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No countable subset works. For any countable set $A\subseteq \mathbb{R}$, choose $b_n\in (n,n+1/2)\setminus A$: each nonempty interval is uncountable. The sequence tends to infinity, so every tailis closed in $\mathbb{R}$ and disjoint from $A$. By part (i), $p(\mathbb{R}\setminus K_n)$ is open in $X_A$. These sets increase and cover $X_A$, since no point belongs to every tail. But any finite selection is contained in its largest member, which still misses every quotient point $p(b_k)$ in that member's remaining tail. The $b_k$ represent distinct singleton classes.

This is an open cover without a finite subcover, provingThe argument does not assume that $A$ is closed, bounded, or discrete. It is the noncompactness after collapsing a countable real subset obstruction.

Let $F$ be a closed subset of the compact space $X$. Express an arbitrary open cover of $F$ in its subspace topology as $\{F\cap U_i:i\in I\}$, where the $U_i$ are open sets of $X$. Since $X\setminus F$ is open, the familyis an open cover of $X$. Take a finite subcover and discard $X\setminus F$ if it occurs. The remaining finitely many $U_i$ still cover $F$, since the discarded set contains no point of $F$. Their traces on $F$ give the required finite subcover. Hence every closed subset of a compact space is compact, with no Hausdorff assumption.

Assume $K$ is compact, and let the closed subsets $F_i$ of $K$ have the finite intersection property. If their total intersection were empty, their complements $K\setminus F_i$ would form an open cover of $K$. A finite subcover would givecontradicting the finite intersection property. Therefore the total intersection is nonempty.

Conversely, assume the stated nonempty-intersection property for every family of closed subsets of $K$. Given an open cover $\{V_i:i\in I\}$ of $K$ in its subspace topology, set $F_i=K\setminus V_i$. Their total intersection is empty. By the contrapositive of the assumed property, this family fails the finite intersection property; hence some finite $J\subseteq I$ satisfiesThis is a finite subcover, proving compactness. Both directions are instances of De Morgan's laws. Thus compactness is equivalent to the finite intersection characterization of compactness.

For completeness, the intersection of the empty family of subsets of $K$ is $K$. If $K$ is empty, no family has the finite intersection property when the empty finite subfamily is included, so the implication remains valid.

A subset $K$ is compact if every open cover has a finite subcover. Explicitly, if $U_i$ are open sets of $X$ and $K\subseteq {\bigcup }_{i\in I}{U}_i$, there are finitely many indices $i_1,\dots ,i_m$ such thatEquivalently, the subspace topology makes $K$ a compact space. An open set in this subspace topology is $K\cap U$ for an ambient open set $U$, which proves equivalence of the two open cover formulations. The empty set is compact: the empty selection is a finite subcover.

For every $k\in K$, the Hausdorff condition gives disjoint open neighbourhoods $U_k$ of $k$ and $V_k$ of $x$. The $U_k$ form an open cover of $K$, so choose a finite subcover $U_{k_1},\dots ,U_{k_m}$. PutThen $U,V$ are open sets, $K\subseteq U$ and $x\in V$. If a point belonged to $U\cap V$, it would belong to some $U_{k_j}$ and to $V_{k_j}$, contrary to their choice. ThusThe finiteness matters: a finite intersection of open sets is open. This is separation of a point and a compact set. If $K=\varnothing$, use $U=\varnothing$, $V=X$.

Every $x\in X\setminus K$ consequently has an open neighbourhood $V$ contained in $X\setminus K$. The complement is open, so a compact subset of a Hausdorff space is closed.

The conclusion can fail without the Hausdorff condition. Take the two-point set $X=\{0,1\}$ with the indiscrete topology $\{\varnothing ,X\}$. The singleton $K=\{0\}$ is compact, since one member of any open cover suffices. It is not closed, because its complement $\{1\}$ is not open. The two points cannot have disjoint open neighbourhoods, so this example is indeed not Hausdorff.

Fix $x_0,y_0\in Z$. For a continuous function $f:Z\times Z\to \mathbb{R}$, every coordinate slice is continuous, because the maps $x\mapsto (x,y)$ and $y\mapsto (x_0,y)$ into the product topology are continuous. Their target $\mathbb{R}$ is Hausdorff. Part (a)(i) givesTherefore every such $f$ is a constant function. This is the two-factor case of continuous maps from products of cofinite spaces to Hausdorff spaces are constant.

The second coordinate projection ${\pi }_2:Z\times \mathbb{R}\to \mathbb{R}$ is continuous and nonconstant. If a homeomorphism $H:Z\times Z\to Z\times \mathbb{R}$ existed, ${\pi }_2\circ H$ would be a nonconstant continuous function, contradicting the preceding conclusion. Hence

For the half-plane, put $H_{+}=[0,\infty )$ as a set, so $K=\mathbb{R}\times H_{+}$. Its two ambient topologies give different answers:In $Z\times Z$, if $K$ were closed, its inverse image under the continuous map $y\mapsto (x_0,y)$ would be closed in $Z$. That inverse image is $H_{+}$, which is neither a finite set nor all of $Z$, so it is not closed in the cofinite topology. Thus $K$ is not closed.

To establish compactness, note that the subspace topology on any subset $A$ of a topological space with the cofinite topology is again cofinite. An open cover of a nonempty such $A$ has a nonempty member omitting at most finitely many points; finitely many more members cover those points. Hence every subset of a cofinite space is compact. In particular both $Z$ and $H_{+}$ with its cofinite subspace topology are compact. The subspace topology on $K\subseteq Z\times Z$ is their product topology, as one checks by tracing basis rectangles. The product of two compact spaces is compact, so $K$ is compact here.

For clarity, the finite-product compactness result needs no Hausdorff condition. Given an open cover of $A\times B$, cover each point by a basis rectangle lying in one covering member. For fixed $a\in A$, use compactness of $B$ to select finitely many rectangles whose second factors cover $B$, and intersect their first factors to get an open neighbourhood $U_a$ of $a$. Those finitely many original members cover $U_a\times B$. Finally finitely many $U_a$ cover $A$, producing a finite subcover of $A\times B$.

In $Z\times \mathbb{R}$, the second coordinate projection hasSince the ray is closed in the usual real line, $K$ is closed. It is not compact: its image under a continuous map under ${\pi }_2$ is $[0,\infty )$, which is not compact. For an explicit obstruction, the open intervals $(-1,n)$, $n\ge 1$, cover this ray but have no finite subcover.

Finally consider the diagonal $W$ in $Z\times \mathbb{R}$. DefineThis is a bijection. The identity function from the usual real line to $Z$ is continuous: each nonempty cofinite open set is the complement of a finite set, and finite sets are closed in the usual real line. Both coordinates of $\phi$ are therefore continuous, so $\phi$ is continuous for the product topology and the subspace topology of $W$. Its inverse is the restricted second coordinate projection, which is continuous. ThereforeFor the second conclusion, $Z$ is compact whereas $W\cong \mathbb{R}$ is not compact; homeomorphisms preserve compactness. The diagonal in a product with cofinite and Euclidean topologies thus inherits the usual topology, despite one cofinite factor.

$X$ is not compact. In the subspace topology on $X=[0,1]$, consider the open coverThe intervals are basic open sets in the ambient Sorgenfrey line, and $\{1\}=X\cap [1,2)$ is open in $X$. Every $x<1$ belongs to one of those intervals, while the singleton covers $1$.

A finite subcover would have to include $\{1\}$. Among its finitely many other members, let $N$ be the largest index. Their union is $[0,1-1/N)$, leaving $1-1/N$ uncovered. Selecting no such member also fails to cover $0$. Hence no finite subcover exists.

Use positive indices $n\ge 1$ and write $p=(0,-1)\in S^1$. SetFor fixed $n$, this function is a homeomorphism from the unit circle to $C_n$: it scales its radius to $1/n$ and translates its centre to $(0,1/n)$. Thus it takes values in the Hawaiian earring $X$ and is surjective onto $X$. It is continuous on each open slice $\{n\}\times S^1$; since these slices form an open cover of $\mathbb{N}\times S^1$, $f$ is continuous on the whole product topology.

Distinct circles $C_n,C_m$ meet only at $o=(0,0)$. To see this directly, their equations areFor $m\ne n$, subtraction gives $y=0$, then $x=0$. Each slice contains exactly one preimage of $o$, namely $(n,p)$. Therefore the only nonsingleton fibre of $f$ is $A$.

Let $q:\mathbb{N}\times S^1\to (\mathbb{N}\times S^1)/A$ be the quotient map collapsing $A$. The universal property of the quotient topology supplies a unique continuous map $\bar{f}$ with $f=\bar{f}\circ q$. The fibre calculation proves injectivity, and surjectivity follows from that of $f$. Hence $\bar{f}$ is the required continuous bijection.

Let $K\subseteq X$ be compact, with $X$ a Hausdorff space. The empty set is closed, so suppose $K\ne \varnothing$. Fix $x\in X\setminus K$. For each $y\in K$, the Hausdorff property supplies disjoint open neighbourhoods $U_y$ of $y$ and $V_y$ of $x$.

The $U_y$ form an open cover of $K$. Choose a finite subcover $U_{y_1},\dots ,U_{y_m}$ and put $V={\bigcap }_{r=1}^m{V}_{y_r}$. This is an open neighbourhood of $x$. If a point of $V$ belonged to $K$, it would belong to some $U_{y_r}$ as well as $V_{y_r}$, a contradiction. Hence $V\subseteq X\setminus K$. Every point of the complement has such an open neighbourhood, so $X\setminus K$ is open and $K$ is closed.

Let $(V_j{)}_{j\in J}$ be an open cover of $f(K)$ by open subsets of $Y$. The continuity of $f$ makes each preimage $f^{-1}(V_j)$ open in $X$, and these preimages cover $K$. By compactness, choose $j_1,\dots ,j_m$ withApplying $f$ gives $f(K)\subseteq {\bigcup }_{r=1}^m{V}_{j_r}$. Thus every open cover of $f(K)$ has a finite subcover, proving the continuous image of a compact space theorem. A cover by sets open in $f(K)$ gives the same argument after expressing them as intersections with ambient open sets.

Give the nonempty subset $A\subseteq X$ its subspace topology. It is compact if every open cover admits a finite subcover. Equivalently, whenever $A\subseteq {\bigcup }_{j\in J}{U}_j$ with each $U_j$ open in $X$, there are $j_1,\dots ,j_m$ such that $A\subseteq U_{j_1}\cup \cdots \cup U_{j_m}$.

It is connected if it has no separation of a topological space: there are no disjoint nonempty open subsets $B,C$ of $A$ with $A=B\cup C$. Here openness is relative to the subspace topology; equivalently, the only clopen sets of $A$ are $\varnothing$ and $A$.

Use the Sierpiński space $X=\{0,1\}$ with topology $\{\varnothing ,\{1\},X\}$. The singleton $A=\{1\}$ is compact: from any open cover, choose one member containing $1$. But $X\setminus A=\{0\}$ is not open, so $A$ is not closed. This also exhibits why the Hausdorff hypothesis in the preceding proof matters.

In the classical convention over an algebraically closed field $k$, a prevariety is an irreducible topological space with a sheaf of rings which has a finite open cover by affine varieties, with compatible local identifications. A prevariety need not be separated. A classical algebraic variety is a separated prevariety: its diagonal morphism has closed image. Some authors allow reducible prevarieties and varieties; the separation criterion and product construction remain the same under that convention.

The product topology of two compact spaces is compact, without a Hausdorff condition. Refine an open cover locally by basis rectangles. For fixed $a\in A$, compactness of $B$ selects finitely many rectangles whose second factors cover $B$. Their first factors have an open finite intersection $U_a$ containing $a$, and the finitely many original covering members cover $U_a\times B$. Use compactness of $A$ to choose finitely many $U_a$ covering $A$. The associated finitely many original members form a finite subcover of $A\times B$. Repeating the argument proves the finite-factor case of Tychonoff's theorem.

A scheme is quasi-compact when its underlying topological space is quasi-compact: every open cover has a finite subcover. Since affine open subsets form a base, this is equivalent to possessing a finite affine open cover. Every affine scheme is quasi-compact.

The sheaf gluing axiom says that sections $s_i$ on an open cover $(U_i)$ that agree on every overlap $U_i\cap U_j$ glue to a unique section on ${\bigcup }_i{U}_i$.

A subcover of an open cover $\{U_i:i\in I\}$ of a subset $K$ is a selection of its members whose union still contains $K$. The selected open sets retain their original sizes; selecting a subcover does not mean replacing them by smaller open sets.

For an invertible sheaf $L$, tensoring sheaves of modules with $L$ is an exact functor: on an open cover trivializing $L$, it is naturally the identity functor. Equivalently, its stalks are free modules of rank one and hence flat modules. Its inverse functor is tensoring with the dual of a sheaf $L^{\vee }$, because $L\otimes L^{\vee }\cong {\mathcal{O}}_X$. This justifies untwisting a surjection onto a positively twisted coherent sheaf.

A topological atlas is a family $\mathcal{A}=\{(U_i,{\phi }_i)\}$ of coordinate charts whose domains form an open cover of a topological manifold. Each point must belong to at least one chart domain; no individual domain needs to cover the entire topological manifold. On overlaps, topological transition maps describe the same points in two coordinate systems.

Every quasi-coherent sheaf on an affine scheme has zero higher sheaf cohomology. Together with the equivalence between modules and quasi-coherent sheaves on an affine scheme, this makes affine varieties the basic acyclic pieces for computing cohomology. An affine open cover of a separated scheme has affine finite intersections, so the acyclic cover theorem identifies its Čech cohomology with sheaf cohomology.