# Mathematics Prelims

## March 18, 2009

### An Aside on Path Connectedness

Filed under: Topology — cjohnson @ 9:31 am

Long ago we talked about topological spaces being connected, which meant we couldn’t partition the space into two disjoint non-empty open sets.  There is a stronger notion that we’ll need when working with the fundamental group called path connectedness.  We say that a space $X$ is path connected if for every pair of points $x, y \in X$ there exists a path in the space, $f : [0, 1] \to X$, such that $f(0) = x$ and $f(1) = y$; we can connect every pair of points with a path.

If a space is not path connected, it can be decomposed into a collection of disjoint subsets with are path connected.  These subsets are called the path components of the space.  More specifically, path connectedness is an equivalence relation on the space and the path components are the equivalence classes.  Clearly every point is path connected to itself (reflexivity); if $x$ and $y$ are connected by path $f$, then the reverse path $\overline{f}$ connects $y$ and $x$ (symmetry); and the concatenation operation from last time gives us transitivity.

We said that path connectedness was a stronger condition than connectedness.  This means that path connectedness implies regular ol’ connectedness.  To see this, suppose that $X$ is path connected.  Let $u, v \in X$ be connected by path $f$.  Let $U$ be an open set containing $u$, and $V$ an open set containing $v$.  Now consider the preimages $f^{-1}(U)$ and $f^{-1}(V)$.  Since $f$ is a path it is continuous, so these are two open subsets of [0, 1].  We must have $f^{-1}(U) \cup f^{-1}(V) = [0, 1]$ since $f(0) = u$ and $f(1) = v$.  We know that [0, 1] is connected, however.  This implies that $U \cup V$ must be connected (otherwise their preimages would be disjoint open sets covering [0, 1]).  Since $U$ and $V$ were arbitrary, we can generalize to get that $X$ is connected.

Notice that this also means that a continuous path must stay inside of one path component.