subset {\mathbb R}\) which has at least two points and isn't an interval is disconnected.

Proof

Meta (Main Idea)

Argue that any set \(E\) with at least two points which isn't an interval must admit points \(a < b < c\) such that both \(a\) and \(c\) belong to the set but \(b\) doesn't. Then take your open sets to be \((-\infty,b)\) and \((b,\infty)\) and show step-by-step that they form a separation of \(E\).

Any set which is not disconnected is called connected.