General Topology/Connected spaces

That is, a space is path-connected if and only if between any two points, there is a path.

Here we have a partial converse to the fact that path-connectedness implies connectedness:

{{proof|First note that path-connected spaces are connected. Then suppose that $$X$$ is connected, fix $$x_0 \in X$$, and define the set
 * $$S := \{y \in X| \exists \gamma:[a,b] \to X: \gamma(a) = x_0, \gamma(b) = y$$.

Note that $$S$$ is open, since if $$y \in S$$, then by local path-connectedness we may pick a path-connected open neighbourhood $$U$$ of $$y$$, so that by applying concatenation, we see that all points in $$U$$ are in $$S$$. On the other hand, $$X \setminus S$$ is open, pretty much by the same argument: If $$z \in X \setminus S$$ and $$U$$ is a path-connected open neighbourhood of $$x$$, then $$U \subseteq X \setminus S$$, since if $$U$$ would contain a point $$y$$ of $$S$$, $$x_0$$ could be joined to $$z$$ by a path, concatenating a path from $$x_0$$ to $$y$$ to one from $$y \to z$$, in contradiction to $$z \notin S$$. Hence $$S$$ is open and closed, and since $$x_0 \in S$$, $$S \neq \emptyset$$, so that $$S = X$$ by connectedness.}}

This theorem has an important application: It proves that manifolds are connected if and only if they are path-connected. Also, later in this book we'll get to know further classes of spaces that are locally path-connected, such as simplicial and CW complexes.

By substituting "connected" for "path-connected" in the above definition, we get:

Exercises

 * 1) Prove that whenever $$X$$ is a connected topological space and $$Y$$ is a topological space and $$f:X \to Y$$ is a continuous function, then $$f(X)$$ is connected with the subspace topology induced on it by $$Y$$.
 * 2) Prove that similarly if $$X$$ is a path-connected top. space, $$Y$$ top. space, $$f:X \to Y$$ continuous, then $$f(X)$$ is path-connected with the subspace topology induced on it by $$Y$$.
 * 3) Prove that a topological space $$X$$ is connected if and only if, when $$\chi_A: X \to \{0,1\}$$ for $$A \subseteq X$$ denotes the indicator function, the only indicator functions which are continuous are the ones where $$A = \emptyset$$ and $$A = X$$; here $$\{0,1\}$$ has the discrete topology.
 * 4) Let $$X := \{(0,0)\} \cup \{(x,y)| \exists n \in \mathbb N: y = 1/n\}$$ with the subspace topology induced by the Euclidean topology of $$\mathbb R^2$$. Prove that $$X$$ is not locally connected by proving that $$(0,0)$$ does not have a connected neighbourhood.