A [topological space](/page/Topological%20Space) can fail to be connected in a way that no metric measurement detects at first glance. The set $[0,1] \cup [2,3] \subsetneq \mathbb{R}$ has no gap inside either interval, but a continuous path trying to move from the first interval to the second must jump across the missing open interval $(1,2)$. Connectedness is the language that detects whether a space has been split into two nonempty open pieces. The basic question is not whether points are close together. It is whether the topology permits a global yes-or-no distinction that is locally invisible. If a space $X$ can be written as two disjoint nonempty open subsets, then a [continuous function](/page/Continuous%20Function) from $X$ to a two-point discrete space records which side a point lies on. Connected spaces are exactly the spaces where no such continuous binary invariant exists. [example: Two Intervals Are Separated] Let $X = [0,1] \cup [2,3]$ with the [subspace topology](/page/Subspace%20Topology) inherited from $\mathbb{R}$. Set $A=[0,1]$ and $B=[2,3]$. Both sets are nonempty because $0 \in A$ and $2 \in B$. They are disjoint: if $x \in A \cap B$, then $0 \le x \le 1$ and $2 \le x \le 3$, which would imply $2 \le x \le 1$, impossible. Also, \begin{align*} A \cup B &= [0,1] \cup [2,3] \\ &= X. \end{align*} To check openness in the subspace topology, compute the two intersections: \begin{align*} X \cap (-1,3/2) &= \bigl([0,1] \cup [2,3]\bigr) \cap (-1,3/2) \\ &= \bigl([0,1]\cap (-1,3/2)\bigr) \cup \bigl([2,3]\cap (-1,3/2)\bigr) \\ &= [0,1] \cup \varnothing \\ &= A, \end{align*} and \begin{align*} X \cap (3/2,4) &= \bigl([0,1] \cup [2,3]\bigr) \cap (3/2,4) \\ &= \bigl([0,1]\cap (3/2,4)\bigr) \cup \bigl([2,3]\cap (3/2,4)\bigr) \\ &= \varnothing \cup [2,3] \\ &= B. \end{align*} Since $(-1,3/2)$ and $(3/2,4)$ are open in $\mathbb{R}$, these equalities show that $A$ and $B$ are open in $X$. Thus $X$ is split into two nonempty disjoint open pieces, so the two intervals are separated even though each interval has no internal gap. [/example] The example suggests the right obstruction. To define connectedness without mentioning distances, intervals, or paths, we first need a purely topological name for a split into two open sides. A separation is not just a partition into two sets; the two pieces must both be topologically visible from within the space. [definition: Separation] Let $(X, \tau)$ be a topological space. A separation of $X$ is an ordered pair $(U,V)$ of subsets of $X$ such that: \begin{align*} U &\in \tau, \\ V &\in \tau, \\ U &\neq \varnothing, \\ V &\neq \varnothing, \\ U \cap V &= \varnothing, \\ U \cup V &= X. \end{align*} [/definition] A separation is the precise failure mode. The primary definition names the spaces in which this failure mode is absent, so that later theorems can speak about continuity preserving a single unbroken whole. ## Definition After naming separations, connectedness is the condition that rules them out. This is the useful direction: instead of trying to prove a space is made of one visible piece by inspecting all its points, we ask whether any proposed two-sided split can exist at all. [definition: Connected Space] A topological space $(X, \tau)$ is connected if there is no separation of $X$. [/definition] This definition is deliberately global. It does not ask whether nearby points can be joined, and it does not mention distances. It asks only whether the topology permits the whole space to be split into two simultaneously visible nonempty sides. ## Detecting Disconnection Connectedness is often used through its failures. Once the primary definition is in place, it is useful to name the opposite condition and then translate separations into simpler tests involving one subset or one continuous binary-valued function. ### Disconnected Spaces A space that admits a separation behaves like two topological worlds placed side by side. Naming that situation lets us discuss examples and counterexamples without carrying the two open pieces every time. [definition: Disconnected Space] A topological space $(X, \tau)$ is disconnected if there exists a separation of $X$. [/definition] The two sides of a separation determine each other by complementation. That observation turns disconnection into a question about a single subset whose presence and absence are both open conditions. ### Clopen Sets A separation can be recorded by either one of its two sides, because the other side is its complement. This motivates a name for subsets whose membership and non-membership are both open conditions. [definition: Clopen Subset] Let $(X, \tau)$ be a topological space. A subset $A \subset X$ is clopen in $X$ if $A$ is open in $X$ and $X \setminus A$ is open in $X$. [/definition] A disconnected space may be hard to split if one searches directly for two open pieces. Clopen subsets isolate the obstruction in one place: a nonempty proper subset whose inside and outside are both topologically visible. The criterion below turns connectedness into the absence of exactly that obstruction. [quotetheorem:4997] The clopen characterisation says that connectedness forbids nonconstant yes-or-no data. To turn that principle into a form usable with continuous maps, we need the target space whose open sets make both answers distinguishable. ### Binary Invariants A binary invariant should be allowed to distinguish $0$ from $1$ without any topological ambiguity. The smallest target with this property is the two-point set with every subset open. [definition: Discrete Two-Point Space] The discrete two-point space is the topological space $(\{0,1\}, \tau)$ where \begin{align*} \tau = \{\varnothing, \{0\}, \{1\}, \{0,1\}\}. \end{align*} [/definition] Once the two labels are separated by open sets in the target, continuity forces each label fibre to be open in the source. A nonconstant binary-valued map would therefore cut the source into two open fibres. The theorem below records the converse as well, so binary invariants become an exact test for connectedness. [quotetheorem:294] This viewpoint is important because many applications of connectedness prove that some quantity cannot change value. Instead of locating all points, one proves that a continuous invariant with discrete values must be constant. ## Intervals and the Intermediate Value Principle The first serious test case is the real line. Connectedness is not an abstract luxury there; it is exactly the topological content behind the [intermediate value theorem](/theorems/629). A continuous real-valued function cannot move from a negative value to a positive value without taking every value in between because intervals cannot be torn into two open pieces. To express the real-line classification, we need the standard order-convex subsets of $\mathbb{R}$. These are the sets with no order gap, and they are precisely the candidates for connected subsets of the line. [definition: Interval] A subset $I \subset \mathbb{R}$ is an interval if for every $a,b \in I$ with $a < b$ and every $t \in \mathbb{R}$ satisfying $a < t < b$, we have $t \in I$. [/definition] The definition includes open intervals, closed intervals, half-open intervals, rays, singletons, and $\mathbb{R}$ itself. On the real line, any missing point between two chosen points acts as a cut: points to the left and right can be separated by open rays. The classification below says this is the only possible obstruction to connectedness in $\mathbb{R}$. [quotetheorem:295] This theorem is the bridge between topological connectedness and the familiar order structure of $\mathbb{R}$. It also prepares the [intermediate value theorem](/theorems/180), since continuous images of connected spaces will become intervals in $\mathbb{R}$. [quotetheorem:629] The theorem explains why a connected domain prevents a continuous scalar function from skipping values. Conversely, when the domain is disconnected, a continuous function can jump between components without violating continuity. [example: A Continuous Function with a Jump Across a Gap] Let $X = [0,1] \cup [2,3] \subsetneq \mathbb{R}$ with the subspace topology, and define \begin{align*} f: X &\to \mathbb{R} \\ x &\mapsto \begin{cases} 0, & x \in [0,1], \\ 1, & x \in [2,3]. \end{cases} \end{align*} The two pieces of $X$ are open in the subspace topology because \begin{align*} X \cap (-1,3/2) &= \bigl([0,1] \cup [2,3]\bigr)\cap (-1,3/2) \\ &= \bigl([0,1]\cap (-1,3/2)\bigr)\cup \bigl([2,3]\cap (-1,3/2)\bigr) \\ &= [0,1]\cup \varnothing \\ &= [0,1], \end{align*} and \begin{align*} X \cap (3/2,4) &= \bigl([0,1] \cup [2,3]\bigr)\cap (3/2,4) \\ &= \bigl([0,1]\cap (3/2,4)\bigr)\cup \bigl([2,3]\cap (3/2,4)\bigr) \\ &= \varnothing \cup [2,3] \\ &= [2,3]. \end{align*} Now let $W \subset \mathbb{R}$ be open. Since $f$ only takes the values $0$ and $1$, membership in $f^{-1}(W)$ is determined by whether $0$ and $1$ lie in $W$: \begin{align*} f^{-1}(W) &= \begin{cases} \varnothing, & 0 \notin W \text{ and } 1 \notin W, \\ [0,1], & 0 \in W \text{ and } 1 \notin W, \\ [2,3], & 0 \notin W \text{ and } 1 \in W, \\ X, & 0 \in W \text{ and } 1 \in W. \end{cases} \end{align*} The sets $\varnothing$ and $X$ are open in $X$, and the computations above show that $[0,1]$ and $[2,3]$ are open in $X$. Therefore $f^{-1}(W)$ is open in $X$ for every open $W \subset \mathbb{R}$, so $f$ is continuous. Its image is \begin{align*} f(X) &= f([0,1]\cup [2,3]) \\ &= f([0,1])\cup f([2,3]) \\ &= \{0\}\cup \{1\} \\ &= \{0,1\}. \end{align*} This image is not an interval: $0,1 \in \{0,1\}$ and $0<1/2<1$, but $1/2 \notin \{0,1\}$. Thus the missing gap in the domain allows a continuous function to jump from $0$ to $1$ without taking any intermediate value. [/example] The lesson is not that continuous functions never jump as formulas. It is that the topology of the domain may hide the jump by removing the points where the transition would have to occur. ## Continuous Images and Constructions Connectedness is useful because it survives many standard constructions. If a space is connected, then any continuous observation of it remains connected. This is why connectedness can be transported from a familiar space to a less familiar one. ### Continuous Images The first permanence principle is that continuous maps cannot create a separation in the image from a connected domain. This is the mechanism behind many intermediate value arguments and parametrized geometric examples. [quotetheorem:296] This theorem is often the fastest way to prove that a complicated set is connected: represent it as the image of an interval, a connected product, or a union assembled from connected pieces. [example: The Unit Circle as a Connected Image] Consider the map \begin{align*} f: [0,1] &\to \mathbb{R}^2 \\ t &\mapsto (\cos(2\pi t), \sin(2\pi t)). \end{align*} The functions $t \mapsto 2\pi t$, $u \mapsto \cos u$, and $u \mapsto \sin u$ are continuous, so both coordinate functions $t \mapsto \cos(2\pi t)$ and $t \mapsto \sin(2\pi t)$ are continuous. Hence $f$ is continuous as a map into $\mathbb{R}^2$. We compute its image. For every $t \in [0,1]$, \begin{align*} \bigl(\cos(2\pi t)\bigr)^2 + \bigl(\sin(2\pi t)\bigr)^2 &= 1 \end{align*} by the identity $\cos^2 u+\sin^2 u=1$, applied with $u=2\pi t$. Therefore \begin{align*} f([0,1]) \subset \{(x,y) \in \mathbb{R}^2 : x^2+y^2=1\}. \end{align*} Conversely, let $(x,y) \in \mathbb{R}^2$ satisfy $x^2+y^2=1$. If $y \ge 0$, set $\theta=\arccos x$. Then $\theta \in [0,\pi]$, \begin{align*} \cos \theta &= x, \\ \sin \theta &= \sqrt{1-\cos^2\theta} \\ &= \sqrt{1-x^2} \\ &= \sqrt{y^2} \\ &= y. \end{align*} If $y<0$, set $\theta=2\pi-\arccos x$. Then $\theta \in (\pi,2\pi]$, \begin{align*} \cos \theta &= \cos(2\pi-\arccos x) \\ &= \cos(\arccos x) \\ &= x, \end{align*} and \begin{align*} \sin \theta &= \sin(2\pi-\arccos x) \\ &= -\sin(\arccos x) \\ &= -\sqrt{1-x^2} \\ &= -\sqrt{y^2} \\ &= y, \end{align*} because $y<0$ implies $\sqrt{y^2}=-y$. In either case, with $t=\theta/(2\pi)$ we have $t \in [0,1]$ and \begin{align*} f(t) &= \left(\cos\left(2\pi \cdot \frac{\theta}{2\pi}\right), \sin\left(2\pi \cdot \frac{\theta}{2\pi}\right)\right) \\ &= (\cos \theta,\sin \theta) \\ &= (x,y). \end{align*} Thus \begin{align*} f([0,1]) &= \{(x,y) \in \mathbb{R}^2 : x^2+y^2=1\} \\ &= S^1. \end{align*} The set $[0,1]$ is an interval, hence connected by *[Connected Subsets of the Real Line](/theorems/295)*. Since $S^1=f([0,1])$ is the continuous image of this connected space, *Continuous Image of a Connected Space* gives that $S^1$ is connected. This proves connectedness by parametrizing the circle, rather than by trying to rule out separations of the circle directly. [/example] ### Gluing Connected Pieces Images are only one construction. We also need ways to build large connected spaces from smaller ones. The guiding principle is that connected pieces may be glued together as long as the gluing prevents a separation from assigning them to different sides. [quotetheorem:298] The common point hypothesis can be weakened to a chain of overlaps, but the common-point version already captures many geometric constructions: cones, stars, and bouquets of spaces. [example: A Cross in the Plane] Let \begin{align*} A &= [-1,1] \times \{0\}, \\ B &= \{0\} \times [-1,1] \end{align*} as subspaces of $\mathbb{R}^2$. The map \begin{align*} h_A: [-1,1] &\to A \\ t &\mapsto (t,0) \end{align*} is continuous, and its inverse is \begin{align*} p_A: A &\to [-1,1] \\ (x,0) &\mapsto x. \end{align*} Both maps are continuous because they are restrictions of coordinate maps and constant-coordinate maps in $\mathbb{R}^2$. Thus $A$ is homeomorphic to $[-1,1]$. Since $[-1,1]$ is an interval, *Connected Subsets of the Real Line* gives that $[-1,1]$ is connected, so $A$ is connected. Similarly, the map \begin{align*} h_B: [-1,1] &\to B \\ t &\mapsto (0,t) \end{align*} is a homeomorphism with inverse \begin{align*} p_B: B &\to [-1,1] \\ (0,y) &\mapsto y. \end{align*} Therefore $B$ is connected. Now compute the intersection. If $(x,y) \in A \cap B$, then $(x,y) \in A$ gives \begin{align*} y=0, \end{align*} and $(x,y) \in B$ gives \begin{align*} x=0. \end{align*} Hence $(x,y)=(0,0)$, so \begin{align*} A \cap B \subset \{(0,0)\}. \end{align*} Conversely, \begin{align*} (0,0) &\in [-1,1]\times\{0\}=A, \\ (0,0) &\in \{0\}\times[-1,1]=B, \end{align*} so \begin{align*} \{(0,0)\} \subset A \cap B. \end{align*} Thus \begin{align*} A \cap B = \{(0,0)\} \neq \varnothing. \end{align*} Since $A$ and $B$ are connected subspaces with a common point, *[Union of Overlapping Connected Sets is Connected](/theorems/298)* implies that $A \cup B$ is connected. This shows that connectedness can survive even when two connected pieces meet at only one point. [/example] ### Products Gluing handles unions, but many geometric spaces are parameter spaces with several independent coordinates. To prove rectangles, boxes, cylinders, and tori connected, we need connectedness to survive Cartesian products. [quotetheorem:299] Product connectedness turns interval connectedness into the connectedness of rectangles, boxes, and many parameter spaces. It also provides a geometric route to connectedness in Euclidean spaces. [example: Rectangles Are Connected] Let $R=[a,b]\times[c,d]\subsetneq \mathbb{R}^2$, where $a<b$ and $c<d$. The sets $[a,b]$ and $[c,d]$ are intervals: if $x_0,x_1\in [a,b]$ and $x_0<t<x_1$, then \begin{align*} a \le x_0 < t < x_1 \le b, \end{align*} so $a\le t\le b$ and hence $t\in [a,b]$; the same argument shows that $[c,d]$ is an interval. Therefore $[a,b]$ and $[c,d]$ are connected by *Connected Subsets of the Real Line*. Since \begin{align*} R=[a,b]\times[c,d], \end{align*} the product theorem *Product of Connected Spaces* gives that $R$ is connected. Now let $g:R\to \mathbb{R}$ be continuous. Because $R$ is connected, *Continuous Image of a Connected Space* implies that $g(R)$ is connected as a subspace of $\mathbb{R}$. Applying *Connected Subsets of the Real Line* again, $g(R)$ is an interval. Thus if $r_0,r_1\in R$ and \begin{align*} g(r_0)<s<g(r_1), \end{align*} then $g(r_0),g(r_1)\in g(R)$ and the interval property forces $s\in g(R)$. Equivalently, there exists $r\in R$ such that $g(r)=s$. Hence a continuous real-valued function on a rectangle cannot take two values while omitting a value between them. [/example] These permanence properties explain why connectedness is common in geometry: many spaces are built from intervals by products, quotients, and controlled gluing. ## Components and Maximal Connected Pieces Disconnected spaces are not useless. They often decompose into maximal connected regions, and those regions are the correct pieces on which continuous real-valued functions have intermediate value behavior. The next definition captures the largest connected subspace containing a point. ### Connected Components A disconnected space may still contain connected regions around a point, but choosing one by hand is not canonical when several connected subsets overlap. The intrinsic construction is to gather all connected subsets containing the point and ask for the largest connected piece forced by that point. [definition: Connected Component] Let $(X, \tau)$ be a topological space and let $x \in X$. The connected component of $x$ in $X$ is the union of all connected subsets of $X$ that contain $x$. [/definition] This definition is meaningful because every singleton $\{x\}$ is connected, so the union is taken over a nonempty family. What still has to be checked is that these pointwise unions behave like genuine pieces rather than overlapping enlargements. The structure theorem below proves that components are maximal connected sets, form a partition, and are closed in the ambient space. [quotetheorem:302] Components partition a space into its largest connected pieces. They need not be open, which is one of the first surprises in general topology. [example: Components of the Rational Line] Let $\mathbb{Q}$ carry the subspace topology inherited from $\mathbb{R}$. We show that every connected subset of $\mathbb{Q}$ has at most one point. Take distinct $p,q \in \mathbb{Q}$ with $p<q$. Choose an irrational number $\alpha$ with \begin{align*} p<\alpha<q. \end{align*} Set \begin{align*} U &= \mathbb{Q}\cap (-\infty,\alpha), \\ V &= \mathbb{Q}\cap (\alpha,\infty). \end{align*} Since $(-\infty,\alpha)$ and $(\alpha,\infty)$ are open in $\mathbb{R}$, the sets $U$ and $V$ are open in $\mathbb{Q}$ by the definition of the subspace topology. They are nonempty because \begin{align*} p \in \mathbb{Q}, \quad p<\alpha \end{align*} imply $p\in U$, and \begin{align*} q \in \mathbb{Q}, \quad \alpha<q \end{align*} imply $q\in V$. The two sets are disjoint. If $x\in U\cap V$, then \begin{align*} x &\in \mathbb{Q}\cap (-\infty,\alpha), \\ x &\in \mathbb{Q}\cap (\alpha,\infty), \end{align*} so \begin{align*} x<\alpha \quad\text{and}\quad \alpha<x, \end{align*} which is impossible. Thus \begin{align*} U\cap V=\varnothing. \end{align*} They also cover $\mathbb{Q}$. If $r\in \mathbb{Q}$, then $r\neq \alpha$ because $\alpha$ is irrational. Hence either $r<\alpha$ or $\alpha<r$, so \begin{align*} r\in U\cup V. \end{align*} Therefore \begin{align*} U\cup V=\mathbb{Q}. \end{align*} Thus $(U,V)$ is a separation of $\mathbb{Q}$ placing $p$ and $q$ on different sides. Now let $C\subset \mathbb{Q}$ be any subset containing both $p$ and $q$. The sets \begin{align*} C\cap U \quad\text{and}\quad C\cap V \end{align*} are open in $C$ by the subspace topology. They are nonempty because $p\in C\cap U$ and $q\in C\cap V$. Also, \begin{align*} (C\cap U)\cap (C\cap V) &= C\cap (U\cap V) \\ &= C\cap \varnothing \\ &= \varnothing, \end{align*} and \begin{align*} (C\cap U)\cup (C\cap V) &= C\cap (U\cup V) \\ &= C\cap \mathbb{Q} \\ &= C. \end{align*} So $C$ is disconnected. Hence no connected subset of $\mathbb{Q}$ contains two distinct points. For each $q\in\mathbb{Q}$, the singleton $\{q\}$ is connected because it cannot be written as a union of two nonempty disjoint subsets. Since no larger connected subset of $\mathbb{Q}$ can contain $q$, the connected component of $q$ is exactly $\{q\}$. Thus the connected components of the rational line are precisely the singleton sets. [/example] ### Total Disconnectedness The rational line shows that having many nearby points is not the same as having connected pieces of positive size. This motivates a term for spaces whose connected components are as small as possible. [definition: Totally Disconnected Space] A topological space $(X, \tau)$ is totally disconnected if every connected component of $X$ is a singleton. [/definition] Total disconnectedness is a strong form of fragmentation, but it is weaker than being discrete. A continuous map can collapse or merge connected pieces, but it cannot send one connected piece across a separation in the target. The compatibility result below makes this obstruction precise by locating the image of a component inside a single component of the codomain. [quotetheorem:4998] This theorem says that components are functorial in a weak but useful sense: a continuous map cannot scatter one connected component across several disconnected pieces of the target. ## Path Connectedness Many spaces encountered in geometry are connected because any two points can be joined by a continuous path. This is stronger than connectedness, and it is often easier to verify. The distinction matters because topology includes spaces where global connectedness exists without any path joining certain points. ### Paths and Path Components A path gives a concrete witness that two points lie in the same moving piece of a space. Before comparing this idea with connectedness, we name the map that records such a motion. [definition: Path] Let $(X, \tau)$ be a topological space. A path in $X$ from $x_0 \in X$ to $x_1 \in X$ is a continuous map \begin{align*} \gamma: [0,1] &\to X \end{align*} such that $\gamma(0)=x_0$ and $\gamma(1)=x_1$. [/definition] A path is a parameterized motion through the space, but the definition only concerns one chosen pair of endpoints. To turn this local notion into a property of the whole space, we ask whether paths are available uniformly between arbitrary pairs of points. This gives a stronger kind of connectedness, one based on explicit routes rather than only on the absence of separations. [definition: Path Connected Space] A topological space $(X, \tau)$ is path connected if for every $x_0,x_1 \in X$ there exists a path in $X$ from $x_0$ to $x_1$. [/definition] [Path connectedness](/page/Path%20Connectedness) supplies connected subsets as images of intervals. If every point can be reached from a basepoint by one of these interval images, then any attempted separation would have to split an interval image, which connected intervals do not allow. The theorem below turns that obstruction into the standard implication from path connectedness to connectedness. [quotetheorem:300] The converse fails. This failure is important because it warns us not to replace connectedness by path connectedness in theorems where only separation arguments are available. [example: The Topologist's Sine Curve] Let \begin{align*} S=\{(x,\sin(1/x)):0<x\le 1\}\subset \mathbb{R}^2, \end{align*} and set \begin{align*} T=\overline{S}. \end{align*} First we identify the closure. If $(x_n,\sin(1/x_n))\in S$ converges to $(x,y)\in \mathbb{R}^2$, then $0<x_n\le 1$, so $0\le x\le 1$. If $x>0$, then eventually $x_n>0$ and continuity of $u\mapsto \sin(1/u)$ at $x$ gives \begin{align*} y &=\lim_{n\to\infty}\sin(1/x_n) \\ &=\sin(1/x), \end{align*} so $(x,y)\in S$. If $x=0$, then $-1\le \sin(1/x_n)\le 1$ for every $n$, hence $-1\le y\le 1$. Thus \begin{align*} \overline{S}\subset S\cup(\{0\}\times[-1,1]). \end{align*} Conversely, every point of $S$ belongs to $\overline{S}$. Now take $y\in[-1,1]$. Choose $\theta\in[-\pi/2,\pi/2]$ with $\sin\theta=y$, and define \begin{align*} x_n=\frac{1}{\theta+2\pi n} \end{align*} for all sufficiently large $n$, so that $\theta+2\pi n>0$ and $x_n\le 1$. Then \begin{align*} x_n&\to 0,\\ \sin(1/x_n) &=\sin(\theta+2\pi n)\\ &=\sin\theta\\ &=y. \end{align*} Therefore \begin{align*} (x_n,\sin(1/x_n))\to (0,y), \end{align*} so $(0,y)\in\overline{S}$. Hence \begin{align*} T=\overline{S}=S\cup(\{0\}\times[-1,1]). \end{align*} The map \begin{align*} g:(0,1]&\to \mathbb{R}^2\\ x&\mapsto (x,\sin(1/x)) \end{align*} is continuous because its coordinate functions $x\mapsto x$ and $x\mapsto \sin(1/x)$ are continuous on $(0,1]$. Since $(0,1]$ is an interval, it is connected by *Connected Subsets of the Real Line*. Therefore $S=g((0,1])$ is connected by *Continuous Image of a Connected Space*, and $T=\overline{S}$ is connected by the standard fact that the closure of a connected subspace is connected. It remains to see why this connected space is not path connected. Suppose, for contradiction, that there is a path \begin{align*} \gamma:[0,1]&\to T\\ t&\mapsto (u(t),v(t)) \end{align*} from a point of $\{0\}\times[-1,1]$ to a point of $S$. Then $u(0)=0$ and $u(1)>0$. Choose $t_1\in[0,1]$ with $u(t_1)>0$, and let \begin{align*} a=\sup\{t\in[0,t_1]:u(t)=0\}. \end{align*} The set $\{t\in[0,t_1]:u(t)=0\}$ is closed because $u$ is continuous, so $u(a)=0$. By the definition of $a$, we have $u(t)>0$ for every $t\in(a,t_1]$. Hence for $t\in(a,t_1]$, the point $\gamma(t)$ lies in $S$, and therefore \begin{align*} v(t)=\sin(1/u(t)). \end{align*} Let $c=u(t_1)>0$. For all sufficiently large $n$, define \begin{align*} r_n^+=\frac{1}{\pi/2+2\pi n}, \qquad r_n^-=\frac{1}{3\pi/2+2\pi n}, \end{align*} so that $0<r_n^+<c$ and $0<r_n^-<c$. Since $u(a)=0$ and $u(t_1)=c$, the intermediate value theorem gives points $s_n,t_n\in[a,t_1]$ such that \begin{align*} u(s_n)&=r_n^+,\\ u(t_n)&=r_n^-. \end{align*} Because $r_n^+,r_n^-\to 0$, these times must satisfy $s_n\to a$ and $t_n\to a$: otherwise, for some $\varepsilon>0$ infinitely many of them would lie in $[a+\varepsilon,t_1]$, where the positive continuous function $u$ has a positive minimum. Now compute the corresponding second coordinates: \begin{align*} v(s_n) &=\sin(1/u(s_n))\\ &=\sin(1/r_n^+)\\ &=\sin(\pi/2+2\pi n)\\ &=1, \end{align*} while \begin{align*} v(t_n) &=\sin(1/u(t_n))\\ &=\sin(1/r_n^-)\\ &=\sin(3\pi/2+2\pi n)\\ &=-1. \end{align*} Since $s_n\to a$ and $t_n\to a$, continuity of $v$ gives \begin{align*} v(a)&=\lim_{n\to\infty}v(s_n)=1,\\ v(a)&=\lim_{n\to\infty}v(t_n)=-1, \end{align*} which is impossible. Therefore no path in $T$ joins the vertical segment $\{0\}\times[-1,1]$ to the oscillating graph $S$. The space $T$ is connected, but its connectedness comes from accumulation, not from paths between all of its points. [/example] [illustration:topologists-sine-curve] ### Local Path Connectedness The topologist's sine curve separates two intuitions: connectedness forbids open separation, while path connectedness asks for parameterized routes. To recover the geometric equivalence between components and path components, we need a local supply of path connected neighbourhoods. [definition: Locally Path Connected Space] A topological space $(X, \tau)$ is locally path connected if for every $x \in X$ and every open neighbourhood $U \subset X$ of $x$, there exists an [open set](/page/Open%20Set) $V \subset X$ such that $x \in V \subset U$ and $V$ is path connected. [/definition] Local path connectedness rules out the infinitely oscillatory obstruction seen above by ensuring that sufficiently small neighbourhoods already contain path connected neighbourhoods. Without that local supply, a connected component can be too thin for paths to detect. The theorem below identifies the situation in which components and path components coincide and become open pieces of the space. [quotetheorem:1058] This theorem explains why in manifolds, graphs, CW complexes, and open subsets of Euclidean space, connectedness is often handled by constructing paths. The local topology supplies enough small path connected neighbourhoods to turn global connected pieces into path components. ## Compact Connected Spaces and Continua Compactness and connectedness interact well. Compactness controls infinite covers, while connectedness prevents separation. Together they produce spaces that behave like generalized intervals: globally controlled and unbroken, without relying on any ambient Euclidean notion of boundedness. ### Continua The standard name for a compact connected [Hausdorff space](/page/Hausdorff%20Space) is a continuum. This term is common in point-set topology and dynamical systems, where such spaces may be far more complicated than intervals. [definition: Continuum] A continuum is a nonempty compact connected Hausdorff topological space. [/definition] The Hausdorff hypothesis ensures that compact subsets behave like closed geometric objects. Continua include intervals, circles, closed disks, and many fractal compact connected sets. [example: The Closed Disk Is a Continuum] Let \begin{align*} D = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \le 1\}. \end{align*} The set is nonempty because \begin{align*} 0^2+0^2=0\le 1, \end{align*} so $(0,0)\in D$. It is closed in $\mathbb{R}^2$ because the function \begin{align*} F:\mathbb{R}^2&\to\mathbb{R}\\ (x,y)&\mapsto x^2+y^2 \end{align*} is continuous and \begin{align*} D=F^{-1}((-\infty,1]). \end{align*} It is bounded because if $(x,y)\in D$, then \begin{align*} x^2+y^2&\le 1,\\ x^2&\le 1,\\ y^2&\le 1, \end{align*} so $-1\le x\le 1$ and $-1\le y\le 1$. Hence \begin{align*} D\subset [-1,1]\times[-1,1]. \end{align*} Therefore $D$ is compact by *Heine-Borel*. Also $D$ is Hausdorff because distinct points of $D$ are distinct points of $\mathbb{R}^2$, can be separated by disjoint open sets in $\mathbb{R}^2$, and intersecting those open sets with $D$ gives disjoint open neighbourhoods in the subspace topology. To show connectedness, fix $(x,y)\in D$ and define \begin{align*} \gamma:[0,1]&\to D\\ t&\mapsto (tx,ty). \end{align*} The coordinate functions $t\mapsto tx$ and $t\mapsto ty$ are continuous, so $\gamma$ is continuous. For each $t\in[0,1]$, \begin{align*} (tx)^2+(ty)^2 &= t^2x^2+t^2y^2\\ &= t^2(x^2+y^2)\\ &\le t^2\cdot 1\\ &= t^2\\ &\le 1, \end{align*} so $\gamma(t)\in D$. Also \begin{align*} \gamma(0)&=(0\cdot x,0\cdot y)=(0,0),\\ \gamma(1)&=(1\cdot x,1\cdot y)=(x,y). \end{align*} Thus every point of $D$ can be joined to $(0,0)$ by a path in $D$, so $D$ is path connected. By *Path Connected Implies Connected*, $D$ is connected. The disk is nonempty, compact, connected, and Hausdorff, so it is a continuum. [/example] ### Real-Valued Functions on Continua A continuum is compact enough for continuous real-valued functions to attain extrema, and connected enough to fill every value between those extrema. The possible obstruction would be a continuous function whose image has a gap or misses its endpoint values. Compactness removes the endpoint problem, while connectedness removes gaps, giving the compact intermediate value principle below. [quotetheorem:4999] This theorem is a compact version of the intermediate value principle. Compactness adds endpoint attainment, while connectedness fills the interval between the endpoints. ## Beyond and Connected Topics Connectedness is the entry point to several larger parts of topology. In algebraic topology, path connectedness controls the basepoint dependence of the fundamental group, while connected components form the zeroth homotopy invariant $\pi_0(X)$. The pages [Cambridge II Algebraic Topology](/page/Cambridge%20II%20Algebraic%20Topology) and [Cambridge III Algebraic Topology](/page/Cambridge%20III%20Algebraic%20Topology) continue this direction by studying how connectedness interacts with homotopy, covering spaces, and homology. In analysis, connectedness is the topological mechanism behind intermediate value arguments and many uniqueness principles. A continuous function into a discrete set is locally constant, and on a connected domain it must be constant. This theme appears already in [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes) and becomes more structural in [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). In point-set topology, connectedness branches into local connectedness, continua, cut points, indecomposable continua, and total disconnectedness. These topics show that connectedness is not merely a property of intervals and manifolds; it also measures subtle global coherence in compact and non-metrizable spaces. ## References Androma, [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes). Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge II Algebraic Topology](/page/Cambridge%20II%20Algebraic%20Topology). Androma, [Cambridge III Algebraic Topology](/page/Cambridge%20III%20Algebraic%20Topology). James R. Munkres, *Topology* (2000). John L. Kelley, *General Topology* (1955). Stephen Willard, *General Topology* (1970).