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A subset $K$ is compact if every cover of $K$ by sets open in the ambient space has a finite subcover. Equivalently, every relatively open cover of the subspace $K$ has a finite subcover. These agree because relatively open sets are intersections with ambient open sets.

If $f(A)=U\cup V$ were a separation, thenwould be disjoint nonempty relatively open sets covering $A$. They are relatively open by continuity of the restriction $f{|}_A:A\to f(A)$. This contradicts connectedness of $A$.This is the continuous image of a connected space theorem; no separation property of the target is required.

A topological space is connected if it has no separation: two disjoint nonempty open sets whose union is the whole space. Equivalently, its only subsets that are both open and closed are the empty set and the whole space.

A subset is connected when its subspace topology has this property. Thus the sets in a proposed separation of a subset need only be relatively open, rather than open in the ambient space.

Every topological subspace $S\subseteq X$ again has the cofinite topology: its nonempty open subsets have finite complements in $S$, and every finite exclusion is obtained by intersecting with a cofinite open subset of $X$. As usual, the empty set is also open; this exception is implicit in the paper's shorthand definition.

If $S$ is infinite, any two nonempty relatively open sets intersect, since their complements are finite. Hence $S$ is hyperconnected space and therefore connected. If $S$ is finite, its cofinite topology is discrete topology. It is then connected exactly when it has at most one point. ThusThere are also infinite disconnected subsets of the product. Choose distinct $a,b\in X$ and put $D=\{a,b\}\times X$. This is infinite. The traces on $D$ of the ambient open setsare respectively $\{a\}\times X$ and $\{b\}\times X$. They are disjoint, nonempty and cover $D$, giving a separation. This answers the infinite-subset question, rather than only producing two isolated product points.

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.

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.

A subset $A$ is connected when it is a connected space in its subspace topology: there is no separation of a topological spaceinto two nonempty disjoint open subsets of $A$. Equivalently, the only clopen sets of $A$ are $\varnothing$ and $A$.

Since the given $A$ is not connected, choose such $P,Q$. Each is the complement of the other in $A$, so both are also closed subsets of $A$. They are compact by the closed subset of a compact space result. We now separate these two compact subsets by ambient open sets.

For each $p\in P$, part (b), applied to $Q$ and $p$, gives disjoint ambient open sets $G_p,H_p$ with $p\in G_p$ and $Q\subseteq H_p$. Choose $p_1,\dots ,p_m$ such that the $G_{p_j}$ cover $P$, using its compactness. DefineThen $P\subseteq U$, $Q\subseteq V$, and $U,V$ are disjoint open sets. Indeed a point of $U$ lies in some $G_{p_j}$, whereas every point of $V$ lies in the corresponding $H_{p_j}$. ConsequentlyThis proves the requested ambient separation, using the more general separation of disjoint compact subsets. The initially given relatively open sets $P,Q$ need not themselves be open in $X$; the finite construction supplies ambient open neighbourhoods.

Write $K={\bigcap }_{i\in I}{K}_i$. Each $K_i$ is closed in $X$, by the compact subset of a Hausdorff space theorem. Therefore $K$ is closed, and is compact because $X$ is compact. The nonempty finite intersections give the finite intersection property, so part (a)(ii) also yields $K\ne \varnothing$.

Suppose $K$ were not connected. Part (c) would give disjoint open sets $U,V$ containing $K$ in their union and each meeting $K$. LetThis is a closed subset of the compact space $X$, and $C\cap {\bigcap }_{i\in I}{K}_i=\varnothing$. Apply the contrapositive of the finite intersection characterization of compactness to the family of closed sets $\{C\}\cup \{K_i:i\in I\}$. Some finite subfamily has empty intersection. Since every finite intersection of just the $K_i$ is nonempty, that subfamily must include $C$. Thus for some finite $J\subseteq I$,But $K\subseteq K_J$ and $K$ meets both $U$ and $V$. Therefore $K_J\cap U$ and $K_J\cap V$ are nonempty disjoint relatively open sets covering $K_J$, contradicting its assumed connectedness. This provesThis is connectedness of an intersection from connected finite intersections. If $I=\varnothing$, its intersection is $X$; the stated hypothesis for the finite set $J=\varnothing$ already makes $X$ nonempty and connected, so that case is included.

A topological space is Hausdorff when any two distinct points have disjoint open neighbourhoods.

Any two nonempty open sets of the infinite cofinite topology intersect: their complements are finite sets, so the complement of their intersection is a finite union of finite sets and cannot be all of $Z$. Thus $Z$ is a hyperconnected space.

If the continuous map $f$ took distinct values at $s,t\in Z$, choose disjoint open neighbourhoods $G,H$ of $f(s),f(t)$ in the Hausdorff space $Y$. By continuity, $f^{-1}(G)$ and $f^{-1}(H)$ would be nonempty disjoint open sets of $Z$, a contradiction. ThereforeThe proof establishes the general fact that continuous maps from an infinite cofinite space to a Hausdorff space are constant.

Write $Q=(Z\times I)/\mathcal{R}$. Since $Q$ is homeomorphic to the circle, it is Hausdorff. For fixed $t\in I$, the functionis continuous, being a coordinate slice of the quotient map. Part (a)(i) therefore makes it constant. Choose $x_0\in Z$ and setThen $g$ is continuous andSince the quotient map $p$ is a surjection, so is $g$. This is Hausdorff factorization through a cofinite product factor: a continuous map into a Hausdorff space cannot distinguish the points of the coordinate with the cofinite topology.

If $g$ were injective, it would be a continuous bijection from the compact closed interval $I$ to the Hausdorff space $Q$. Part (a)(ii) would make it a homeomorphism, contradicting the fact that the circle and interval are not homeomorphic.

To recall that obstruction explicitly, remove the interior point $1/2$ from $I$. The resulting topological subspace is separated by the nonempty relatively open sets $[0,1/2)$ and $(1/2,1]$, so it is not connected. The circle with any one point removed is connected: after a rotation, it is the image under a continuous map of the connected open interval $(0,2\pi )$ under $t\mapsto (\cos t,\sin t)$. A homeomorphism and its restriction preserve connectedness, giving a contradiction.

Thus $g$ is not injective. Choose distinct $a,b\in I$ with $g(a)=g(b)$. ThenIn particular the requested $x,a,b$ exist. The argument proves the stronger conclusion that the same two interval coordinates work for every $x$.

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.

The interval $[0,1)$ is a basic open set contained in $[0,1]$. The remaining point $1$ is not an interior point, by the preceding argument. Thus

Let $U$ be open in the Euclidean topology on the real line, and let $x\in U$. There is an $\epsilon >0$ with $(x-\epsilon ,x+\epsilon )\subseteq U$. The basic open set $[x,x+\epsilon )$ contains $x$ and is contained in $U$. Taking the union of these basic neighbourhoods over $x\in U$ shows that $U\in \mathcal{T}$. ThusIn fact this is a strictly finer topology: $[0,1)$ is open in $\mathcal{T}$ but not in the Euclidean topology, since no usual open ball about $0$ is contained in it.

$X$ 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.

TakeIt is continuous for the Euclidean metric, since $|g(x)-g(y)|=|x-y|$. But the basic open set $[0,1)$ of the target Sorgenfrey line has preimageThis is not open in the source Sorgenfrey line: every basic neighbourhood of $0$ contains positive points. Thus $g$ is not continuous for the lower limit topology.

Use the Heaviside step function with the value at the jump chosen as follows:For any open set in the target Sorgenfrey line, its preimage is one of $\varnothing$, $\mathbb{R}$, $(-\infty ,0)$, and $[0,\infty )$. All four are open in the source Sorgenfrey line, so $f$ is continuous. In the Euclidean topology, the preimage of $(1/2,3/2)$ is $[0,\infty )$, which is not open. Hence the same function is not continuous as a usual real-to-real map.

Let $q_S:S^2\to S^2/{\sim }_S$ and $q_D:D^2\to D^2/{\sim }_D$ be the quotient maps. The continuous mapis a homeomorphism onto the upper hemisphere; its inverse function is the restriction of the projection map $(x,y,z)\mapsto (x,y)$. Every antipodal pair on $S^2$ meets that hemisphere. Moreover, two points of this hemisphere are antipodal only if both lie on the equator. ThusBy the universal property of the quotient topology, $q_S\circ h$ induces a continuous bijection

To justify continuity of the inverse function, the source is compact, being a continuous image of a compact space $D^2$; the Heine-Borel theorem gives compactness of $D^2$. The target is Hausdorff, as follows. For distinct equivalence classes $[u]\ne [v]$, putIn $S^2$, take the union $U$ of the two relative open balls of radius $\delta /3$ about $u,-u$, and similarly take $V$ about $v,-v$. Every distance between their respective centres is at least $\delta$, so the triangle inequality gives $U\cap V=\varnothing$. These open sets contain entire antipodal pairs, hence $q_S^{-1}(q_S(U))=U$ and similarly for $V$. The quotient topology therefore makes $q_S(U)$ and $q_S(V)$ disjoint open neighbourhoods of the two classes.

The compact-to-Hausdorff continuous bijection theorem now gives the required homeomorphism $H$.

False, with “disjoint union” understood as a topological disjoint union. Take the rational numbers $\mathbb{Q}$ with the usual subspace topology from the real line.

Every connected subset of $\mathbb{Q}$ has at most one point. Indeed, if it contained $r<s$, choose an irrational number $t\in (r,s)$. The two relatively open setspartition $\mathbb{Q}$ and separate any subset containing both $r$ and $s$. Hence the connected components of $\mathbb{Q}$ are singletons.

If a topological disjoint union of nonempty connected spaces were homeomorphic to $\mathbb{Q}$, the image of each summand would be connected, hence a singleton. Every summand would therefore be a one-point topological space. Their topological disjoint union is a discrete space, since each summand is open. But $\mathbb{Q}$ is not a discrete space: for each $r\in \mathbb{Q}$, the distinct rational numbers $r+1/n$ approach $r$, so $r$ is not an isolated point. This is a contradiction. The set decomposition into connected components always exists; giving those pieces the topological disjoint union topology need not recover the original topology.

Write $Z={\coprod }_{n\in \mathbb{N}}{X}_n$, where each $X_n$ is a labelled copy of the topological space $X$, with its topological disjoint union topology. Define the bijectionFor a subset $U\subseteq \mathbb{N}\times X$, let $U_n=\{x\in X:(n,x)\in U\}$. By definition of the topological disjoint union, $F^{-1}(U)$ is open in $Z$ exactly when each $U_n$ is open in $X$.

The same condition characterizes open sets in the product topology on $\mathbb{N}\times X$. If $U$ is open, the map $x\mapsto (n,x)$ is continuous, so each $U_n$ is open. Conversely, if each $U_n$ is open, thenis open, since the discrete topology makes every $\{n\}$ open in $\mathbb{N}$. Hence $F$ and its inverse function are continuous, proving the required homeomorphism. This also covers the case $X=\varnothing$.