Relation between components and path-components of ##X##

[Terrell]

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, $f([a,b])=P$.

 

[WWGD]

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, $f([a,b])=P$.

Can you define also local-path-connectedness, please? EDIT: Maybe too, you can use the fact that $\mathbb R^n$ satisfies , is an example of this, i.e., the path-components agree with the connected components. You seem to be assuming path-connected implies connected $P \subset f([a,b])$ which is true but it would be nice to have an argument for it. And maybe you can specify whether C is a path-component or a connected component. And it is not just "some" component, it is the component containing $x_1, x_2$, I think this is essential here. It seems too, that you could make a proof by showing both $P \subset C$ and $C \subset P$ using local-path connectedness, so that you showed path-components of locally-path connected spaces agree with connected components of those spaces.

 

[Terrell]

You seem to be assuming path-connected implies connected P⊂f([a,b])P⊂f([a,b]) P \subset f([a,b]) which is true but it would be nice to have an argument for it.

I meant $C$ is a connected component of $X$. I assumed that if the path $f$ in $P$ is inside the connected subspace $f([a,b]) \subset C$, then $P \subset C$; which I think now is insufficient. Is this what you were trying to point out that is deficient? Thanks!

 

[WWGD]

I meant $C$ is a connected component of $X$. I assumed that if the path $f$ in $P$ is inside the connected subspace $f([a,b]) \subset C$, then $P \subset C$; which I think now is insufficient. Is this what you were trying to point out that is deficient? Thanks!

Yes, partly. A disconnection would not allow the path to remain within the space; would be "broken" up by the disconnection. So, yes, the path component of an element is contain within its connected component. EDIT: I was trying to throw in some ideas. I guess you know of the standard example of the topologist's sine curve Sin(1/x) as an example of a connected but not path-connected space?

 

[Terrell]

Yes, partly. A disconnection would not allow the path to remain within the space; would be "broken" up by the disconnection. So, yes, the path component of an element is contain within its connected component. EDIT: I was trying to throw in some ideas. I guess you know of the standard example of the topologist's sine curve Sin(1/x) as an example of a connected but not path-connected space?

I haven't studied examples, yet. Thanks as you were helpful for pointing this out!