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

• Terrell
In summary: I am not sure if it is that simple, but these are some ideas I had when reading your attempt at a proof.In summary, the theorem states that in a topological space, the path components of the space are contained within the connected components of the space. If the space is also locally path-connected, then the path components and the connected components are the same. The attempt at a proof begins by defining the terms path and path-connectedness, as well as components and locally connectedness. The author then shows that the path component is a subset of a connected subspace and that this subspace is contained within a component of the space. When considering local-path connectedness, the author shows that any point in the connected subspace can
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.

## 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##.

Terrell said:

## 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.

## 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.

Last edited:
Terrell
WWGD said:
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
Terrell said:
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?

Last edited:
Terrell
WWGD said:
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!

WWGD

## What is the difference between components and path-components of X?

The components of X are the maximal connected subsets of X, while the path-components of X are the maximal subsets of X that can be connected by a path. In other words, the components of X are determined by connectedness, while the path-components are determined by path-connectedness.

## How are the components and path-components related in a topological space?

In a topological space, the path-components are always contained within the components. This means that every path-component is also a component, but not every component is a path-component.

## Can a topological space have infinitely many components and path-components?

Yes, a topological space can have infinitely many components and path-components. For example, the real line has infinitely many components and path-components, as any interval on the real line is both a component and a path-component.

## What is the significance of studying the relation between components and path-components of X?

Understanding the relation between components and path-components can provide insight into the structure of a topological space. It can also help in distinguishing between different topological spaces, as the number and arrangement of components and path-components can vary between spaces.

## How can the relation between components and path-components be used in practical applications?

The relation between components and path-components can be used in various fields such as physics, engineering, and computer science. For example, in physics, the components and path-components of a physical system can provide information about its stability and behavior. In computer science, this relation can be used in data analysis and network optimization.

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