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Two disjoint connected sets can have a connected union when one contains a contact point of the other's closure. To prove the displayed criterion, suppose the union had a separation into disjoint nonempty relatively open sets $P,Q$. Each connected set lies wholly on one side; if both lie on the same side, the other side is empty. Otherwise assume $A\subset P$ and $B\subset Q$. Choose $b\in B\cap \overline{A}$. Since $Q$ is relatively open in the union and contains $b$, some ambient open set containing $b$ meets the union only in $Q$. It must nevertheless meet $A$ by the definition of closure, a contradiction.

For example, the open unit disc centred at zero and the closed unit disc centred at two are disjoint, but their union contains the common contact point one through the second disc and is connected. In this example it is even path-connected: the real segment from zero to one lies in the first disc except for its endpoint, and that endpoint belongs to the second disc. Omitting the contact point from both discs instead gives a disconnected union.

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 ${\mathbb{R}}^2$.

The open sets $U$ and $X\setminus C_{\lambda }$ cover $X$. A finite subcover exists by compactness, and taking complements yieldsfor a finite index set. This is the finite intersection characterization of compactness in a neighborhood form; the finite index set may be empty when $U=X$.

The Šura-Bura lemma states that in a compact Hausdorff space the connected component $C(x)$ of a point equals its quasi-component, the intersection of all clopen neighborhoods of that point. It also follows that each component has a neighborhood basis of clopen sets. Here is a proof of both assertions.

Let $Q$ be the intersection of the clopen neighborhoods of $x$. Every connected set containing $x$ lies in each such neighborhood, so $C(x)\subseteq Q$. The set $Q$ is closed and compact. If $Q$ were disconnected, write it as the disjoint union of nonempty relatively closed sets $A,B$, choosing $x\in A$. Both are compact and closed in $X$. Normality gives disjoint open sets $O_A,O_B$ containing them. By the finite-intersection result, a finite intersection $V$ of clopen neighborhoods of $x$ satisfies $Q\subseteq V\subseteq O_A\cup O_B$. Then $V\cap O_A$ is clopen in $V$, hence clopen in $X$, and contains $x$ while excluding $B$. This contradicts the definition of $Q$. Thus $Q$ is connected and equals $C(x)$. Given any open neighborhood of the component, the same finite-intersection argument supplies a clopen neighborhood inside it. This proves the neighborhood-basis form as well.

For an explicit noncompact counterexample to the Šura-Bura lemma, put $p=(0,0)$, $q=(1,0)$, and ${\epsilon }_m=1/(m+2)$ for $m\ge 1$. Let $A_m$ be the polygonal arc with successive verticesand give $Z=\{p,q\}\cup {\bigcup }_{m\ge 1}{A}_m$ its subspace topology from the plane. The arcs are pairwise disjoint: a higher arc has its vertical sides outside the horizontal span of any lower arc. Each $A_m$ is compact and relatively open in $Z$, so is clopen. For relative openness, all sufficiently high arcs have vertical sides close to $x=0,1$ and horizontal sides above a fixed bounded neighborhood of $A_m$; the remaining finitely many compact disjoint arcs have positive distance from it.

A connected subset containing $p$ cannot meet any of these clopen arcs, so it lies in $\{p,q\}$ and is just $\{p\}$. But any clopen neighborhood of $p$ contains the left endpoints of all sufficiently high arcs. Connectedness of each arc forces the entire arc into that neighborhood, and closedness then forces the limit $q$ of their right endpoints into it. Conversely the clopen set $Z\setminus A_m$ excludes every point of $A_m$. ThereforeThe compact set containing the endpoints has lost the connection supplied by arcs escaping to arbitrarily large height; compactness of the ambient space is essential to the lemma.

Now let $D={\bigcap }_n{C}_n$ for the decreasing compact connected sets in the compact Hausdorff setting. Every finite intersection is one of the nonempty $C_n$, so the finite intersection property gives $D\ne \varnothing$. It is closed and hence compact. If it were disconnected, its two nonempty compact pieces would have disjoint open neighborhoods $U,V$ in $X$. Since $D\subseteq U\cup V$, the initial finite-intersection result and the decreasing property give $C_N\subseteq U\cup V$ for some $N$. This set meets both neighborhoods because it contains $D$, contradicting its connectedness. ThusThe argument does not need connectedness of the whole ambient space.

For the requested failure without ambient compactness, add the central ray $L=\{(1/2,t):t\ge 1\}$ to the previous example and put $X=Z\cup L$. For $n\ge 1$ defineEach graph formed by the displayed arcs and ray is connected: each arc meets the ray at its top. Adding $p,q$, which are limit points of the graph, preserves connectedness because the closure of a connected set is connected. Its relative closure in $X$ adds exactly these two points, so $C_n$ is closed in $X$. To check that no omitted point is a further limit, a bounded neighborhood away from the lines $x=0,1$ meets only finitely many arcs; the deleted initial arcs are disjoint compact sets, and the deleted ray segment lies strictly below the remaining ray. All other planar limit points on $x=0,1$ are outside $X$.

The sets are decreasing and nonempty, and $X=C_1$ is connected and Hausdorff. Every individual arc and every point of finite height on the ray disappears from sufficiently late sets. Consequentlya nonempty compact disconnected set. Reindexing by $n+1$ gives the source's indexing by $\omega$. The graph is unbounded, so the ambient subspace is not compact. This is an explicit decreasing closed connected sets with disconnected compact intersection example.

Write $C^{\prime }=C\setminus \{(1,1)\}$ and $V=\{1\}\times [0,1)$. Let $L$ and $R$ be the portions of $C^{\prime }$ with first coordinate respectively less than and greater than one. Each is path-connected: move along a rational tooth to its corresponding open half of the top line, then horizontally there. By the density of the rational numbers, every point of $V$ is a limit point of both $L$ and $R$.

Use the elementary fact that adjoining any collection of limit points to a connected set preserves connectedness. Indeed, in a proposed separation, the connected subset lies on one side; relative openness would prevent any point on the other side from being a limit point of it. Therefore $L\cup V$ and $R\cup V$ are connected. They intersect in $V$, so their union $C^{\prime }$ is connected.

There is, however, no path from $(1,0)$ to $(0,1)$. If $\gamma (t)=(x(t),y(t))$ were one, let $t_0$ be the first time at which $y(t_0)=1$. Such a first time exists by continuity and compactness of $[0,1]$, and $t_0>0$. For $0\le t<t_0$, membership in the comb forces $x(t)\in \mathbb{Q}$. The continuous image of a connected space is connected, and a connected subset of $\mathbb{Q}$ is a singleton: an irrational number between any two distinct rational values would separate it. Hence $x(t)=1$ for $t<t_0$. By continuity, $\gamma (t_0)=(1,1)$, which was deleted.

A connected set is one whose relative topology admits no decomposition into two disjoint nonempty open sets. A path-connected set permits, between any two points, a continuous map $\gamma :[0,1]\to A$ with the prescribed endpoints.

Path connectedness implies connectedness, because a continuous image of an interval is connected. The converse can fail, for example for the topologist's sine curve. On open subsets of Euclidean spaces ${\mathbb{R}}^n$, the two properties are equivalent; more generally this holds in locally path-connected spaces.

A homogeneous hyperelastic material has a stored energy per reference volume $W(F)$ depending on the local deformation gradient but not explicitly on the material point. Its first Piola-Kirchhoff response isA superposed rigid rotation changes the observer's description without changing the energy of the deformation. Material frame indifference is thereforeThis left-rotation invariance is distinct from a material isotropy assumption on right rotations. Differentiate it in an arbitrary matrix direction $H$:The Frobenius inner product then gives $Q^T\hat{T}(QF)=\hat{T}(F)$, proving $\hat{T}(QF)=Q\hat{T}(F)$.

For the converse, that stress covariance makes the derivative in $F$ of $W(QF)-W(F)$ zero. The positive-determinant matrices form a connected set, as also follows from polar decomposition, so this difference is a constant $c(Q)$ independent of $F$. Composition gives $c(Q_1{Q}_2)=c(Q_1)+c(Q_2)$. For fixed $F$, continuity of $W$ and compactness of $\mathrm{SO}(3)$ make $c$ bounded. But $c(Q^n)=nc(Q)$ for every positive integer, forcing $c(Q)=0$. Thus stress covariance and frame indifference of the energy are equivalent. Compactness removes the possible additive energy constant in the converse.

The quasi-component of a point is the intersection of its clopen neighborhoods. Its connected component is always contained in it: a connected set meeting a clopen set cannot also meet its complement. Equality need not hold in noncompact Hausdorff spaces. The Šura-Bura lemma proves equality in compact Hausdorff spaces.

The displayed subspace of the Euclidean plane is path-connected: each vertical rational tooth joins the horizontal line at height one. Delete $(r,1)$ for rational $r$. The residual tooth $V=\{r\}\times [0,1)$ cannot be joined by a path to the top line. Before a path first reaches height one, its first coordinate is a continuous rational-valued function on an interval and must therefore be constant; it would reach the deleted point.

The punctured space remains connected. Its portions $L,R$ on either side of $x=r$ are path-connected, and $V\subseteq \overline{L}\cap \overline{R}$ by the density of the rational numbers. Adjoining limit points to a connected set preserves connectedness: a separation would place the connected subset on one side and prevent the other side from containing its limit points. Hence $L\cup V$ and $R\cup V$ are connected and intersect in $V$. Their union is connected. This example distinguishes connectedness from path-connected using the loss of a single attachment point.