## 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. $\square$

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 $(V_i)$ is a refinement of $(U_j)$ is each $V_i$ is contained in some $U_j$.) 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.

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