- #1

Terrell

- 317

- 26

## Homework Statement

Theorem: If ##X## is a topological space, each path component of ##X## lies in a component of ##X##. If ##X## is locally path connected, then the components and the path components of ##X## are the same.

I need help locating errors in my proof. Please help.

## Homework Equations

Definition (

**path and path-connected**). Given points ##x## and ##y## of the space ##X##, a path in ##X## from ##x## to ##y## is a continuous map ##f:[a,b]\to X## of some closed interval in the real line into ##X##, such that ##f(a)=x## and ##f(b)=y##. A space ##X## is said to be path connected if every pair of points of ##X## can be joined by a path ##X##.

Definition (

**components**).

Given ##X##, define an equivalence relation on ##X## by setting ##x \sim y## if there is a connected subspace of ##X## containing both ##x## and ##y##. The equivalence classes are the components (or the "connected components") of ##X##.

Definition(

**locally connected**). A space ##X## is said to be locally connected at ##x## if for every neighborhood ##U## of ##x##, there is a connected neighborhood ##V## of ##x## contained in ##Y##. If ##X## is locally connected at each of its points, it is said simply to be locally connected.

## The Attempt at a Solution

Let ##X## be a topological space and ##P## be a path component of ##X##. Note ##\forall x_1,x_2 \in P##, there exist a continuous function ##f:[a,b]\subset\Bbb{R}\to X## such that ##f(a)=x_1## and ##f(b)=x_2##. Since ##[a,b]## is connected, then ##f([a,b])## is also a connected subspace of ##X## and ##x_1,x_2 \in f([a,b])##. Thus, ##P\subseteq f([a,b])##. Since connnected subspaces lie entirely within a component, let ##f([a,b])\subset C## where ##C## is some component of ##X##. Hence, ##P\subset C##.

Moreover, what if we also assume that ##X## is locally path-connected. So far we have ##x_1,x_2 \in P## and ##f(a)=x_1, f(b)=x_2 \in f([a,b])##. Let ##x'\in f([a,b])##. We want to show ##f([a,b])\subset P##. Then, by local path-connectedness of ##X##, for every open set ##O'## in ##X## such that ##x' \in O'## imply there exists a path-connected neighborhood ##\mathscr{P'}## such that ##x'\in \mathscr{P'}\subset O'##. Since ##x'\in f([a,b])##, then ##\exists c\in [a,b]\subset\Bbb{R}(f(c)=x')##. Since ##f## is continuous, then ##g=f|_{[a,c]}## and ##h=f|_{[c,b]}## are continuous and by definition of a path, ##g(a)=x_1##, ##g(c)=x'##, ##h(c)=x'##, and ##h(b)=x_2## imply there exist paths between ##x_1## and ##x'## and also ##x'## and ##x_2##; denoted, ##x_1\sim x'## and ##x'\sim x_2## where ##\sim## relates two point linked by a path. Since ##\forall\alpha\in\mathscr{P'}(\alpha\sim x')##, then, by transitivity of ##\sim##, ##\alpha\sim x_1## and ##\alpha\sim x_2##. By the symmetry of ##\sim##, ##\mathscr{P'}\subset P##, ##P\subset\mathscr{P'}##, ##f([a,b])\subset\mathscr{P'}##, and ##\mathscr{P'}\subset f([a,b])##. Therefore, ##f([a,b])=P##.