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Suppose $X,Y$ are nonempty connected spaces, and $A⊊X$, $B⊊Y$ are proper subsets. Then $C=(X\times Y)\setminus (A\times B)$, with its subspace topology from the product topology, is a connected space. No closed set or open set hypothesis on $A,B$ is needed.

Choose $x_0\in X\setminus A$, $y_0\in Y\setminus B$. The two topological subspaces $\{x_0\}\times Y$ and $X\times \{y_0\}$ are homeomorphic to the factors, intersect at $(x_0,y_0)$, and lie in $C$. Their union is a connected subset $H$. Every $(x,y)\in C$ has $x\notin A$ or $y\notin B$. In the first case its entire vertical copy $\{x\}\times Y$ lies in $C$ and meets $H$; in the second case the horizontal copy $X\times \{y\}$ does so. Thus $C$ is a connected union with a connected hub.

Both properness assumptions are needed for this general result. Taking $X=Y=\mathbb{R}$, $A=X$ and $B=\{0\}$ gives the two disjoint open half-planes $y<0$ and $y>0$, a separation of a topological space of the complement.

Let $X$ be a compact Hausdorff space and let $K_i$ be compact subsets such that all finite intersections are nonempty and connected. Their total intersection $K$ is nonempty by the finite intersection characterization of compactness, and compact because it is closed in $X$.

If $K$ were not connected, its separation of a topological space would divide it into two nonempty compact subsets. Separation of disjoint compact subsets gives disjoint ambient open sets $U,V$ containing the two pieces. The family consisting of $X\setminus (U\cup V)$ and all $K_i$ has empty total intersection, so some finite subfamily has empty intersection. Hence a finite intersection $K_J$ of the $K_i$ lies in $U\cup V$. Since $K\subseteq K_J$ meets both open sets, they disconnect $K_J$, a contradiction. The hypothesis on all finite intersections is stronger than requiring each individual $K_i$ to be connected.

A topological space $X$ is connected when it has no separation of a topological space: no partition of a set into two nonempty open subsets $U,V$ satisfyingHere the union covers all of $X$, and the intersection is the empty set, so no point belongs to both pieces. For a subset $X$ of a larger topological space, "open" means open in the subspace topology: each piece is the intersection of $X$ with an ambient open set. Both pieces must be open; merely dividing a drawing into two colours does not establish a separation of a topological space. ^[ref]

For example, the closed interval $I=[0,1]$ with its usual subspace topology from the real line is connected. The proposed split $[0,1/2)\cup [1/2,1]$ fails: the second subset is not open in $I$, since every neighbourhood of $1/2$ in $I$ contains points less than $1/2$. More generally, a connected subset of the real line is precisely a real interval. In contrast, for the unionthe open intervals $(-3,0)$ and $(0,3)$ exhibit $U,V$ as nonempty open subsets in the subspace topology of $D$. Their intersection is empty and their union is $D$, so $D$ is not connected. Although $U,V$ are closed intervals in the real line, they are also open within $D$.

Why is connectedness useful? A continuous image of a connected space is connected: a proposed separation of a topological space in the image would pull back through the continuous map to a forbidden separation of a topological space in its source. Thus a continuous map from a real interval to the real line cannot skip values between two attained values. This is the intermediate value theorem. For a computed example, the continuous function $f(x)=x^2$ on $[0,1]$ has image $[0,1]$; the value $1/2$ occurs at $x=1/\sqrt{2}\approx 0.707107$. Connectedness guarantees an input attaining that value even when we have no formula for it.

Path connectedness is a stronger condition: it supplies a continuous path between every pair of points. The topologist's sine curve below shows that a connected space need not be a path-connected space.

A subset $A$ of a topological space $X$ is a connected subset when $A$, with the subspace topology inherited from $X$, is a connected space. Thus connectedness is tested inside $A$: a forbidden separation of a topological space consists of nonempty subsetswhere $O_1,O_2$ are open sets in $X$. The ambient open sets need not be disjoint outside $A$.

For example, in the real line with its usual topology, $[0,2]$ is a connected subset, whereas $\{0,2\}$ is not: its two singleton sets are open in its subspace topology. A subset of a connected space need not be connected. Conversely, $[-2,-1]$ is a connected subset of the disconnected topological space $[-2,-1]\cup [1,2]$. The ambient topological space need not itself be connected.

A finite set of distinct points in the real line inherits the discrete topology: a sufficiently small open interval around each point isolates that point. Its only nonempty connected subsets are singleton sets. This depends on the topology: a nonempty finite set with the indiscrete topology is a connected space, since its only open sets are itself and the empty set. Here, as in the definition of connected space, the empty set is also connected because it admits no separation of a topological space.^[ref]

The distinction matters when finding connected components: we search among connected subsets of the specified topological space, rather than judging whether the surrounding drawing looks like one piece.

In the usual subspace topology from the real line, a subset $A\subseteq \mathbb{R}$ is connected exactly when it has the defining property of a real interval:Every real number between two members must be present. Thus open intervals, closed intervals, intervals with only one endpoint included, and unbounded real intervals are connected. Singleton sets are connected too. With the no-separation of a topological space convention used here, the empty set satisfies the same property; some texts reserve the word connected for nonempty topological spaces.^[ref]

Removing an interior point changes the answer:These two real intervals are exactly its connected components and its path components. Removing an endpoint instead leaves a real interval, for example $[-2,2]\setminus \{2\}=[-2,2)$, which stays connected. This is why an interior point of a closed interval is a cut point, whereas an endpoint is not. A nonempty finite set of points in the real line is connected exactly when it is a singleton set.

Proof outline: if $a,b\in A$ but $c\in (a,b)$ is missing, the subsetsgive a separation of a topological space, since the two open intervals make both pieces open in $A$. Conversely, a nonempty real interval contains the continuous path $t\mapsto (1-t)a+tb$ between any two of its points, so it is path-connected and hence connected. Together with continuous image of a connected space, this characterization gives the intermediate value theorem: a continuous function on a real interval cannot jump over a value lying between two attained values.

For a set $X$ with its cofinite topology, a subset $S$ is a connected subset exactly when $S$ is empty, a singleton, or infinite. Indeed, its subspace topology is the cofinite topology on $S$: the traces of open sets are precisely $S\setminus F$ for finite $F\subseteq S$, together with the empty set. For infinite $S$, two nonempty open sets intersect, because their complements have finite union. Hence $S$ is a hyperconnected space and therefore a connected space. A finite $S$ has the discrete topology, which admits a separation of a topological space as soon as $S$ has at least two points.

If $f:X\to Y$ is a continuous map and $X$ is a connected space, then its image $f(X)$, with the subspace topology from $Y$, is a connected subset of $Y$. The whole codomain $Y$ need not be connected; the assertion concerns the values actually attained. No injectivity is required.^[ref]

For example, the continuous function $f(x)=x^2$ maps the connected closed interval $[-2,2]$ onto the connected closed interval $[0,4]$, even though opposite inputs have the same output. The same continuous function maps the disconnected union $[-2,-1]\cup [1,2]$ onto $[1,4]$. Thus a connected image does not imply a connected source: a continuous map can merge different connected components.

For a continuous function $r:[0,1]\to \mathbb{R}$ representing a continuous reading, $r(0)=0$ and $r(1)=1$ force some input with $r(t)=1/2$. Its image is a connected subset of the real line, hence a real interval containing both endpoint readings. For the synthetic model $r(t)=t^2$, the missing-value test is explicit: $r(1/\sqrt{2})=1/2$. A step reading equal to $0$ for $t<1/2$ and $1$ for $t\ge 1/2$ skips $1/2$, but fails continuity at $t=1/2$.

Proof outline: a separation of a topological space of $f(X)$ into nonempty open subsets pulls back under $f$ to nonempty open subsets of $X$. Their intersection is empty and their union is $X$, contradicting connectedness. This also shows why a continuous map sends each connected component of its source into a single connected component of its target.

A topological space is disconnected space if it can be written as the union of two disjoint nonempty open subsets. Such a pair is a separation of a topological space. Each of the subsets is also closed, and conversely a nontrivial subset that is both open and closed gives a separation of a topological space with its complement.

The set $A$ is disconnected space if it has a separation of a topological space: two nonempty disjoint subsets $U,V$ open in $A$ with $A=U\cup V$. Each is then also closed in $A$, being the other's complement.

The set $A$ is connected if no such separation of a topological space exists. Equivalently its only subsets that are both open and closed are $\varnothing$ and $A$. A separation of a topological space uses openness in $A$ rather than necessarily in the entire complex plane.

For $x$ in a compact Hausdorff space $K$, its quasi-component $Q$ is the intersection of all clopen sets containing $x$. It is closed and therefore compact. Suppose $Q=A\bigsqcup B$ is a separation of a topological space into two nonempty relatively closed sets, with $x\in A$. The sets $A,B$ are compact and hence closed in $K$. Normality of a compact Hausdorff space gives disjoint open neighborhoods $U,V$ of $A,B$.

The compact set $K\setminus (U\cup V)$ is disjoint from $Q$. For each point $y$ of that set choose a clopen $C_y$ containing $x$ but not $y$. Finitely many complements cover $K\setminus (U\cup V)$, so their corresponding intersection $C$ is clopen, contains $Q$, and is contained in $U\cup V$. Now $C\cap U$ is clopen in $C$, hence clopen in $K$, and contains $x$ while missing $B$. This contradicts the definition of $Q$. Thus $Q$ is connected. Any connected set containing $x$ lies in every clopen neighborhood of $x$, so the connected component of $x$ equals $Q$.

For the second assertion, $D$ is closed in the ambient Hausdorff space $X$, and is itself compact Hausdorff. Let $C$ be a connected component of $D$ disjoint from ${\partial }_{X}D$. Choose $p\in C$. By the first assertion $C$ is the quasi-component of $p$ in $D$. For each $y\in {\partial }_{X}D$ choose a clopen subset of $D$ containing $p$ but not $y$. Compactness of ${\partial }_{X}D\subseteq D$ gives a finite intersection $H$ of these sets which contains $C$ and avoids the whole boundary. Thus $H\subseteq {\mathrm{int}}_{X}D$. It is closed in $X$, because it is closed in the closed set $D$, and is open in $X$, because it is relatively open in $D$ and contained in its interior. It is nonempty and proper, contradicting connectedness of $X$. Hence every connected component of $D$ meets its boundary.

Finally suppose a closed subset $A$ of the square has the point-separating clopen property specified in the paper. Every nonempty relatively open subset of the square contains a nontrivial line segment: it contains an interior point of the square and a small ball about that point. If such a segment lay in $A$, a clopen subset of $A$ separating its endpoints would induce a separation of a topological space of the connected segment. This is impossible. Therefore $A$ contains no nonempty relatively open subset of the square. Being closed, it is nowhere dense in the square, and also nowhere dense in ${\mathbb{R}}^2$.

A topological space is connected if it is not the union of two disjoint nonempty open sets, or equivalently has no nontrivial subset that is both an open set and a closed set.

Let $K_x$ be the union of all connected subspaces containing $x$. A union of connected subsets with a common point is connected: if a separation of a topological space existed, every member containing $x$ would lie wholly in the same side. Hence $K_x$ is connected and contains every connected subspace through $x$, so it is the connected component of $x$.

If $K_x\cap K_y\ne \varnothing$, their union is connected and maximality gives $K_x=K_y$. Otherwise they are disjoint. Since every $x$ belongs to $K_x$, the connected components partition $X$.

Suppose for contradiction that the continuous image of a connected space $Y=f(X)$ is disconnected. Then there are disjoint nonempty sets $U,V$, open in the subspace topology on $Y$, with $Y=U\cup V$. By continuity, $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint open subsets of $X$; by surjectivity both are nonempty, and their union is $X$. This is a separation of a topological space of $X$, contradicting that $X$ is a connected space. Hence $Y$ is connected.

A metric space is connected when it has no separation of a topological space: it cannot be written as the union of two disjoint nonempty open subsets. Equivalently, the only subsets that are both open and closed are the empty set and the whole space.The open sets here are those of the metric topology.

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.

The usual real line is connected. In contrast,is a separation of a topological space. Both pieces are nonempty and disjoint; the first is open by part (c), and the second is the open union ${\bigcup }_{n\ge 1}[0,n)$. Thus the Sorgenfrey line is not connected. Since a homeomorphism preserves connectedness, the two topological spaces are not homeomorphic. Merely showing that their topologies differ would not by itself exclude some other homeomorphism between them.

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 $A$.

The intended hypothesis is that $A$ is not connected. The original PDF has the same grammatical ambiguity as the transcription. If only $X$ were assumed not connected, the assertion would fail for the discrete space $X=\{0,1\}$ with its usual metric and the closed subset $A=\{0\}$: two disjoint sets cannot both meet that singleton.

Under the intended reading, choose a separation of a topological space $A=B\bigsqcup C$. Both $B$ and $C$ are nonempty and closed in $A$, because each is the complement of the other relatively open part. Since $A$ is closed in $X$, both are closed in $X$.

For a nonempty subset $F$ of the metric space $(X,d)$, the distance to a set satisfiesIndeed, the triangle inequality gives $d(x,F)\le d(x,z)+d(z,F)$, and swapping $x,z$ proves the bound. Thus the two distance to a set functions below are continuous. SetThese are disjoint open sets. For $b\in B$, we have $d(b,B)=0$ and $d(b,C)>0$, since $C$ is closed and $b\notin C$. Hence $B\subseteq U$. Similarly $C\subseteq V$. Therefore $A\subseteq U\cup V$ and both $U$ and $V$ meet $A$, as required. The construction is a concrete instance of the normality of every metric space.

Removing $1/2$ from the closed interval $I=[0,1]$ leaves the separation of a topological spaceThus $1/2$ is a cut point of $I$.

In contrast, removing any point $p$ from the square $Q=[0,1{]}^2$ leaves a path-connected space. To prove this, take distinct $u,v\in Q\setminus \{p\}$. Choose an interior point $w$ outside the two lines through $(u,p)$ and $(v,p)$. Such a choice exists: a small circle in the interior meets each of those lines in at most two points. Since $Q$ is a convex set, the line segments $[u,w]$ and $[w,v]$ stay in $Q$; our choice of $w$ ensures that both avoid $p$. Their concatenation is a continuous path from $u$ to $v$ in $Q\setminus \{p\}$.

The product topology on $I\times I$ agrees with its usual subspace topology in the Euclidean plane. If $h:Q\to I$ were a homeomorphism, its restriction would give a homeomorphismThe domain is connected, since path connectedness implies connectedness, while the target is not connected. This contradicts the preservation of connectedness by a homeomorphism.

Suppose $K\setminus \{a\}$ were compact. By the compact subset of a Hausdorff space theorem, it would be closed in $X$, hence closed in $K$. The singleton $\{a\}$ is also closed in the Hausdorff space $X$: each other point has an open neighbourhood avoiding $a$.

Consequently both $\{a\}$ and $K\setminus \{a\}$ are clopen in $K$. They are disjoint, nonempty, and cover $K$, giving a separation of a topological space in contradiction to the connectedness of $K$. Thus $K\setminus \{a\}$ is not compact.

True. The canonical projection of a quotient set $p:X\to X/\sim$ is a continuous surjection. Its image is compact by the continuous image of a compact space theorem.

For connectedness, if $U,V$ formed a separation of a topological space of $X/\sim$, then the preimages $p^{-1}(U),p^{-1}(V)$ would be disjoint nonempty open sets covering $X$, contradicting its connectedness. This proves the continuous image of a connected space theorem in this case. Hence the quotient space is both compact and connected.