Under the ultrafilter lemma, Frolík's internal open-cover characterisation of Čech-completeness

Given: The objects, hypotheses, and choice principles stated above.

A Tychonoff space $X$ is Čech-complete when there is a Hausdorff compactification $(K,i)$ of $X$ (def-compactification-of-a-tychonoff-space) for which $i[X]$ is a $G_{\delta }$ subset of $K$ (def-g-delta-and-f-sigma-in-a-topological-space). The definition asks for one compactification; thm-cech-completeness-is-independent-of-compactification proves the equivalent every-compactification form. (Čech-complete spaces as $G_{\delta }$ subspaces of Hausdorff compactifications).

Assume the ultrafilter lemma and Dependent Choice. A Tychonoff space is a $G_{\delta }$ subset of some Hausdorff compactification if and only if it is a $G_{\delta }$ subset of every Hausdorff compactification. (Under the ultrafilter lemma and Dependent Choice, a Tychonoff space is $G_{\delta }$ in some Hausdorff compactification exactly when it is $G_{\delta }$ in every one).

Let $(X,\mathcal{T})$ be a topological space (def-topological-space). For a family $\mathcal{A}$ of subsets of $X$ write $\bigcap \mathcal{A}:=\{x\in X:x\in A\text{ for every }A\in \mathcal{A}\},$ so that $\bigcap \mathrm{\varnothing }=X$, matching the convention for the empty finite intersection in def-finite-intersection-property. Then: 1. $(X,\mathcal{T})$ is compact (def-compact-space) if and only if every family $\mathcal{A}$ of closed subsets of $X$ with the finite intersection property (def-finite-intersection-property) satisfies $\bigcap \mathcal{A}\mathrm{\ne }\mathrm{\varnothing }$. 2. Equivalently: $(X,\mathcal{T})$ is compact if and only if every family of closed subsets of $X$ that is contained in some filter on $X$ (def-filter) has nonempty intersection, a family of subsets of $X$ lying in a filter exactly when it has the finite intersection property (lem-fip-generates-filter). No choice principle is used in either direction: complementation is a canonical bijection, so no member of a family ever has to be selected. (A space is compact exactly when every family of closed subsets with the finite intersection property has nonempty intersection).

Let $(X,\mathcal{T})$ be a topological space (def-topological-space). The following implications hold, and each is proved by an earlier item of this page. 1. Perfectly normal implies completely normal, assuming the Axiom of Countable Choice (def-countable-choice). 2. Completely normal implies normal, and perfectly normal implies normal. 3. Normal together with $T_1$ implies $T_3$, that is regular together with $T_1$. 4. Completely regular implies regular, and Tychonoff implies $T_3$. 5. Regular together with $T_1$ implies Urysohn, which implies Hausdorff, which implies $T_1$, which implies $T_0$. 6. Metrizable implies every property named above: a metrizable space is perfectly normal, completely normal, normal, Tychonoff, completely regular, $T_3$, regular, Urysohn, Hausdorff, $T_1$ and $T_0$, with no choice principle used. Reading the numbered axioms in order, clauses 1 to 5 give $T_6\Rightarrow T_5\Rightarrow T_4\Rightarrow T_3\Rightarrow T_{2\frac{1}{2}}\Rightarrow T_2\Rightarrow T_1\Rightarrow T_0,$ the first arrow under ${\mathrm{A}\mathrm{C}}_{\omega }$, together with $T_{3\frac{1}{2}}\Rightarrow T_3$. This is the whole of the classical chain that this page proves, and it is one arrow short of the classical chain. The implication $T_4\Rightarrow T_{3\frac{1}{2}}$ — a normal $T_1$ space is completely regular — is Urysohn's lemma and is not available at this point in the reading order. Its absence is recorded, with what would license it, in this page's conventions remark; it is deliberately not asserted here, and no clause above may be read as giving it. (The implications proved on this page: perfectly normal gives completely normal under countable choice, and completely normal gives normal; normal with $T_1$ gives $T_3$; completely regular gives regular; regular with $T_1$ gives Urysohn, hence Hausdorff, hence $T_1$, hence $T_0$; and metrizable gives every one of them).

Let $X$ be a set. A family $\mathcal{B}\subseteq \mathcal{P}(X)$ is a filter base on $X$ when it satisfies: - (B1) nonemptiness: $\mathcal{B}\mathrm{\ne }\mathrm{\varnothing }$; - (B2) properness: $\mathrm{\varnothing }\mathrm{\notin }\mathcal{B}$; - (B3) downward directedness: for all $B_1,B_2\in \mathcal{B}$ there is $B_3\in \mathcal{B}$ with $B_3\subseteq B_1\cap B_2$. (Filter base and the filter it generates).

Let $X$ be a compact Hausdorff topological space. Then $X$ is regular, and $X$ is normal (A compact Hausdorff space is regular and normal, hence $T_3$ and $T_4$).