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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.

  1. We say that a topological space X is quasi-compact if every open covering of X has a finite subcover.

  2. 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.

  3. 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.

  1. If E \subset X is quasi-compact, then it is closed.

  2. 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.

  1. If X is quasi-compact, then f(X) is quasi-compact.

  2. 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

  1. X is nonempty,

  2. X is quasi-compact, and

  3. 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

  1. X is quasi-compact,

  2. X has a basis for the topology consisting of quasi-compact opens, and

  3. 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

  1. X is quasi-compact,

  2. X has a basis for the topology consisting of quasi-compact opens, and

  3. the intersection of two quasi-compact opens is quasi-compact.

For a subset T \subset X the following are equivalent:

  1. T is an intersection of open and closed subsets of X, and

  2. 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.

  1. The space X is quasi-compact.

  2. 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


Comments (3)

Comment #6508 by Patrick Rabau on

Definition 005A (1) An equivalent definition of quasicompact is to require that every open cover has a finite subcover, instead of a finite refinement. (A family is a refinement of is each is contained in some .) It's easily seen to be the same thing, but the definition by subcovers is more usual and is easier to work with. And that's what is actually used in all the proofs in this section. So maybe replace the definition.

Comment #9799 by Patrick Rabau on

Definition 005A (3) The notion of retrocompact set is not intrinsic to the set, it's relative to the enclosing space . So to emphasize that, we can write "We say a subset is retrocompact in if ...".


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