Connectedness

Let $X$ be a topological space. A separation of $X$ is a pair of disjoint non-empty open sets whose union is $X.$ $X$ is disconnected if it has a separation and connected otherwise, and a subset $A\subseteq X$ is connected if it is connected as a subspace. Equivalently, $X$ is connected if and only if the only subsets of $X$ that are both open and closed are $\varnothing$ and $X.$

A subset of $\mathbb{R}$ is connected if and only if it is an interval. The rational numbers are disconnected, since $\left\{x\in\mathbb{Q}:x<\sqrt{2}\right\}$ and $\left\{x\in\mathbb{Q}:x>\sqrt{2}\right\}$ separate them, and a set with the discrete topology is connected only if it has at most one point. Connected sets have the following properties:

• the closure of a connected set is connected, as is any set between a connected set and its closure;
• the union of connected sets with a common point is connected;
• the product of finitely many connected spaces is connected;
• the continuous image of a connected set is connected.

The last property states that if $f:X\rightarrow Y$ is continuous and $A\subseteq X$ is connected then $f(A)$ is connected. Since the connected subsets of $\mathbb{R}$ are the intervals, a continuous real-valued function on a connected space takes every value between any two of its values, which extends the intermediate value theorem. Connectedness is a topological property, so an interval is not homeomorphic to a circle, which stays connected when a point is removed.

Let $x\in X.$ The connected component of $x$ is the union of all connected subsets of $X$ containing $x,$ which is the largest connected set containing $x.$ The connected components partition $X$ into closed sets, and $X$ is connected if and only if it has exactly one component. The components of $\mathbb{Q}$ are its single points.

A path in $X$ from $x$ to $y$ is a continuous function $\gamma:\left[0,1\right]\rightarrow X$ with $\gamma(0)=x$ and $\gamma(1)=y.$ $X$ is path-connected if any two of its points are joined by a path, and the path component of $x$ is the set of points joined to $x$ by a path. A subset of $\mathbb{R}^n$ that contains the line segment $\left\{\left(1-t\right)\underline{x}+t\,\underline{y}:t\in\left[0,1\right]\right\}$ between any two of its points is path-connected by those segments, so balls, intervals and $\mathbb{R}^n$ are path-connected, as are spheres and $\mathbb{R}^n$ with a point removed when $n\geq 2.$ The continuous image of a path-connected set is path-connected.

A path-connected space is connected, since the image of each path is connected and the paths from a fixed point share that point. The converse is false: the topologist’s sine curve, the closure in $\mathbb{R}^2$ of the graph of $x\mapsto\sin\frac{1}{x}$ on $\left(0,1\right],$ is connected but no path joins the graph to the segment $\left\{0\right\}\times\left[-1,1\right]$ in its closure. An open connected subset of $\mathbb{R}^n$ is path-connected, and any two of its points are joined by a path made of finitely many line segments, so the two notions agree for the open sets on which vector fields are defined.