The Stacks project

Proof. This is immediate from the definition. $\square$

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. $\square$

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. $\square$

Proof. The implication (a) $\Rightarrow $ (b) is Lemma 5.12.3. The implication (b) $\Rightarrow $ (a) is Lemma 5.12.4. $\square$

Proof. Omitted. $\square$

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. $\square$

Proof. Consider the set

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. $\square$

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. $\square$

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 $. $\square$

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 $. $\square$

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. $\square$

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. $\square$

Proof. The conditions imply immediately that $X$ has a finite covering by Noetherian subsets, and hence is Noetherian by Lemma 5.9.4. $\square$

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. $\square$