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The circle $S^1$ and the closed interval $[0,1]$ are both compact and connected, so those two properties alone do not distinguish them. Instead examine connectedness after removing a point. Every punctured circle is connected: after rotating the missing point to $(1,0)$, the continuous map $t\mapsto (\cos t,\sin t)$ from the connected open interval $(0,2\pi )$ maps onto it. Continuous maps preserve connectedness. ^[ref]

If a homeomorphism $h:S^1\to [0,1]$ existed, choose $p=h^{-1}(1/2)$. Its restriction would give a homeomorphism between $S^1\setminus \{p\}$ and $[0,1]\setminus \{1/2\}$. But the latter splits into two nonempty disjoint open sets in its subspace topology, $[0,1/2)$ and $(1/2,1]$, contradicting connectedness of the punctured circle. Removing an endpoint of $[0,1]$ leaves a connected space; the proof deliberately chooses the preimage of an interior point. This illustrates how a simple connectedness test can rule out every possible homeomorphism.

The cofinite topology on a set $X$ declares the empty set and all cofinite subsets to be open sets: a nonempty open set may exclude only a finite set of points. There is no assumed metric. On an infinite set, this gives a particularly small collection of open sets while still making every singleton a closed set. On a finite set, every subset has finite complement, so the cofinite topology is the discrete topology.^[ref]

For an original example, use $X={\mathbb{N}}_{>0}$ and chooseBoth are open sets, and their intersection is $U\cap V=X\setminus \{1,2,3,5,7,8\}$. Thus $4,6,9,10,\dots$ are in both. Their finite exclusions cannot separate the infinite remainder. The finite intersection of any finite collection of nonempty open sets is again a nonempty open set, since its complement is a finite union of finite sets. An arbitrary nonempty union is open because its complement is contained in the finite complement of any one nonempty member.

The cofinite topology is useful for separating hypotheses that familiar metric spaces package together. An infinite cofinite topological space is a T1 space, because its singletons are closed, but is not a Hausdorff space, because two nonempty open sets cannot be disjoint sets. It is a hyperconnected space and hence a connected space. Moreover, every subset of a cofinite space is compact, and sequence convergence in the cofinite topology can have many limits. These are compatible properties, rather than contradictions: the usual uniqueness and closedness results use the Hausdorff space hypothesis.^[ref]

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.

A closed filter base in a compact space has nonempty intersection. If the space is Hausdorff and every base member is connected, the intersection is connected. Otherwise normality separates its two closed components by disjoint open sets. Compactness of the complement supplies finitely many base members whose intersection lies in that union; directedness then gives a single base member meeting both sides, contradicting its connectedness.

In a locally path-connected space, every path component is open: a path-connected open neighbourhood of a point lies within that point's path component. The complement of a path component is a union of the other open path components, hence is also open. In a nonempty connected space, a nonempty subset that is both open and closed must be the whole space. Thus a connected space that is locally path-connected is path-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.

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 path-connected space $S$ has, for every pair $z_0,z_1\in S$, a continuous path $\gamma :[0,1]\to S$ with $\gamma (0)=z_0$ and $\gamma (1)=z_1$. Every path-connected space is a connected space, because a separation of the target would pull a joining path apart and contradict connectedness of an interval.

A connected space $S$ cannot be expressed as $S=A\cup B$ with $A,B$ disjoint nonempty relatively open sets. Equivalently, the only subsets simultaneously relatively open and closed are $\varnothing$ and $S$. A decomposition into such $A,B$ is a separation of the topological space.

Constructing the subsheaf. On an integral scheme, every nonempty open subset is an irreducible topological space and hence a connected space. A locally constant function from such an open subset to $F$ is constant. Thus the constant sheaf $\underline{F}$ has sections $F$ on every nonempty open subset, and the structure sheaf embeds in it by restriction to the generic point. Empty members of the given open cover may be discarded.

For every open subset $W\subseteq X$, defineThis is a subsheaf of the sheaf of modules $\underline{F}$: its condition is local, and compatible sections glue in $\underline{F}$, with the regularity conditions gluing in ${\mathcal{O}}_X$. On $U_i\cap U_j$, the hypothesis says that $f_j/f_i$ is a section of the sheaf of units of the structure sheaf. Therefore multiplication by it preserves ${\mathcal{O}}_X$, andIt follows that $\mathcal{L}{|}_{U_i}=f_i^{-1}{\mathcal{O}}_{U_i}$: the condition with index $i$ implies every other condition on an overlap, and the converse is part of the definition. Multiplication by $1/f_i$ is an isomorphism ${\mathcal{O}}_{U_i}\to \mathcal{L}{|}_{U_i}$, with inverse multiplication by $f_i$. Hence $\mathcal{L}$ is an invertible sheaf, or equivalently a line bundle, andThese local descriptions also give uniqueness as a subsheaf of $\underline{F}$. This construction is the divisor line bundle for the local rational equations $f_i$.

The tensor product. Refine the two open covers to $W_{ij}=U_i\cap V_j$. The ratiosare sections of the sheaf of units of the structure sheaf on the corresponding overlaps, so the products give a line bundle $\mathcal{M}\subseteq \underline{F}$ by the construction just proved. Multiplication in $F$ defines a sheaf morphismOn each $W_{ij}$ it sends the local generator $(1/f_i)\otimes (1/g_j)$ to the local generator $1/(f_i{g}_j)$. It is thus an isomorphism on an open cover, and hence an isomorphism globally. In particular,

The sign for projective twists. Write homogeneous coordinates $[x_0:\cdots :x_n]$ on projective space ${\mathbb{P}}_{\mathbb{Z}}^n$, and let $U_i=D_{+}(x_i)$ be its standard affine open cover. For any integer $m$, take the nonzero elements of the function fieldTheir ratios are $f_i/f_j=(x_j/x_i{)}^m$, which are sections of the sheaf of units of the structure sheaf on $U_i\cap U_j$, including when $m<0$. The twisting sheaf on projective space $\mathcal{O}(m)$ has local generator $x_i^m$ on $U_i$. Divide a rational section of $\mathcal{O}(m)$ by $x_0^m$ to embed it in the constant sheaf of the function field; the local generator becomes $(x_i/x_0{)}^m=1/f_i$. This identifies the constructed line bundle with $\mathcal{O}(m)$ and fixes the sign explicitly.

For $n\ge 1$, $\mathrm{Pic}({\mathbb{P}}_{\mathbb{Z}}^n)\cong \mathbb{Z}$, with generator $[\mathcal{O}(1)]$; the displayed data gives all its powers, with $\mathcal{O}(m)\otimes \mathcal{O}(r)\cong \mathcal{O}(m+r)$ by the product rule above. If $n=0$, projective space is $\mathrm{Spec}\mathbb{Z}$, its Picard group is zero, and the same formula gives only the trivial line bundle.

In an irreducible topological space, any two nonempty open subsets intersect. Let $W\subseteq X$ be a nonempty open subset. Two nonempty subsets open in $W$ are also open in $X$, and hence intersect; thus $W$ is itself an irreducible topological space. If $W$ were disconnected, a separation into two disjoint nonempty open subsets would contradict this intersection property. Hence $W$ is a connected space. The empty open subset is connected as well under the usual convention.

Discard empty members of $\mathcal{U}$ as in (b). Every finite intersection $U_{i_0}\cap \cdots \cap U_{i_q}$ is nonempty by repeated use of the intersection property and is an open subset, hence a connected space. A locally constant function on this intersection is constant. It follows that the constant sheaf hasand every restriction map between these nonempty intersections is the identity on $A$. The whole Čech cochain complex is therefore the same as for the indexed open cover of a point having the same index set. The cochain homotopy constructed in (b) applies component by component and provesThe argument concerns the constant sheaf of locally constant functions; arbitrary coefficient sheaves on an irreducible topological space need not have the same Čech cohomology.

A connected space admits no separation into two disjoint nonempty open subsets whose union is the space. A path-connected space has, for every pair of points, a continuous path from $[0,1]$ joining them. If a path-connected space had a separation $U,V$, choose its endpoints in different pieces. Their inverse images under the path would separate the connected interval $[0,1]$, a contradiction. Thus path connectedness implies connectedness.

For the specified subspace, first consider $Y=I\cup {\bigcup }_{n\ge 1}{J}_n$. Every point of a tooth $J_n$ can move vertically to its base in $I$, then horizontally to the origin; points already on $I$ use only the horizontal segment. Concatenating these paths shows that $Y$ is path-connected and hence connected.

Every point $(0,t)$ of $A$ is a limit of $(1/n,t)\in Y$. Consequently $Y$ is dense in $X=Y\cup A$ in the subspace topology. To prove $X$ connected, suppose $X=U\cup V$ were a separation. Restricting to the connected $Y$ puts all of $Y$ in one piece, say $U$. But the other nonempty relatively open piece $V$ must meet dense $Y$, a contradiction. Therefore $\boxed{X\text{ is connected}}$. This uses density, not an unproved path from the added vertical segment to the teeth.

Use the one-sided product bound for an entire function by consideringThis is an entire function, and $f^2=(u^2-v^2)+2iuv$ gives $|F|=e^{2uv}$. If $uv\le M$, then $|F|\le e^{2M}$ everywhere. The Liouville theorem therefore makes $F$ constant.

Because an exponential never vanishes, differentiating yields $0=F^{\prime }=-2if{f}^{\prime }F$, and hence $(f^2{)}^{\prime }=2f{f}^{\prime }=0$. Thus $f^2$ is constant. If that constant is zero, $f=0$; otherwise $f$ takes values in the two square roots of one nonzero constant. Continuity on the connected space $\mathbb{C}$ prevents switching between those two values. Therefore $f$ is constant.

Every point has arbitrarily small path-connected open neighbourhoods. Indeed, in a coordinate chart homeomorphic to ${\mathbb{R}}^n$, take the inverse image of a sufficiently small open Euclidean ball. The ball is convex, hence path-connected, and its inverse image is path-connected by the homeomorphism. If the supplied neighbourhood is not itself open, first take an open neighbourhood within it and a ball contained in its chart image. This proves that $X$ is a locally path-connected space.

Fix $x_0\in X$ and let $A$ consist of the points joined to $x_0$ by a path. For $x\in A$, a path-connected open neighbourhood $U_x$ is contained in $A$: concatenate a path from $x_0$ to $x$ with one from $x$ to any point of $U_x$. Hence $A$ is open. For $x\notin A$, the same kind of neighbourhood cannot meet $A$, since a meeting point would provide a path to $x$. Hence $X\setminus A$ is also open. Since $X$ is a connected space and $A$ is nonempty, $A=X$. This proves $X$ is path-connected, including the general principle that connected locally path-connected spaces are path-connected. The empty-space case, if allowed, is vacuous.

Set $h=f-g$, an entire function. By the Bolzano-Weierstrass theorem, a subsequence of the distinct zeros $z_i$ converges to a point $a$ in the closed unit disc. An analytic function has a convergent Taylor series in a neighbourhood of each point. If the first nonzero coefficient of $h$ at $a$ had index $m$, thenwith $k$ analytic and nonzero in a sufficiently small disc. The zero $a$ would then be isolated, contradicting the accumulating distinct zeros. All Taylor coefficients at $a$ vanish, so $h$ vanishes on a disc.

To justify continuation across the whole plane, let $Z$ be the set of points at which $h$ vanishes on a neighbourhood. This set is nonempty and open. It is also closed: if points of $Z$ tend to $b$, continuity of each derivative gives $h^{(m)}(b)=0$ for every $m$, and the Taylor series at $b$ gives a zero neighbourhood. Since $\mathbb{C}$ is a connected space, $Z=\mathbb{C}$. This proves the identity theorem here and gives $\boxed{f\equiv g}$.

Without boundedness, the distinct zeros need not have an accumulation point in the plane. For example,are distinct entire functions agreeing at all the indicated positive integers.

The strong maximum principle for harmonic functions states that a $C^2$ harmonic function on a connected domain which attains its supremum at an interior point is constant. Applying it to $-u$ gives the corresponding assertion for an attained infimum.

Let $M=u(x_0)={\sup }_{\Omega }u$. The ball mean value property for harmonic functions gives, for every sufficiently small ball about $x_0$,The integrand is continuous and nonnegative. If it were positive at any point, it would be bounded below by a positive number on a smaller ball, contradicting the zero integral. Therefore $u=M$ on the whole ball. The set $A=\{x\in \Omega :u(x)=M\}$ is relatively closed by continuity and relatively open by this same argument at each of its points. It is nonempty. Connected space forces $A=\Omega$, proving the principle. If the open set is disconnected, constancy is asserted on the relevant connected component.

True. For nonempty connected spaces $X,Y$, choose $x_0\in X$, $y_0\in Y$. For each $x\in X$, the subspace $(X\times \{y_0\})\cup (\{x\}\times Y)$ is connected: its two connected pieces meet at $(x,y_0)$. The union of these subspaces over $x$ is all of $X\times Y$, and every piece contains the common connected subset $X\times \{y_0\}$. Their union is therefore connected. The assertion is also true if a factor is empty, since the empty product space is connected. This proves product of connected spaces.

An algebraic group is connected when its underlying Zariski topology is a connected space. An unipotent algebraic group is a linear algebraic group whose elements, in a faithful matrix realization, satisfy that $u-I$ is nilpotent. A reductive algebraic group, in the connected convention, is a connected linear algebraic group whose unipotent radical is trivial: it has no nontrivial connected normal unipotent algebraic subgroup.

Let $E={\mathrm{End}}_Q(X)$. It is a finite-dimensional associative algebra, and ${\mathrm{Aut}}_Q(X)=E^{\times }$. The latter is the nonempty open subset where the determinant on ${⨁}_i{X}_i$ is nonzero. Since the affine space $E$ is irreducible, this open subset is irreducible and hence connected. It is a linear algebraic group: within $\mathrm{GL}({⨁}_i{X}_i)$ it is cut out by the linear equations of preserving the vertex summands and commuting with the arrow maps. Thus the automorphism group is connected.

By the Krull–Schmidt theorem, write $X\cong {⨁}_{a=1}^s{M}_a^{\oplus m_a}$, with pairwise nonisomorphic indecomposables $M_a$. The Fitting lemma makes each ${\mathrm{End}}_Q(M_a)$ a local algebra. Its residue division algebra is $k$: every element of a finite-dimensional division algebra over the algebraically closed field has an eigenvalue for left multiplication, and subtracting that scalar gives a noninvertible element, hence zero.

The standard radical description of the endomorphism algebra of a Krull-Schmidt decomposition therefore givesConcretely, maps between nonisomorphic summands lie in the radical, while on each isotypic block one reduces all entries modulo the local radical. The off-diagonal rule follows because a composition $M_a\to M_b\to M_a$ cannot be a unit for $a\ne b$: it would split $M_a$ off the indecomposable $M_b$, forcing an isomorphism. The finite-dimensional radical description then identifies the kernel of this block reduction with $J(E)$. Put $U=1+J(E)$. The Jacobson radical of the finite-dimensional algebra is nilpotent, so every $1+j$ is a unit and is unipotent on ${⨁}_i{X}_i$. Also $U$ is closed, being the translate of the linear subspace $J(E)$, and normal, being the kernel ofThis map has an explicit algebraic section: after fixing the direct-sum decomposition, a matrix $B_a$ acts on the $m_a$ copies by $B_a\otimes 1_{M_a}$. These block actions give a subgroup $L\cong {\prod }_a{\mathrm{GL}}_{m_a}(k)$ with trivial intersection with $U$. Every unit is uniquely a product of an element of $U$ and one of $L$. HenceThis Levi decomposition of a quiver automorphism group holds in every characteristic; the section is constructed directly from the multiplicity spaces.