Existence and basic properties of irreducible components

Let $T\subseteq X$ be irreducible and let $\bar{T}$ be its closure in $X$ [F2]. Suppose $\bar{T}=Z_1\cup Z_2$ with $Z_1,Z_2$ closed in the subspace $\bar{T}$. By [F5] each $Z_i$ is the trace $Z_i=\bar{T}\cap F_i$ of a closed $F_i\subseteq X$, so each intersection $T\cap Z_i$ is closed in $T$; and $T=(T\cap Z_1)\cup (T\cap Z_2)$. Since $T$ is irreducible and nonempty by [F1], $T=T\cap Z_i$ for at least one $i$, that is, $T\subseteq Z_i$. The closure of $T$ computed inside the subspace $\bar{T}$ equals $\bar{T}$ by [F4], and it is contained in $Z_i$ because $Z_i$ is closed in $\bar{T}$ and contains $T$; hence $Z_i=\bar{T}$. So $\bar{T}$ admits no decomposition into two proper closed subsets and is irreducible, which is clause 1.

Let $T\subseteq X$ be irreducible [F1] and let $P$ be the poset of those irreducible $T^{\mathrm{\prime }}$ with $T\subseteq T^{\mathrm{\prime }}\subseteq X$, ordered by inclusion; $P$ is nonempty because $T\in P$. Every chain in $P$ has an upper bound in $P$: for a chain $\mathcal{C}\subseteq P$ put $E:={\bigcup }_{T^{\mathrm{\prime }}\in \mathcal{C}}{T}^{\mathrm{\prime }}$, so that $T\subseteq E$, and $E$ is irreducible, because if $E=Z_1\cup Z_2$ with $Z_i$ closed in $E$ [F5], then for every $T^{\mathrm{\prime }}\in \mathcal{C}$ the pair $(T^{\mathrm{\prime }}\cap Z_1,T^{\mathrm{\prime }}\cap Z_2)$ is a decomposition of the irreducible nonempty set $T^{\mathrm{\prime }}$ into closed subsets [F1], so $T^{\mathrm{\prime }}\subseteq Z_1$ or $T^{\mathrm{\prime }}\subseteq Z_2$; if every $T^{\mathrm{\prime }}$ is contained in $Z_1$ then $E=Z_1$, and otherwise some $T_0^{\mathrm{\prime }}\in \mathcal{C}$ satisfies $T_0^{\mathrm{\prime }}\subseteq ̸Z_1$, whence $T_0^{\mathrm{\prime }}\subseteq Z_2$, and for any $T^{\mathrm{\prime }}\in \mathcal{C}$ either $T^{\mathrm{\prime }}\subseteq T_0^{\mathrm{\prime }}$, which gives $T^{\mathrm{\prime }}\subseteq Z_2$, or $T_0^{\mathrm{\prime }}\subseteq T^{\mathrm{\prime }}$, in which case $T^{\mathrm{\prime }}\subseteq ̸Z_1$ and irreducibility of $T^{\mathrm{\prime }}$ gives $T^{\mathrm{\prime }}\subseteq Z_2$; so $E=Z_2$ in that case too. Thus every chain in $P$ has an upper bound, and Zorn's lemma [F7], which is a consequence of the Axiom of Choice [F9] assumed in the statement, produces a maximal element $C\in P$ [F8]. Then $C$ is an irreducible component: it is irreducible and contains $T$, and if $D$ is irreducible with $C\subseteq D\subseteq X$ then $T\subseteq D$, so $D\in P$ and maximality of $C$ in $P$ gives $D=C$.

Let $C\subseteq X$ be irreducible [F1] and let $W\subseteq X$ be closed with $X=C\cup W$ and $C\subseteq ̸W$. Then $C\setminus W\mathrm{\ne }\mathrm{\varnothing }$, and $X\setminus W=C\setminus W\subseteq C$ because $X=C\cup W$; in particular the set whose closure is taken in clause 6 is nonempty. Put $Z^{\mathrm{\prime }}:=\overline{X\setminus W}$ [F2]. Then $Z^{\mathrm{\prime }}$ is closed in $X$ by [F3], and $C\subseteq Z^{\mathrm{\prime }}\cup W$, so $C=(C\cap Z^{\mathrm{\prime }})\cup (C\cap W)$ with $C\cap Z^{\mathrm{\prime }}$ and $C\cap W$ closed in $C$ [F5]. Since $C$ is irreducible and nonempty [F1], one of the two sets equals $C$; the alternative $C=C\cap W$ would give $C\subseteq W$, which is excluded, so $C=C\cap Z^{\mathrm{\prime }}$ and $C\subseteq Z^{\mathrm{\prime }}$. Now suppose $Z^{\mathrm{\prime }}=E_1\cup E_2$ with $E_1,E_2$ closed in $Z^{\mathrm{\prime }}$ and $E_i\mathrm{\ne }{Z}^{\mathrm{\prime }}$. Each $E_i$ is a trace of a closed subset of $X$ [F5], hence closed in $X$ because $Z^{\mathrm{\prime }}$ is closed in $X$ [F3]; and $C=(C\cap E_1)\cup (C\cap E_2)$ with each $C\cap E_i$ closed in $C$, so irreducibility and nonemptiness of $C$ [F1] give $C\subseteq E_i$ for some $i$. Then $X\setminus W\subseteq C\subseteq E_i$ with $E_i$ closed in $X$, so $Z^{\mathrm{\prime }}=\overline{X\setminus W}\subseteq E_i$ by [F3], whence $E_i=Z^{\mathrm{\prime }}$, contradicting $E_i\mathrm{\ne }{Z}^{\mathrm{\prime }}$. Therefore $Z^{\mathrm{\prime }}$ is irreducible, and together with the nonemptiness and the inclusion $X\setminus W\subseteq C$ this is clause 6.

Let $C\subseteq X$ be an irreducible component [F10]. Then $C$ is irreducible, so its closure $\bar{C}$ is irreducible by [step 1.1], and $C\subseteq \bar{C}$. Maximality in [F10] applied to the irreducible subset $\bar{C}$ gives $\bar{C}=C$, and $\bar{C}$ is closed by [F3]; hence $C$ is a closed subset of $X$, which is clause 2.

A singleton subset $\{x\}\subseteq X$ is irreducible: it is nonempty, and in any decomposition $\{x\}=F_1\cup F_2$ into closed subsets the point $x$ lies in $F_1$ or in $F_2$, so the corresponding $F_i$ equals $\{x\}$ [F1]. Applying [step 1.2] to $T=\{x\}$ produces an irreducible component of $X$ containing $x$; hence every point of $X$ lies in an irreducible component and $X$ is the union of its irreducible components, which completes clause 3. If moreover $X$ is nonempty and irreducible, then $X$ is itself an irreducible subset of $X$ contained in no larger irreducible subset, so $X$ is an irreducible component by [F10]; and every irreducible component $C\subseteq X$ satisfies $C=X$ by maximality in [F10]. Thus $X$ is the unique irreducible component of $X$, which is clause 4.

Let $X=X_1\cup \cdots \cup X_n$ with each $X_i$ irreducible and closed in $X$, and suppose no $X_i$ is contained in ${\bigcup }_{j\mathrm{\ne }i}{X}_j$. Let $C\subseteq X$ be an irreducible component. Then $C=(C\cap X_1)\cup \cdots \cup (C\cap X_n)$, and each $C\cap X_i$ is closed in $C$ because $X_i$ is closed in $X$ [F5]; since $C$ is irreducible and nonempty [F1], $C=C\cap X_i$ for some $i$, that is $C\subseteq X_i$, and maximality in [F10] applied to the irreducible subset $X_i$ gives $C=X_i$. Conversely, for a given $i$ the set $X_i$ is contained in an irreducible component $C$ by [step 2.2], and $C=X_j$ for some $j$ by what was just proved; then $X_i\subseteq X_j$, and $j\mathrm{\ne }i$ would put $X_i$ inside the union of the other members, so $j=i$ and $X_i=C$ is an irreducible component. Hence the irreducible components of $X$ are exactly $X_1,\dots ,X_n$, which is clause 5.

Clause 1 is [step 1.1], clause 2 is [step 2.1], clause 3 is [step 1.2] together with [step 2.2], clause 4 is [step 2.2], clause 5 is [step 3.1] and clause 6 is [step 1.3], so all six clauses are proved. The Axiom of Choice [F9] is used at exactly one point, in [step 1.2], through Zorn's lemma [F7] applied to the poset of irreducible subsets containing a fixed irreducible subset; the verification of the chain condition there is a direct computation with unions and closed subsets [F6], and the remaining arguments use no choice principle. ∎