A [connected](/page/Connectedness) space cannot be split into two disjoint open pieces. But this algebraic condition — the absence of a separation — does not guarantee that you can "walk" continuously from one point to another. The topologist's sine curve is connected (it cannot be partitioned into two open halves), yet no continuous path joins the oscillatory graph to the vertical segment at the origin. The fundamental difficulty is that connectedness is a *global* condition on the open-set structure, while the ability to travel between points is a *local-to-global* geometric property that requires building continuous curves. **Path connectedness** is the stronger, more geometric notion: a space is path-connected if every pair of points can be joined by a continuous map from the unit interval. This condition is strictly stronger than connectedness — every path-connected space is connected, but not conversely — and it is the notion that underpins most geometric and algebraic-topological constructions. The [fundamental group](/page/Fundamental%20Group), covering space theory, and the classification of surfaces all begin with paths. Homotopy theory is, at its foundation, the study of when two paths can be continuously deformed into one another. The gap between the two notions is closed in many natural settings. For [open subsets of Euclidean space](/page/Connectedness), connectedness and path connectedness are equivalent — a fact that depends critically on the local convexity of $\mathbb{R}^n$. More generally, the two notions coincide for any space that is locally path-connected, a condition that captures precisely the local structure needed to piece together short paths into global ones. [example: The Topologist's Sine Curve Is Not Path-Connected] The canonical example demonstrating the gap between connectedness and path connectedness is the **topologist's sine curve**. Define $S \subset \mathbb{R}^2$ by: \begin{align*} S := \left\{\left(x, \sin\frac{1}{x}\right) : x \in (0, 1]\right\} \cup \left(\{0\} \times [-1, 1]\right). \end{align*} The graph $\Gamma := \{(x, \sin(1/x)) : x \in (0, 1]\}$ is the continuous image of the connected interval $(0, 1]$, hence connected. Every point of the vertical segment $I := \{0\} \times [-1, 1]$ is a limit of points of $\Gamma$: for any $y \in [-1, 1]$, the function $\sin$ takes the value $y$ infinitely often on $(0, \varepsilon)$ for every $\varepsilon > 0$, so there exist $x_k \to 0^+$ with $\sin(1/x_k) = y$. Therefore $S = \overline{\Gamma}$, and since the [closure of a connected set is connected](/page/Connectedness), $S$ is connected. However, $S$ is **not** path-connected. Suppose for contradiction that $\gamma: [0, 1] \to S$ is a continuous path with $\gamma(0) = (0, 0) \in I$ and $\gamma(1) = (1, \sin 1) \in \Gamma$. Write $\gamma(t) = (\gamma_1(t), \gamma_2(t))$. Since $\gamma_1$ is continuous and $\gamma_1(0) = 0$, the set $\{t \in [0, 1] : \gamma_1(t) = 0\}$ is closed; let $t_0$ be its supremum. Then $\gamma_1(t_0) = 0$, and for $t > t_0$ we have $\gamma_1(t) > 0$, so $\gamma_2(t) = \sin(1/\gamma_1(t))$. As $t \to t_0^+$, we have $\gamma_1(t) \to 0^+$, so $\sin(1/\gamma_1(t))$ oscillates between $-1$ and $1$ without converging. But continuity of $\gamma_2$ requires $\gamma_2(t) \to \gamma_2(t_0)$, a contradiction. The essential mechanism is that the graph oscillates with unbounded frequency near $x = 0$: any path approaching the vertical segment from the graph side must traverse infinitely many oscillations in finite "time," which is incompatible with continuity. This phenomenon cannot occur in an open subset of $\mathbb{R}^n$, where every point has a convex (hence path-connected) neighbourhood. [/example] ## Definition The definition of path connectedness requires two preliminary notions: what a path is, and what it means to concatenate paths. [definition: Path] Let $X$ be a [topological space](/page/Topology). A **path** in $X$ is a [continuous](/page/Continuity) map $\gamma: [0, 1] \to X$. The point $\gamma(0)$ is the **initial point** (or **start point**) and $\gamma(1)$ is the **terminal point** (or **end point**). We say $\gamma$ is a path **from** $\gamma(0)$ **to** $\gamma(1)$. For points $x, y \in X$, a path from $x$ to $y$ is a continuous map $\gamma: [0, 1] \to X$ with $\gamma(0) = x$ and $\gamma(1) = y$. [/definition] The choice of $[0, 1]$ as the domain is a convention. Any closed interval $[a, b]$ with $a < b$ would serve equally well, since $[a, b]$ is homeomorphic to $[0, 1]$ via the affine map $t \mapsto (t - a)/(b - a)$. The unit interval is chosen for definiteness. [remark: Paths vs. Curves] A path is a *map*, not a *set*. The **image** $\gamma([0, 1]) \subset X$ is sometimes called the **trace** or **arc** of the path, but this is a subset of $X$, not a path itself. Two paths can have the same image but differ as maps — for instance, a path traversed at constant speed and the same path traversed with a pause halfway through are different paths (as functions) with identical images. This distinction becomes critical in the [fundamental group](/page/Fundamental%20Group), where paths are considered up to homotopy, not just up to their images. [/remark] Paths can be composed end-to-end, reversed, and degenerated to a point. These operations give the collection of paths an algebraic structure that is not quite a group (concatenation is not associative on the nose, and reversal is not a strict inverse), but becomes one after passing to homotopy classes. [definition: Path Operations] Let $X$ be a topological space. **Concatenation.** If $\gamma: [0, 1] \to X$ is a path from $x$ to $y$ and $\delta: [0, 1] \to X$ is a path from $y$ to $z$, their **concatenation** $\gamma * \delta: [0, 1] \to X$ is defined by: \begin{align*} (\gamma * \delta)(t) := \begin{cases} \gamma(2t) & \text{if } 0 \le t \le \tfrac{1}{2}, \\ \delta(2t - 1) & \text{if } \tfrac{1}{2} \le t \le 1. \end{cases} \end{align*} This is continuous by the [pasting lemma](/page/Continuity): the two pieces agree at $t = 1/2$ (where $\gamma(1) = y = \delta(0)$), and each piece is continuous on a closed subset of $[0, 1]$. **Reverse.** If $\gamma: [0, 1] \to X$ is a path from $x$ to $y$, the **reverse path** $\bar{\gamma}: [0, 1] \to X$ is defined by $\bar{\gamma}(t) := \gamma(1 - t)$. This is a path from $y$ to $x$. **Constant path.** For any $x \in X$, the **constant path at $x$** is $c_x: [0, 1] \to X$ defined by $c_x(t) := x$ for all $t$. [/definition] With these ingredients, we can state the central definition. [definition: Path-Connected Space] A topological space $X$ is **path-connected** if for every pair of points $x, y \in X$, there exists a path from $x$ to $y$: a continuous map $\gamma: [0, 1] \to X$ with $\gamma(0) = x$ and $\gamma(1) = y$. A subset $A \subset X$ is **path-connected** if $A$ is path-connected in the [subspace topology](/page/Topology) inherited from $X$. [/definition] [remark: The Relation "Connected by a Path" Is an Equivalence Relation] For a topological space $X$, define the relation $x \sim y$ if there exists a path in $X$ from $x$ to $y$. This is an equivalence relation: - **Reflexivity:** the constant path $c_x$ is a path from $x$ to $x$. - **Symmetry:** if $\gamma$ is a path from $x$ to $y$, then $\bar{\gamma}$ is a path from $y$ to $x$. - **Transitivity:** if $\gamma$ is a path from $x$ to $y$ and $\delta$ is a path from $y$ to $z$, then $\gamma * \delta$ is a path from $x$ to $z$. The equivalence classes of this relation are the **path components** of $X$, and $X$ is path-connected if and only if it has exactly one path component. [/remark] ## Path Components The equivalence relation "connected by a path" partitions any topological space into maximal path-connected subsets. Understanding the structure of these pieces — and how they compare to the [connected components](/page/Connectedness) of the same space — clarifies the precise relationship between the two notions of "one piece." [definition: Path Component] Let $X$ be a topological space and $x \in X$. The **path component** of $x$ is the set: \begin{align*} P_x := \{y \in X : \text{there exists a path in } X \text{ from } x \text{ to } y\}. \end{align*} Equivalently, $P_x$ is the equivalence class of $x$ under the relation "connected by a path." [/definition] Each path component is path-connected (by transitivity of the path relation) and is the largest path-connected subset containing $x$: if $A \subset X$ is path-connected and $x \in A$, then $A \subset P_x$. The fundamental comparison between path components and connected components is as follows. Since path connectedness implies connectedness, every path component is contained in a connected component. But a connected component may contain several path components. [quotetheorem:1055] Unlike connected components, path components need not be closed. The topologist's sine curve illustrates this: the graph $\Gamma = \{(x, \sin(1/x)) : x \in (0, 1]\}$ is a path component of $S$ (it is path-connected, and no path joins it to the vertical segment), but $\Gamma$ is not closed in $S$ — its closure is all of $S$. [explanation: Why Path Components Can Fail to Be Closed] Connected components are always closed, because the closure of a connected set is connected. The analogous statement for path components fails: the closure of a path-connected set need not be path-connected. The topologist's sine curve provides the definitive counterexample. The graph $\Gamma = \{(x, \sin(1/x)) : x \in (0, 1]\}$ is path-connected (as the continuous image of the path-connected interval $(0, 1]$), but its closure $S = \overline{\Gamma}$ is not path-connected, as shown in the opening example. Therefore the path component $\Gamma$ has closure $S \supsetneq \Gamma$, and in particular $\Gamma$ is not closed. This failure has structural consequences. Path components do not generally form a "nice" partition of the space: they may be neither open nor closed. The situation improves dramatically when the space is [locally path-connected](/page/Path%20Connectedness) — a condition discussed below — in which case path components coincide with connected components. [/explanation] ## The Asymmetry: Path Connectedness Is Strictly Stronger The fundamental asymmetry of the theory is that path connectedness is strictly stronger than connectedness. Every path-connected space is connected, but the converse fails. Understanding *why* the implication goes in this direction — and *exactly where* the converse breaks down — is essential. [quotetheorem:300] The idea is clean: if $X$ were disconnected, there would be a continuous surjection $f: X \to \{0, 1\}$ (where $\{0, 1\}$ has the discrete topology). Pick any $a \in f^{-1}(0)$ and $b \in f^{-1}(1)$. Since $X$ is path-connected, there exists a path $\gamma: [0, 1] \to X$ from $a$ to $b$. The composition $f \circ \gamma: [0, 1] \to \{0, 1\}$ is a continuous map from the connected interval $[0, 1]$ to the discrete space $\{0, 1\}$, with $(f \circ \gamma)(0) = 0$ and $(f \circ \gamma)(1) = 1$. But a continuous map from a connected space to a discrete space must be constant — contradiction. The converse fails because connectedness does not constrain the *local* geometry of the space. A connected space may have points where the topology is so wild that no continuous curve can navigate the local structure. The topologist's sine curve is the classical example: the vertical segment $\{0\} \times [-1, 1]$ is a limit of the oscillating graph, but the oscillation frequency is unbounded, preventing any continuous path from reaching it. [example: The Deleted Comb Space] Path connectedness is not merely a strengthening of connectedness for "nice" spaces — it can fail even in simple subsets of $\mathbb{R}^2$. Define the **comb space** $Y \subset \mathbb{R}^2$ by: \begin{align*} Y := ([0, 1] \times \{0\}) \cup \bigcup_{n=1}^{\infty} \left\{\frac{1}{n}\right\} \times [0, 1]. \end{align*} This consists of a horizontal base $[0, 1] \times \{0\}$ together with vertical "teeth" at $x = 1/n$, each of height $1$. The comb space $Y$ is path-connected: every tooth meets the base, so any point $(1/n, t)$ can be joined to the origin by travelling down the tooth to $(1/n, 0)$ and then along the base to $(0, 0)$. Now form the **deleted comb space** by adjoining the single limit point $(0, 1)$: \begin{align*} X := Y \cup \{(0, 1)\}. \end{align*} The point $(0, 1)$ is a limit of the tooth tips $(1/n, 1) \to (0, 1)$ as $n \to \infty$, so $X \subset \overline{Y}$. Since $Y$ is connected and $Y \subset X \subset \overline{Y}$, the [closure theorem](/page/Connectedness) guarantees that $X$ is connected. But $X$ is not path-connected: no continuous path in $X$ joins $(0, 1)$ to any point of $Y$. Suppose $\gamma: [0, 1] \to X$ is such a path with $\gamma(0) = (0, 1)$. Write $\gamma(t) = (\gamma_1(t), \gamma_2(t))$. The only points of $X$ with first coordinate $0$ are $(0, 0)$ and $(0, 1)$, so as the path leaves $(0, 1)$, it must immediately jump to a tooth $\{1/n\} \times [0, 1]$. But reaching $(0, 1)$ from the tooth tips requires $\gamma_1(t) \to 0$, which forces the path to visit teeth with arbitrarily large index, and travelling between teeth requires passing through the base at height $\gamma_2 = 0$. Thus $\gamma_2$ would oscillate between values near $1$ (at the tooth tips) and $0$ (on the base) infinitely often as $t \to 0^+$, contradicting the continuity of $\gamma_2$ at $t = 0$. [/example] ## Preservation Under Continuous Maps and Products Path connectedness, like [connectedness](/page/Connectedness), is preserved under continuous surjections and products. These preservation results are among the most frequently used tools for establishing path connectedness of complicated spaces. [quotetheorem:1056] The proof is immediate: for any $y_1, y_2 \in Y$, choose preimages $x_1 \in f^{-1}(y_1)$ and $x_2 \in f^{-1}(y_2)$. Since $X$ is path-connected, there exists a path $\gamma: [0, 1] \to X$ from $x_1$ to $x_2$. Then $f \circ \gamma: [0, 1] \to Y$ is a path from $y_1$ to $y_2$. This result immediately implies that quotient spaces of path-connected spaces are path-connected (since the quotient map is a continuous surjection), and that retracts of path-connected spaces are path-connected. [example: The $n$-Sphere Is Path-Connected for $n \ge 1$] The sphere $S^n = \{x \in \mathbb{R}^{n+1} : |x| = 1\}$ is path-connected for $n \ge 1$. The normalisation map $f: \mathbb{R}^{n+1} \setminus \{0\} \to S^n$ defined by $f(x) = x/|x|$ is a continuous surjection. Since $\mathbb{R}^{n+1} \setminus \{0\}$ is path-connected for $n \ge 1$ (given $a, b \in \mathbb{R}^{n+1} \setminus \{0\}$, the straight-line segment from $a$ to $b$ avoids the origin unless $b = -ta$ for some $t > 0$; in that case, choose any $c$ not on the line through $a$ and $b$ — possible since $n + 1 \ge 2$ — and concatenate the segment from $a$ to $c$ with the segment from $c$ to $b$), the image $S^n$ is path-connected. For $n = 0$, $S^0 = \{-1, 1\}$ is a two-point discrete space, which is not path-connected (and not connected). [/example] Products of path-connected spaces are path-connected, for arbitrary index sets. The proof is more direct than the corresponding result for connectedness, because paths in a product can be built coordinate-by-coordinate. [quotetheorem:1057] Given two points $x, y \in \prod_{\alpha \in A} X_\alpha$, choose for each $\alpha \in A$ a path $\gamma_\alpha: [0, 1] \to X_\alpha$ from $x(\alpha)$ to $y(\alpha)$. Define $\gamma: [0, 1] \to \prod_{\alpha \in A} X_\alpha$ by $\gamma(t)(\alpha) = \gamma_\alpha(t)$. By the universal property of the product topology, $\gamma$ is continuous if and only if each coordinate function $\pi_\alpha \circ \gamma = \gamma_\alpha$ is continuous. Since each $\gamma_\alpha$ is continuous by assumption, $\gamma$ is a path from $x$ to $y$. This argument is *simpler* than the proof that arbitrary products of connected spaces are connected. The connected case requires a density argument (showing that the union of finitely-constrained slices is dense), whereas the path-connected case is entirely constructive. This is one of the advantages of path connectedness over connectedness: it often has more direct proofs because paths provide explicit witnesses. [explanation: Path Connectedness Under Quotients] If $X$ is path-connected and $\sim$ is any equivalence relation on $X$, the quotient space $X / {\sim}$ is path-connected. The quotient map $q: X \to X / {\sim}$ is a continuous surjection, so this follows immediately from the continuous image theorem. Important examples include: - The **torus** $T^2 = \mathbb{R}^2 / \mathbb{Z}^2$ is path-connected (as a quotient of path-connected $\mathbb{R}^2$). - The **projective space** $\mathbb{R}P^n = S^n / \{x \sim -x\}$ is path-connected for $n \ge 1$ (as a quotient of path-connected $S^n$). - The **wedge sum** $X \vee Y = (X \sqcup Y) / \{x_0 \sim y_0\}$ is path-connected whenever $X$ and $Y$ are path-connected (the quotient of the disjoint union, which is not path-connected, but the identification of the basepoints creates a bridge). The quotient result does **not** hold for connectedness in the reverse direction: the quotient of a disconnected space can be connected (the wedge sum example above shows this). [/explanation] ## Convex Sets and Path Connectedness in $\mathbb{R}^n$ In Euclidean space, the most elementary path-connected sets are the convex ones. Convexity provides a canonical choice of path — the straight-line segment — and this geometric simplicity makes convex sets the building blocks for path-connectedness arguments in $\mathbb{R}^n$. [definition: Convex Set] A subset $C \subset \mathbb{R}^n$ is **convex** if for every $x, y \in C$ and every $t \in [0, 1]$, the point $(1 - t)x + ty$ lies in $C$. Equivalently, $C$ contains the entire line segment between any two of its points. [/definition] Every convex set is path-connected: the map $\gamma: [0, 1] \to C$ defined by $\gamma(t) = (1 - t)x + ty$ is a continuous path from $x$ to $y$ lying entirely in $C$. This path is the **straight-line path** or **affine path** from $x$ to $y$. [example: Standard Convex Subsets of $\mathbb{R}^n$] The following sets are convex, hence path-connected: - Open balls $B(x_0, r) = \{x \in \mathbb{R}^n : |x - x_0| < r\}$ and closed balls $\overline{B}(x_0, r)$. - Half-spaces $\{x \in \mathbb{R}^n : \langle a, x \rangle < c\}$ for $a \in \mathbb{R}^n \setminus \{0\}$ and $c \in \mathbb{R}$. - The entire space $\mathbb{R}^n$ and any affine subspace. - Intersections of convex sets (though not unions — the union of two disjoint intervals in $\mathbb{R}$ is not convex). The convexity of open balls is the key geometric fact underlying the equivalence of connectedness and path connectedness for open subsets of $\mathbb{R}^n$: every point in an open set has a convex neighbourhood. [/example] The most important result connecting path connectedness to connectedness in Euclidean space is that for open sets, the two notions are equivalent. This was stated on the [Connectedness](/page/Connectedness) page; we reproduce it here because it is central to working with path connectedness. [quotetheorem:301] The "if" direction is the general fact that path connectedness implies connectedness. The "only if" direction uses the local convexity of $\mathbb{R}^n$. The argument is a model instance of the **clopen argument** (see [Standard Arguments in Connectedness](/page/Connectedness)): fix $a \in U$ and let $W = \{x \in U : \text{there exists a path in } U \text{ from } a \text{ to } x\}$. One shows that $W$ is both open and closed in $U$: - **$W$ is open:** If $x \in W$, choose $r > 0$ with $B(x, r) \subset U$. For any $y \in B(x, r)$, the straight-line path from $x$ to $y$ lies in $B(x, r) \subset U$, so concatenating the path from $a$ to $x$ with this segment gives a path from $a$ to $y$. Thus $B(x, r) \subset W$. - **$W$ is closed in $U$:** If $x \in U \setminus W$, choose $r > 0$ with $B(x, r) \subset U$. If any $y \in B(x, r)$ were in $W$, we could reach $x$ from $a$ via $y$ and a straight-line segment, placing $x \in W$ — contradiction. So $B(x, r) \subset U \setminus W$. Since $U$ is connected and $W$ is non-empty (it contains $a$), $W = U$. The hypothesis that $U$ is open is essential: the topologist's sine curve is a closed connected subset of $\mathbb{R}^2$ that is not path-connected. The argument fails for non-open sets because the key step — finding a ball $B(x, r) \subset U$ — requires $U$ to be open. [example: Star-Shaped Sets Are Path-Connected] A set $A \subset \mathbb{R}^n$ is **star-shaped** with respect to a point $a \in A$ if for every $x \in A$, the segment $\{(1-t)a + tx : t \in [0, 1]\}$ is contained in $A$. Every convex set is star-shaped with respect to any of its points, but star-shaped sets need not be convex: the set \begin{align*} A := \{(x, y) \in \mathbb{R}^2 : |y| \le 1 + x,\; x \ge -1\} \end{align*} is star-shaped with respect to the origin (every segment from $0$ to a point of $A$ stays in $A$) but not convex (the points $(-1, 0)$ and $(0, 1)$ are both in $A$, but the midpoint $(-1/2, 1/2)$ satisfies $|1/2| = 1/2$ and $1 + (-1/2) = 1/2$, so $(-1/2, 1/2)$ is on the boundary; for a non-convex example, take instead the union of two triangles sharing only a vertex). Every star-shaped set is path-connected: for any $x, y \in A$, concatenate the straight-line path from $x$ to $a$ with the straight-line path from $a$ to $y$. Both segments lie in $A$ by the star-shaped condition. Star-shaped domains appear frequently in complex analysis: Cauchy's theorem holds on star-shaped domains (not just simply connected ones), and the existence of antiderivatives for holomorphic functions is often first proved on star-shaped domains before being extended. [/example] ## Local Path Connectedness The topologist's sine curve is connected but not path-connected because its local structure near the vertical segment is pathological: every neighbourhood of $(0, 0)$ in $S$ is disconnected, with infinitely many oscillating "strands" that prevent path construction. The resolution is a local condition — **local path connectedness** — that ensures the space has a well-behaved local geometry compatible with building paths. [definition: Locally Path-Connected Space] A topological space $X$ is **locally path-connected** if for every point $x \in X$ and every open set $U$ containing $x$, there exists a path-connected open set $V$ with $x \in V \subset U$. Equivalently, $X$ is locally path-connected if and only if $X$ has a [basis](/page/Topology) consisting of path-connected open sets. [/definition] Local path connectedness is independent of both connectedness and path connectedness: - $\mathbb{R}^n$ is connected, path-connected, and locally path-connected. - The topologist's sine curve is connected but neither path-connected nor locally path-connected. - The discrete space $\{0, 1\}$ is locally path-connected (each singleton is open and path-connected) but not connected. - $\mathbb{Q}$ (with the subspace topology from $\mathbb{R}$) is neither connected nor path-connected nor locally path-connected: every open interval around a rational contains irrationals, and the subspace $\mathbb{Q} \cap (a - \varepsilon, a + \varepsilon)$ is totally disconnected. The power of local path connectedness lies in its interaction with connectedness: in a locally path-connected space, the two notions of "one piece" collapse into one. [quotetheorem:1058] The key argument for (1) is that in a locally path-connected space, every path component is open. To see this: if $P$ is a path component and $x \in P$, choose a path-connected open neighbourhood $V$ of $x$ (by local path connectedness). Every point $y \in V$ is connected to $x$ by a path in $V$, hence in $X$, so $y \in P$. Thus $V \subset P$, and $P$ is open. Since path components are open, they are also closed (the complement of a path component is a union of other path components, each open, hence the complement is open). So each path component is clopen. But connected components are the maximal connected subsets, and each path component is connected (as it is path-connected) and clopen. A clopen connected subset must be an entire connected component, so path components equal connected components. This result explains why the equivalence of connectedness and path connectedness for open subsets of $\mathbb{R}^n$ is true: open subsets of $\mathbb{R}^n$ are locally path-connected (every point has an open ball neighbourhood, and open balls are convex hence path-connected). The theorem above gives the result immediately. [example: Topological Manifolds Are Locally Path-Connected] A [topological manifold](/page/Manifold) $M$ of dimension $n$ is, by definition, a Hausdorff, second-countable space in which every point has a neighbourhood homeomorphic to an open subset of $\mathbb{R}^n$. Since open subsets of $\mathbb{R}^n$ are locally path-connected (open balls form a path-connected basis), and local path connectedness is preserved by homeomorphisms, every chart domain in $M$ is locally path-connected. Since local path connectedness is a local property (it depends only on the structure of small neighbourhoods), $M$ is locally path-connected. Consequently, for topological manifolds, connectedness and path connectedness are equivalent. A manifold is connected if and only if it is path-connected. This is used without further comment in essentially all of differential geometry and algebraic topology. [/example] [example: The Components of a Locally Path-Connected Space Are Clopen] Consider $\mathrm{GL}_n(\mathbb{R}) = \{A \in \mathbb{R}^{n \times n} : \det A \neq 0\}$. This is an open subset of $\mathbb{R}^{n^2}$, hence locally path-connected. Its connected components — equivalently, its path components — are: \begin{align*} \mathrm{GL}_n^+(\mathbb{R}) &:= \{A : \det A > 0\}, \\ \mathrm{GL}_n^-(\mathbb{R}) &:= \{A : \det A < 0\}. \end{align*} Each is open (as the preimage of an open set under the continuous map $\det$) and path-connected. To see that $\mathrm{GL}_n^+(\mathbb{R})$ is path-connected, one shows that every matrix $A$ with $\det A > 0$ can be connected by a continuous path to the identity $I$. This uses the fact that elementary row operations (which are continuous paths in $\mathrm{GL}_n(\mathbb{R})$) can reduce $A$ to $I$ without the determinant changing sign. The connected component of the identity, $\mathrm{GL}_n^+(\mathbb{R})$, is a normal subgroup of $\mathrm{GL}_n(\mathbb{R})$ (it is the kernel of the group homomorphism $A \mapsto \operatorname{sgn}(\det A)$), and the quotient $\mathrm{GL}_n(\mathbb{R}) / \mathrm{GL}_n^+(\mathbb{R}) \cong \{-1, +1\}$ is the group of connected components. This interaction between topology and algebra — where the identity component is always a normal subgroup — is a general feature of [topological groups](/page/Topological%20Group). [/example] ## Paths as the Foundation of Algebraic Topology Path connectedness is not merely a strengthening of connectedness — it is the entry point to an entire algebraic framework. The fundamental construction of algebraic topology — the [fundamental group](/page/Fundamental%20Group) $\pi_1(X, x_0)$ — is built from paths. Understanding what paths can and cannot tell us about a space motivates the passage from point-set topology to algebraic topology. The basic idea is that while a single path tells us whether two points can be joined, the *space of all paths* between two points carries much richer information. Two paths from $x$ to $y$ may or may not be "essentially the same" — that is, continuously deformable into each other. The equivalence classes of paths under this deformation (called **homotopy**) form an algebraic structure that detects "holes" in the space. [definition: Path Homotopy] Let $X$ be a topological space and let $\gamma, \delta: [0, 1] \to X$ be two paths with the same endpoints: $\gamma(0) = \delta(0) = x$ and $\gamma(1) = \delta(1) = y$. A **path homotopy** from $\gamma$ to $\delta$ is a continuous map $H: [0, 1] \times [0, 1] \to X$ satisfying: \begin{align*} H(s, 0) &= \gamma(s) \quad \text{for all } s \in [0, 1], \\ H(s, 1) &= \delta(s) \quad \text{for all } s \in [0, 1], \\ H(0, t) &= x \quad \text{for all } t \in [0, 1], \\ H(1, t) &= y \quad \text{for all } t \in [0, 1]. \end{align*} If such an $H$ exists, we say $\gamma$ and $\delta$ are **path-homotopic** and write $\gamma \simeq \delta$ (rel endpoints). [/definition] The parameter $s$ traces the path, and $t$ controls the deformation: at $t = 0$ we have $\gamma$, at $t = 1$ we have $\delta$, and at each intermediate $t$ we have a path from $x$ to $y$. In a convex subset of $\mathbb{R}^n$, any two paths with the same endpoints are path-homotopic: the **straight-line homotopy** $H(s, t) = (1-t)\gamma(s) + t\delta(s)$ is a continuous deformation (the convexity of the domain guarantees that $H(s, t)$ remains in the set for all $s, t$). In a space with "holes," this is no longer possible — a path going around the hole cannot be deformed into a path that does not, without leaving the space. [example: Two Non-Homotopic Paths in $\mathbb{R}^2 \setminus \{0\}$] Let $X = \mathbb{R}^2 \setminus \{0\}$ and consider two paths from $(1, 0)$ to $(-1, 0)$: \begin{align*} \gamma(s) &:= (\cos(\pi s), \sin(\pi s)), \\ \delta(s) &:= (\cos(\pi s), -\sin(\pi s)). \end{align*} The path $\gamma$ traverses the upper semicircle of $S^1$, while $\delta$ traverses the lower semicircle. Both paths lie in $X$ (they avoid the origin, since they stay on the unit circle). These two paths are **not** path-homotopic in $X$. The intuition is that any deformation from $\gamma$ to $\delta$ would have to sweep through the origin (which has been removed). The rigorous proof uses the winding number: the concatenation $\gamma * \bar{\delta}$ is a loop around the origin with winding number $1$, and a path homotopy from $\gamma$ to $\delta$ (rel endpoints) would yield a null-homotopy of $\gamma * \bar{\delta}$, contradicting the fact that the winding number is a homotopy invariant. This example illustrates the fundamental principle: path connectedness tells you that paths *exist*, but the fundamental group tells you how many *essentially different* paths there are. The space $\mathbb{R}^2 \setminus \{0\}$ is path-connected (it has one "piece"), but it has fundamental group $\pi_1(\mathbb{R}^2 \setminus \{0\}) \cong \mathbb{Z}$, indicating infinitely many distinct ways to loop around the missing point. [/example] The full development of the fundamental group — the proof that path homotopy classes of loops form a group under concatenation, the dependence on the basepoint, the relationship to covering spaces — belongs to the [Fundamental Group](/page/Fundamental%20Group) page. Here, the essential point is that paths are the *raw material* of algebraic topology: they are the morphisms in the fundamental groupoid, they define the transport maps in fibre bundles, and they underlie every construction from homotopy groups to singular homology. ## Standard Arguments Involving Paths Working with path connectedness involves a small set of recurring techniques. The methods below are the most common tools for establishing or using path connectedness, and they appear throughout topology, geometry, and analysis. ### Proving Path Connectedness: The Clopen Argument To show that a connected space $X$ is path-connected, fix a point $a \in X$ and let $W$ be the set of all points reachable from $a$ by a path. Show that $W$ is both open and closed. If $X$ is connected, $W = X$. This technique requires $X$ to have enough local path-connected structure to make $W$ open. It works for: - Open subsets of $\mathbb{R}^n$ (use convexity of balls). - Topological manifolds (use chart homeomorphisms to reduce to $\mathbb{R}^n$). - CW complexes (use the fact that cells are path-connected and attachment maps are continuous). ### Proving Path Connectedness: The Continuous Surjection Method To show $Y$ is path-connected, exhibit a continuous surjection from a path-connected space onto $Y$. This is often the quickest method and was used above for $S^n$ and quotient spaces. ### Constructing Paths: Concatenation of Local Paths Given a connected, locally path-connected space $X$ and two points $x, y \in X$, one often constructs a path by: 1. Covering a connected set containing $x$ and $y$ with path-connected open sets. 2. Choosing a finite chain $V_1, V_2, \ldots, V_k$ of these open sets with $x \in V_1$, $y \in V_k$, and $V_i \cap V_{i+1} \neq \varnothing$. 3. Picking points $z_i \in V_i \cap V_{i+1}$ and concatenating local paths: $x \to z_1 \to z_2 \to \cdots \to z_{k-1} \to y$. The existence of such a finite chain follows from compactness when the connected set is compact, and from the Lindelof property when the space is second-countable. ### Detecting Non-Path-Connectedness: The Oscillation Argument To show that a space $X$ is *not* path-connected, suppose $\gamma: [0, 1] \to X$ is a path connecting two candidate points and derive a contradiction from the continuity of $\gamma$. The standard technique is to show that a coordinate function of $\gamma$ must oscillate infinitely many times in a finite parameter interval, contradicting uniform continuity (since $[0, 1]$ is compact, every continuous function on $[0, 1]$ is uniformly continuous). This was the method used for both the topologist's sine curve and the deleted comb space. The pattern is: 1. Identify a coordinate $\gamma_i$ that must take widely separated values infinitely often as the parameter $t$ approaches some limit point $t_0$. 2. Uniform continuity of $\gamma_i$ on $[0, 1]$ implies the oscillation $\omega_{\gamma_i}(\delta) = \sup_{|s - t| < \delta} |\gamma_i(s) - \gamma_i(t)| \to 0$ as $\delta \to 0$. 3. The infinite oscillation at $t_0$ forces $\omega_{\gamma_i}(\delta) \ge c > 0$ for all $\delta$, a contradiction. ### Using Path Connectedness: Lifting Paths In covering space theory and fibre bundle theory, the fundamental technique is **path lifting**: given a continuous map $p: E \to B$ (e.g., a covering map) and a path $\gamma$ in the base $B$, construct a path $\tilde{\gamma}$ in the total space $E$ with $p \circ \tilde{\gamma} = \gamma$. The existence and uniqueness of lifts depends on the algebraic properties of the fundamental groups involved. Path connectedness of the base $B$ ensures that the fibre $p^{-1}(b)$ is "uniformly accessible" from any point of $E$: for any $e_1, e_2 \in E$ with $p(e_1)$ and $p(e_2)$ in the same path component of $B$, there is a lifted path from $e_1$ to a point in the fibre over $p(e_2)$. The full theory is developed on the [Covering Space](/page/Covering%20Space) page. ## References - Munkres, J. R., *Topology* (2nd edition, 2000). - Willard, S., *General Topology* (1970). - Hatcher, A., *Algebraic Topology* (2002). - Lee, J. M., *Introduction to Topological Manifolds* (2nd edition, 2011). - Steen, L. A. and Seebach, J. A., *Counterexamples in Topology* (1978).