5.12 Quasi-compact spaces and maps
The phrase “compact” will be reserved for Hausdorff topological spaces. And many spaces occurring in algebraic geometry are not Hausdorff.
Definition 5.12.1. Quasi-compactness.
We say that a topological space X is quasi-compact if every open covering of X has a finite subcover.
We say that a continuous map f : X \to Y is quasi-compact if the inverse image f^{-1}(V) of every quasi-compact open V \subset Y is quasi-compact.
We say a subset Z \subset X is retrocompact if the inclusion map Z \to X is quasi-compact.
In many texts on topology a space is called compact if it is quasi-compact and Hausdorff; and in other texts the Hausdorff condition is omitted. To avoid confusion in algebraic geometry we use the term quasi-compact. The notion of quasi-compactness of a map is very different from the notion of a “proper map”, since there we require (besides closedness and separatedness) the inverse image of any quasi-compact subset of the target to be quasi-compact, whereas in the definition above we only consider quasi-compact open sets.
Lemma 5.12.2. A composition of quasi-compact maps is quasi-compact.
Proof.
This is immediate from the definition.
\square
Lemma 5.12.3. A closed subset of a quasi-compact topological space is quasi-compact.
Proof.
Let E \subset X be a closed subset of the quasi-compact space X. Let E = \bigcup V_ j be an open covering. Choose U_ j \subset X open such that V_ j = E \cap U_ j. Then X = (X \setminus E) \cup \bigcup U_ j is an open covering of X. Hence X = (X \setminus E) \cup U_{j_1} \cup \ldots \cup U_{j_ n} for some n and indices j_ i. Thus E = V_{j_1} \cup \ldots \cup V_{j_ n} as desired.
\square
Lemma 5.12.4. Let X be a Hausdorff topological space.
If E \subset X is quasi-compact, then it is closed.
If E_1, E_2 \subset X are disjoint quasi-compact subsets then there exists opens E_ i \subset U_ i with U_1 \cap U_2 = \emptyset .
Proof.
Proof of (1). Let x \in X, x \not\in E. For every e \in E we can find disjoint opens V_ e and U_ e with e \in V_ e and x \in U_ e. Since E \subset \bigcup V_ e we can find finitely many e_1, \ldots , e_ n such that E \subset V_{e_1} \cup \ldots \cup V_{e_ n}. Then U = U_{e_1} \cap \ldots \cap U_{e_ n} is an open neighbourhood of x which avoids V_{e_1} \cup \ldots \cup V_{e_ n}. In particular it avoids E. Thus E is closed.
Proof of (2). In the proof of (1) we have seen that given x \in E_1 we can find an open neighbourhood x \in U_ x and an open E_2 \subset V_ x such that U_ x \cap V_ x = \emptyset . Because E_1 is quasi-compact we can find a finite number x_ i \in E_1 such that E_1 \subset U = U_{x_1} \cup \ldots \cup U_{x_ n}. We take V = V_{x_1} \cap \ldots \cap V_{x_ n} to finish the proof.
\square
Lemma 5.12.5. Let X be a quasi-compact Hausdorff space. Let E \subset X. The following are equivalent: (a) E is closed in X, (b) E is quasi-compact.
Proof.
The implication (a) \Rightarrow (b) is Lemma 5.12.3. The implication (b) \Rightarrow (a) is Lemma 5.12.4.
\square
The following is really a reformulation of the quasi-compact property.
Lemma 5.12.6. Let X be a quasi-compact topological space. If \{ Z_\alpha \} _{\alpha \in A} is a collection of closed subsets such that the intersection of each finite subcollection is nonempty, then \bigcap _{\alpha \in A} Z_\alpha is nonempty.
Proof.
We suppose that \bigcap _{\alpha \in A} Z_{\alpha } = \emptyset . So we have \bigcup _{\alpha \in A} (X\setminus Z_{\alpha })=X by complementation. As the subsets Z_\alpha are closed, \bigcup _{\alpha \in A} (X \setminus Z_{\alpha }) is an open covering of the quasi-compact space X. Thus there exists a finite subset J \subset A such that X = \bigcup _{\alpha \in J} (X\setminus Z_{\alpha }). The complementary is then empty, which means that \bigcap _{\alpha \in J} Z_{\alpha } = \emptyset . It proves there exists a finite subcollection of \{ Z_{\alpha }\} _{\alpha \in J} verifying \bigcap _{\alpha \in J} Z_{\alpha }=\emptyset , which concludes by contraposition.
\square
Lemma 5.12.7. Let f : X \to Y be a continuous map of topological spaces.
If X is quasi-compact, then f(X) is quasi-compact.
If f is quasi-compact, then f(X) is retrocompact.
Proof.
If f(X) = \bigcup V_ i is an open covering, then X = \bigcup f^{-1}(V_ i) is an open covering. Hence if X is quasi-compact then X = f^{-1}(V_{i_1}) \cup \ldots \cup f^{-1}(V_{i_ n}) for some i_1, \ldots , i_ n \in I and hence f(X) = V_{i_1} \cup \ldots \cup V_{i_ n}. This proves (1). Assume f is quasi-compact, and let V \subset Y be quasi-compact open. Then f^{-1}(V) is quasi-compact, hence by (1) we see that f(f^{-1}(V)) = f(X) \cap V is quasi-compact. Hence f(X) is retrocompact.
\square
Lemma 5.12.8. Let X be a topological space. Assume that
X is nonempty,
X is quasi-compact, and
X is Kolmogorov.
Then X has a closed point.
Proof.
Consider the set
\mathcal{T} = \{ Z \subset X \mid Z = \overline{\{ x\} } \text{ for some }x \in X\}
of all closures of singletons in X. It is nonempty since X is nonempty. Make \mathcal{T} into a partially ordered set using the relation of inclusion. Suppose Z_\alpha , \alpha \in A is a totally ordered subset of \mathcal{T}. By Lemma 5.12.6 we see that \bigcap _{\alpha \in A} Z_\alpha \not= \emptyset . Hence there exists some x \in \bigcap _{\alpha \in A} Z_\alpha and we see that Z = \overline{\{ x\} }\in \mathcal{T} is a lower bound for the family. By Zorn's lemma there exists a minimal element Z \in \mathcal{T}. As X is Kolmogorov we conclude that Z = \{ x\} for some x and x \in X is a closed point.
\square
Lemma 5.12.9. Let X be a quasi-compact Kolmogorov space. Then the set X_0 of closed points of X is quasi-compact.
Proof.
Let X_0 = \bigcup U_{i, 0} be an open covering. Write U_{i, 0} = X_0 \cap U_ i for some open U_ i \subset X. Consider the complement Z of \bigcup U_ i. This is a closed subset of X, hence quasi-compact (Lemma 5.12.3) and Kolmogorov. By Lemma 5.12.8 if Z is nonempty it would have a closed point which contradicts the fact that X_0 \subset \bigcup U_ i. Hence Z = \emptyset and X = \bigcup U_ i. Since X is quasi-compact this covering has a finite subcover and we conclude.
\square
Lemma 5.12.10. Let X be a topological space. Assume
X is quasi-compact,
X has a basis for the topology consisting of quasi-compact opens, and
the intersection of two quasi-compact opens is quasi-compact.
For any x \in X the connected component of X containing x is the intersection of all open and closed subsets of X containing x.
Proof.
Let T be the connected component containing x. Let S = \bigcap _{\alpha \in A} Z_\alpha be the intersection of all open and closed subsets Z_\alpha of X containing x. Note that S is closed in X. Note that any finite intersection of Z_\alpha 's is a Z_\alpha . Because T is connected and x \in T we have T \subset S. It suffices to show that S is connected. If not, then there exists a disjoint union decomposition S = B \amalg C with B and C open and closed in S. In particular, B and C are closed in X, and so quasi-compact by Lemma 5.12.3 and assumption (1). By assumption (2) there exist quasi-compact opens U, V \subset X with B = S \cap U and C = S \cap V (details omitted). Then U \cap V \cap S = \emptyset . Hence \bigcap _\alpha U \cap V \cap Z_\alpha = \emptyset . By assumption (3) the intersection U \cap V is quasi-compact. By Lemma 5.12.6 for some \alpha ' \in A we have U \cap V \cap Z_{\alpha '} = \emptyset . Since X \setminus (U \cup V) is disjoint from S and closed in X hence quasi-compact, we can use the same lemma to see that Z_{\alpha ''} \subset U \cup V for some \alpha '' \in A. Then Z_\alpha = Z_{\alpha '} \cap Z_{\alpha ''} is contained in U \cup V and disjoint from U \cap V. Hence Z_\alpha = U \cap Z_\alpha \amalg V \cap Z_\alpha is a decomposition into two open pieces, hence U \cap Z_\alpha and V \cap Z_\alpha are open and closed in X. Thus, if x \in B say, then we see that S \subset U \cap Z_\alpha and we conclude that C = \emptyset .
\square
Lemma 5.12.11. Let X be a topological space. Assume X is quasi-compact and Hausdorff. For any x \in X the connected component of X containing x is the intersection of all open and closed subsets of X containing x.
Proof.
Let T be the connected component containing x. Let S = \bigcap _{\alpha \in A} Z_\alpha be the intersection of all open and closed subsets Z_\alpha of X containing x. Note that S is closed in X. Note that any finite intersection of Z_\alpha 's is a Z_\alpha . Because T is connected and x \in T we have T \subset S. It suffices to show that S is connected. If not, then there exists a disjoint union decomposition S = B \amalg C with B and C open and closed in S. In particular, B and C are closed in X, and so quasi-compact by Lemma 5.12.3. By Lemma 5.12.4 there exist disjoint opens U, V \subset X with B \subset U and C \subset V. Then X \setminus U \cup V is closed in X hence quasi-compact (Lemma 5.12.3). It follows that (X \setminus U \cup V) \cap Z_\alpha = \emptyset for some \alpha by Lemma 5.12.6. In other words, Z_\alpha \subset U \cup V. Thus Z_\alpha = Z_\alpha \cap V \amalg Z_\alpha \cap U is a decomposition into two open pieces, hence U \cap Z_\alpha and V \cap Z_\alpha are open and closed in X. Thus, if x \in B say, then we see that S \subset U \cap Z_\alpha and we conclude that C = \emptyset .
\square
Lemma 5.12.12. Let X be a topological space. Assume
X is quasi-compact,
X has a basis for the topology consisting of quasi-compact opens, and
the intersection of two quasi-compact opens is quasi-compact.
For a subset T \subset X the following are equivalent:
T is an intersection of open and closed subsets of X, and
T is closed in X and is a union of connected components of X.
Proof.
It is clear that (a) implies (b). Assume (b). Let x \in X, x \not\in T. Let x \in C \subset X be the connected component of X containing x. By Lemma 5.12.10 we see that C = \bigcap V_\alpha is the intersection of all open and closed subsets V_\alpha of X which contain C. In particular, any pairwise intersection V_\alpha \cap V_\beta occurs as a V_\alpha . As T is a union of connected components of X we see that C \cap T = \emptyset . Hence T \cap \bigcap V_\alpha = \emptyset . Since T is quasi-compact as a closed subset of a quasi-compact space (see Lemma 5.12.3) we deduce that T \cap V_\alpha = \emptyset for some \alpha , see Lemma 5.12.6. For this \alpha we see that U_\alpha = X \setminus V_\alpha is an open and closed subset of X which contains T and not x. The lemma follows.
\square
Lemma 5.12.13. Let X be a Noetherian topological space.
The space X is quasi-compact.
Any subset of X is retrocompact.
Proof.
Suppose X = \bigcup U_ i is an open covering of X indexed by the set I which does not have a refinement by a finite open covering. Choose i_1, i_2, \ldots elements of I inductively in the following way: Choose i_{n + 1} such that U_{i_{n + 1}} is not contained in U_{i_1} \cup \ldots \cup U_{i_ n}. Thus we see that X \supset (X \setminus U_{i_1}) \supset (X \setminus U_{i_1} \cup U_{i_2}) \supset \ldots is a strictly decreasing infinite sequence of closed subsets. This contradicts the fact that X is Noetherian. This proves the first assertion. The second assertion is now clear since every subset of X is Noetherian by Lemma 5.9.2.
\square
Lemma 5.12.14. A quasi-compact locally Noetherian space is Noetherian.
Proof.
The conditions imply immediately that X has a finite covering by Noetherian subsets, and hence is Noetherian by Lemma 5.9.4.
\square
Lemma 5.12.15 (Alexander subbase theorem). Let X be a topological space. Let \mathcal{B} be a subbase for X. If every covering of X by elements of \mathcal{B} has a finite refinement, then X is quasi-compact.
Proof.
Assume there is an open covering of X which does not have a finite refinement. Using Zorn's lemma we can choose a maximal open covering X = \bigcup _{i \in I} U_ i which does not have a finite refinement (details omitted). In other words, if U \subset X is any open which does not occur as one of the U_ i, then the covering X = U \cup \bigcup _{i \in I} U_ i does have a finite refinement. Let I' \subset I be the set of indices such that U_ i \in \mathcal{B}. Then \bigcup _{i \in I'} U_ i \not= X, since otherwise we would get a finite refinement covering X by our assumption on \mathcal{B}. Pick x \in X, x \not\in \bigcup _{i \in I'} U_ i. Pick i \in I with x \in U_ i. Pick V_1, \ldots , V_ n \in \mathcal{B} such that x \in V_1 \cap \ldots \cap V_ n \subset U_ i. This is possible as \mathcal{B} is a subbasis for X. Note that V_ j does not occur as a U_ i. By maximality of the chosen covering we see that for each j there exist i_{j, 1}, \ldots , i_{j, n_ j} \in I such that X = V_ j \cup U_{i_{j, 1}} \cup \ldots \cup U_{i_{j, n_ j}}. Since V_1 \cap \ldots \cap V_ n \subset U_ i we conclude that X = U_ i \cup \bigcup U_{i_{j, l}} a contradiction.
\square
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