A map of Hausdorff compactifications carries the larger remainder onto the smaller remainder

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

A Hausdorff compactification of a space $X$ is a pair $(K,i)$ in which $K$ is compact (def-compact-space) and Hausdorff (def-hausdorff-space), and $i:X\to K$ is an embedding with dense image (def-homeomorphism-and-open-maps, def-dense-top). We identify $X$ with $i[X]$ only after naming $i$; the density condition is a condition on that named image. (A Hausdorff compactification as a dense embedding into a compact Hausdorff space).

Let $(X,{\mathcal{T}}_X)$ and $(Y,{\mathcal{T}}_Y)$ be topological spaces (def-topological-space), and let $\mathbb{R}$ carry its usual topology, the metric topology of $d_{\mathbb{R}}(s,t)=\mathrm{\mid }s-t\mathrm{\mid }$ (lem-real-line-is-a-metric-space, def-metric-topology, def-metrizable-space). Then: 1. Continuous images. If $f:X\to Y$ is continuous (def-continuous-map-top) and $(X,{\mathcal{T}}_X)$ is compact (def-compact-space), then $f[X]$ is a compact subset of $Y$. More generally, if $K\subseteq X$ is a compact subset of $X$ then $f[K]$ is a compact subset of $Y$. 2. Extreme values. If $(X,{\mathcal{T}}_X)$ is compact and nonempty and $g:X\to \mathbb{R}$ is continuous, then $g[X]$ has a maximum and a minimum (def-max-min): there are $x_{\max },x_{\min }\in X$ with $g(x_{\min })\le g(x)\le g(x_{\max })\text{for every }x\in X\mathrm{.}$ 3. Compact to Hausdorff. If $(X,{\mathcal{T}}_X)$ is compact, $(Y,{\mathcal{T}}_Y)$ is Hausdorff (def-hausdorff-space) and $f:X\to Y$ is a continuous bijection, then $f$ is a homeomorphism (def-homeomorphism-and-open-maps). Nonemptiness in claim 2 is a hypothesis and not an oversight: for $X=\mathrm{\varnothing }$ the image is empty and has neither a maximum nor a minimum. No choice principle is used: the one selection made below is over a finite index set, where lem-finite-choice is a theorem of ZF. (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism).

Let $(X,\mathcal{T})$ be a Hausdorff topological space (def-hausdorff-space, def-topological-space), with compact subsets as in def-compact-space. Then: 1. A point and a disjoint compact set are separated. If $K\subseteq X$ is compact and $x\in X\setminus K$, there are $U,V\in \mathcal{T}$ with $x\in U,K\subseteq V,U\cap V=\mathrm{\varnothing }\mathrm{.}$ 2. Two disjoint compact sets are separated. If $K,L\subseteq X$ are compact and $K\cap L=\mathrm{\varnothing }$, there are $U,V\in \mathcal{T}$ with $L\subseteq U,K\subseteq V,U\cap V=\mathrm{\varnothing }\mathrm{.}$ 3. Compact implies closed. Every compact subset of $X$ is closed in $X$. 4. In a compact Hausdorff space the two classes coincide. If in addition $(X,\mathcal{T})$ is compact, then a subset of $X$ is compact if and only if it is closed. The proof is written choice-free, and that is not a stylistic preference. The textbook argument says "for each $y\in K$ choose disjoint open $U_y,V_y$", which is a selection over an arbitrary index set and therefore an appeal to the full Axiom of Choice. What is done below instead is to take the family of all open $V$ that admit some open $U\ni x$ disjoint from them — a family cut out by a formula, with nothing selected — extract a finite subcover from it, and only then make finitely many selections, which lem-finite-choice supplies as a theorem of ZF. (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones).