Appendix E. Topology（拓扑）
In Section 17 and in the sections that deal with algebraic geometry, we need to use some notions from topology. In this appendix, we give a brief description of these notions.
1 Topological Spaces
Let \(X\) be a set. A topology on \(X\) is a collection \(\mathcal{T}\) of subsets of \(X\) that satisfy the following properties:
- \(X \in \mathcal{T}\) and \(\varnothing \in \mathcal{T}\),
- If \(U, V \in \mathcal{T}\), then \(U \cap V \in \mathcal{T}\),
- If \(\{U_i\}\) is a collection of subsets of \(X\) such that each \(U_i \in \mathcal{T}\), then \(\bigcup_i U_i \in \mathcal{T}\).
A set with a topology on it is called a topological space. The elements of a topology are called open sets. A subset \(C\) of \(X\) is called closed if \(X - C\) is open. We can define a topology by specifying which are the closed sets. The closed sets of a topology on \(X\) satisfy the following properties.
- Both \(X\) and \(\varnothing\) are closed sets.
- If \(A\) and \(B\) are closed sets, then \(A \cup B\) is closed.
- If \(\{A_i\}\) is a collection of closed sets, then \(\bigcap_i A_i\) is closed.
These properties follow immediately from the definition of a topology and the DeMorgan laws of set theory.
Example 1.1 The standard topology on \(\mathbb{R}\) is defined as follows. A nonempty subset \(U\) of \(\mathbb{R}\) is open, provided that for every \(x \in U\) there is a positive number \(\delta\) such that the open interval \((x-\delta, x+\delta)\) is contained in \(U\). An easy exercise shows that this does make \(\mathbb{R}\) into a topological space.
Example 1.2 Recall that a metric space is a set \(X\) together with a function \(d\) from \(X \times X\) to the nonnegative real numbers such that (i) \(d(x,x) = 0\) for all \(x \in X\), and if \(d(x,y) = 0\), then \(x = y\), (ii) \(d(x,y) = d(y,x)\) for all \(x, y \in X\), and (iii) \(d(x,y) + d(y,z) \geq d(x,z)\) for all \(x, y, z \in X\). The function \(d\) is called a metric. We can use \(d\) to put a topology on \(X\). A nonempty subset \(U\) of \(X\) is defined to be open, provided that for every \(x \in U\) there is a positive number \(\delta\) such that the open ball
centered at \(x\) with radius \(\delta\) is contained in \(U\). This topology is called the metric space topology. The standard topology on \(\mathbb{R}\) is an example of this construction. For another example, if \(X = \mathbb{R}^n\), then we obtain a topology on \(\mathbb{R}^n\), since we have a distance function on \(\mathbb{R}^n\).
Example 1.3 If \(X\) is a topological space and \(Y\) is a subset of \(X\), then we can put a topology on \(Y\). We define a subset \(V\) of \(Y\) to be open if there is an open subset of \(X\) with \(V = Y \cap U\). It is straightforward to show that \(Y\) is indeed a topological space. This topology on \(Y\) is called the subspace topology.
Example 1.4 Let \(X\) be a set. The discrete topology on \(X\) is the topology for which every subset of \(X\) is open.
Example 1.5 Let \(X\) be a set. We define a topology on \(X\) by defining a proper subset of \(X\) to be closed if it is finite. The definition of a topology is easy to verify in this case. Note that a nonempty subset is open exactly when its complement is finite. This topology is called the finite complement topology on \(X\).
There are often more efficient ways to describe a topology than to list all of the closed sets. If \(X\) is a topological space, a basis for the topology on \(X\) is a collection of open subsets such that every open set is a union of elements from the basis. For example, the collection of open intervals forms a basis for the standard topology on \(\mathbb{R}\). Similarly, the collection of open balls forms a basis for the metric topology on a metric space. A collection \(\mathcal{C}\) of sets forms a basis for a topology on \(X\) provided that, given any two sets \(U\) and \(V\) in \(\mathcal{C}\), for any \(x \in U \cap V\) there is a set \(W\) in \(\mathcal{C}\) such that \(x \in W\) and \(W \subseteq U \cap V\). The proof of this fact is left to Problem 1.
Example 1.6 Let \(R\) be a commutative ring, and let \(I\) be an ideal of \(R\). The \(I\)-adic topology on \(R\) is defined as follows. A nonempty subset of \(R\) is open if it is the union of sets of the form \(a + I^n\) for some \(a \in R\) and \(n \geq 0\). We set \(I^0 = R\) for this definition. In other words, \(\{a + I^n : a \in R, n \geq 0\}\) is a basis for this topology. The only nontrivial thing to verify to see that this does define a topology is that the intersection of two open sets is open. If \(\bigcup_i (a_i + I^{n_i})\) and \(\bigcup_j (b_j + I^{m_j})\) are open sets, then their intersection is \(\bigcup_{i,j} (a_i + I^{n_i}) \cap (b_j + I^{m_j})\). It then suffices to show that \((a + I^n) \cap (b + I^m)\) is open for any \(a, b \in R\) and \(n, m \geq 0\). To prove this, we can assume that \(n \geq m\), so \(I^m \subseteq I^n\). If this intersection is empty, there is nothing to prove. If not, let \(c \in (a + I^n) \cap (b + I^m)\). Then \(c + I^n = a + I^n\) and \(c + I^m = b + I^m\), so
an open set.
Example 1.7 Here is an example that arises in algebraic geometry. Let \(R\) be a commutative ring, and let \(X = \mathrm{spec}(R)\) be the set of all prime ideals of \(R\). If \(S\) is a subset of \(R\), we set \(Z(S) = \{P \in X : S \subseteq P\}\). We define the Zariski topology on \(X\) by defining a subset of \(X\) to be closed if it is of the form \(Z(S)\) for some subset \(S\) of \(R\). We verify that this is a topology on \(X\). First, note that \(R = Z(\{0\})\) and \(\varnothing = Z(\{1\})\). Next, it is easy to see that \(\bigcup_i Z(S_i) = Z(\bigcap_i S_i)\). Finally, we show that \(Z(S) \cup Z(T) = Z(ST)\), where \(ST = \{st : s \in S, t \in T\}\). Let \(P \in Z(ST)\). If \(P \notin Z(S)\), then there is an \(s \in S\) with \(s \notin P\). Since \(st \in P\) for all \(t \in T\), we see that \(T \subseteq P\), since \(P\) is a prime ideal. Thus, \(P \in Z(T)\). Therefore, \(Z(ST) \subseteq Z(S) \cup Z(T)\). For the reverse inclusion, let \(P \in Z(S) \cup Z(T)\). Then \(S \subseteq P\) or \(T \subseteq P\). Since \(P\) is an ideal, in either case we have \(ST \subseteq P\), so \(P \in Z(ST)\). We point out the relation between the Zariski topology on \(\mathrm{spec}(R)\) and the Zariski topology that we define in Section 21. We require some concepts from Section 21 in order to do this. Let \(C\) be an algebraically closed field, let \(V\) be a variety in \(C^n\), and let \(R = C[V]\) be the coordinate ring of \(V\). Then \(V\) is homeomorphic to the subspace of \(\mathrm{spec}(R)\) consisting of all maximal ideals of \(R\). This is mostly a consequence of the Nullstellensatz.
Example 1.8 Let \(X\) and \(Y\) be topological spaces. Then the product \(X \times Y\) can be given a topology in the following way. We define a subset of \(X \times Y\) to be open if it is a union of sets of the form \(U \times V\), where \(U\) is an open subset of \(X\) and \(V\) is an open subset of \(Y\); that is, the collection \(\mathcal{C}\) of these subsets is a basis for the topology. It is easy to verify that this collection does satisfy the requirement to be a basis. If \((x,y) \in (U \times V) \cap (U' \times V')\), then \((U \cap U') \times (V \cap V')\) is a basic open set that contains \((x,y)\) and is contained in \((U \times V) \cap (U' \times V')\). This topology on \(X \times Y\) is called the product topology. More generally, if \(X_1, \ldots, X_n\) is a collection of topological spaces, then we get a similar topology on \(X_1 \times \cdots \times X_n\).
Example 1.9 Let \(I\) be a set, and let \(\{X_i\}_{i \in I}\) be a collection of topological spaces. We can generalize the previous construction to define the product topology on \(\prod_i X_i\). If \(I\) is infinite, then we need an extra step in the definition. Consider the set \(\mathcal{S}\) of all subsets of \(\prod_i X_i\) of the form \(\prod_i U_i\), where \(U_i\) is open in \(X_i\) and \(U_i = X_i\) for all but finitely many \(i\). If \(I\) is finite, then \(\mathcal{S}\) is the basis described in the previous example. If \(I\) is not finite, then we let \(\mathcal{C}\) be the collection of all sets that are finite intersections of elements of \(\mathcal{S}\). It is not hard to show that \(\mathcal{C}\) does form a basis for a topology on \(\prod_i X_i\), and we call this the product topology on \(\prod_i X_i\). It is true that \(\mathcal{S}\) also forms a basis for a topology on \(X\), the box topology, but this topology is not as useful as the product topology.
2 Topological Properties
There are various properties of topological spaces that we need to discuss. Let \(X\) be a topological space. Then \(X\) is called Hausdorff if for every two distinct points \(x, y \in X\), there are disjoint open sets \(U\) and \(V\) with \(x \in U\) and \(y \in V\). For example, if \(X\) is a metric space, then we see that the metric space topology is Hausdorff. If \(x, y \in X\) are distinct points, let \(\delta = \frac{1}{2} d(x,y)\). Then the open balls \(B(x,\delta)\) and \(B(y,\delta)\) are disjoint open sets containing \(x\) and \(y\), respectively. The finite complement topology on an infinite set \(X\) is not Hausdorff, since any two nonempty open sets must have a nonempty intersection. If \(R\) is an integral domain, then we show that the Zariski topology on \(\mathrm{spec}(R)\) is not Hausdorff either. We note that the zero ideal is prime and that \((0) \notin Z(S)\) for any \(S\) unless \(Z(S) = \mathrm{spec}(R)\). Consequently, \((0)\) is contained in any nonempty open set. Therefore, any two nonempty open sets have a nonempty intersection, so \(\mathrm{spec}(R)\) is not Hausdorff.
The next concept we discuss is compactness. If \(X\) is a topological space, then an open cover of \(X\) is a collection of open sets whose union is \(X\). If \(\{U_i\}\) is an open cover of \(X\), then a finite subcover is a finite subset of the collection whose union is also \(X\). The space \(X\) is called compact if every open cover of \(X\) has a finite subcover.
Example 2.1 The space \(\mathbb{R}\) is not compact, since \(\{(a, a+1) : a \in \mathbb{R}\}\) is an open cover of \(\mathbb{R}\) that does not have a finite subcover. Subspaces of \(\mathbb{R}^n\) may be compact. Recall that a subset \(Y\) of \(\mathbb{R}^n\) is bounded if \(Y\) is contained in an open ball \(B(0,\delta)\) for some \(\delta\). The Heine–Borel theorem says that a subset of \(\mathbb{R}^n\) is compact if and only if it is closed and bounded.
Example 2.2 Let \(R\) be a commutative ring. The Zariski topology on \(\mathrm{spec}(R)\) is compact, as we now show. Suppose that \(\{U_i\}\) is an open cover of \(\mathrm{spec}(R)\). If \(Z(S_i)\) is the complement of \(U_i\), then \(\bigcap_i Z(S_i) = Z(\bigcup S_i) = \varnothing\). We first point out that if \(I_i\) is the ideal generated by \(S_i\), then \(Z(I_i) = Z(S_i)\) and \(Z(\bigcup S_i) = Z(\sum_i I_i)\). The ideal \(\sum_i I_i\) cannot be a proper ideal, since if it is, then it is contained in a maximal ideal, and so \(Z(\sum_i I_i) \neq \varnothing\). Thus, \(\sum_i I_i = R\), so there is a finite subcollection \(I_1, \ldots, I_n\) and elements \(r_i \in I_i\) such that \(r_1 + \cdots + r_n = 1\). Then \(\sum_{i=1}^n I_i = R\), and so there is no prime ideal that contains each \(I_i\). Consequently, \(\bigcap_{i=1}^n Z(I_i) = \varnothing\), so \(\bigcup_{i=1}^n U_i = \mathrm{spec}(R)\). We have found a finite subcover of \(\{U_i\}\), so \(\mathrm{spec}(R)\) is compact.
Example 2.3 Let \(\{X_i\}\) be a collection of compact topological spaces. Then the product \(\prod_i X_i\) is compact in the product topology. This nontrivial fact is the Tychonoff theorem and can be found in Chapter 5 of Munkres [22].
Let \(X\) be a topological space, and let \(S\) be a subset of \(X\). The closure \(\overline{S}\) of \(S\) is defined to be the intersection of all closed sets that contain \(S\). Since \(X\) is closed, the closure is a closed set that contains \(S\). The main property about this concept is given in the following proposition. The simple proof is left to Problem 4.
Proposition 2.4 Let \(X\) be a topological space, and let \(S\) be a subset of \(X\).
- If \(C\) is any closed set that contains \(S\), then \(\overline{S} \subseteq C\).
- If \(U\) is an open set with \(U \cap \overline{S} \neq \varnothing\), then \(U \cap S \neq \varnothing\).
One consequence of this proposition is that an element \(x \in X\) is in the closure of a subset \(S\), provided that for any open set \(U\) that contains \(x\), we have \(U \cap S \neq \varnothing\). This is a useful way to determine when an element is in \(\overline{S}\).
If \(X\) is a topological space and \(Y\) is a subset of \(X\), then \(Y\) is dense in \(X\) if \(\overline{Y} = X\). For example, any set \(S\) is dense in its closure \(\overline{S}\). The open interval \((0,1)\) is dense in \([0,1]\). If \(R\) is a commutative ring, then we show that any nonempty open subset of \(\mathrm{spec}(R)\) is dense in \(\mathrm{spec}(R)\). If \(U\) is an open set, then \(\overline{U}\) is a closed subset of \(\mathrm{spec}(R)\), and \(U \cap (\mathrm{spec}(R) - \overline{U}) = \varnothing\). However, we have seen that any two nonempty open sets in \(\mathrm{spec}(R)\) have a nonempty intersection. This forces \(\overline{U} = \mathrm{spec}(R)\), so \(U\) is dense in \(\mathrm{spec}(R)\).
We have not yet discussed functions between topological spaces. If \(X\) and \(Y\) are topological spaces, then a function \(f : X \to Y\) is called continuous if \(f^{-1}(V)\) is open in \(X\) for any open set \(V\) in \(Y\). If \(X\) and \(Y\) are subsets of \(\mathbb{R}\), then this definition of continuity is equivalent to the limit definition given in calculus; see Problem 6.
Let \(X\) be a topological space, and let \(\sim\) be an equivalence relation on \(X\). We let \(X^*\) be the set of equivalence classes, and for \(x \in X\) we denote the equivalence class of \(x\) by \(\overline{x}\). We have a natural surjective function \(\pi : X \to X^*\) given by \(\pi(x) = \overline{x}\). We define the quotient topology on \(X^*\) as follows. A subset \(Y\) of \(X^*\) is defined to be open if \(\pi^{-1}(Y)\) is open in \(X\). It is a simple exercise to show that this does define a topology on \(X^*\) and that \(\pi\) is continuous. Moreover, the quotient topology is the topology on \(X^*\) that has the fewest open sets for which \(\pi\) is continuous.
We end this appendix with a concept that will arise in Section 17. A topological space \(X\) is called connected if \(X\) is not the union of two disjoint closed sets. For example, \(\mathbb{R}\) is a connected set, while the subspace \([0,1] \cup [2,3]\) is not connected. On the other extreme, a space \(X\) is called totally disconnected if the only connected subsets of \(X\) are singleton sets. A space with the discrete topology is totally disconnected. The topology on a Galois group we define in Section 17 is totally disconnected.