nLab locally connected topological space

Redirected from "locally connected topological spaces".
Contents

Context

Topology

topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory

Introduction

Basic concepts

Universal constructions

Extra stuff, structure, properties

Examples

Basic statements

Theorems

Analysis Theorems

topological homotopy theory

Contents

Idea

It is not generally true that a topological space is the disjoint union space (coproduct in Top) of its connected components. The spaces such that this is true for all open subspaces are the locally connected topological spaces.

Definition

Definition

(locally connected topological space)

A topological space XX is locally connected if every point has a neighborhood basis of connected open subsets.

Proposition

(alternative characterizations of local connectivity)

For XX a topological space, then the following are equivalent:

  1. XX is locally connected (def. );

  2. every connected component of every open subspace of XX is open;

  3. every open subset, as a topological subspace, is the disjoint union space (coproduct in Top) of its connected components.

In particular, in a locally connected space, every connected component SS is a clopen subset; hence connected components and quasi-components coincide.

Proof

\,

1) \Rightarrow 2)

Assume XX is locally connected, and let UXU \subset X be an open subset with U 0UU_0 \subset U a connected component. We need to show that U 0U_0 is open.

Consider any point xU 0x \in U_0. Since then also xUx \in U, the defintion of local connectedness, def. , implies that there is a connected open neighbourhood U x,0U_{x,0} of XX. Observe that this must be contained in U 0U_0, for if it were not then U 0U x,0U_0 \cup U_{x,0} were a larger open connected open neighbourhood, contradicting the maximality of the connected component U 0U_0.

Hence U 0=xU 0U x,0U_0 = \underset{x \in U_0}{\cup} U_{x,0} is a union of open subsets, and hence itself open.

2) \Rightarrow 3)

Now assume that every connected component of every open subset UU is open. Since the connected components generally consitute a cover of XX by disjoint subsets this means that now they for an open cover by disjoint subsets. But by forming intersections with the cover this implies that every open subset of UU is the disjoint union of open subsets of the connected components (and of course every union of open subsets of the connected components is still open in UU), which is the definition of the topology on the disjoint union space of the connected components.

3) \Rightarrow 1)

Finally assume that every open subspace is the disjoint union of its connected components. Let xx be a point and U x{x}U_x \supset \{x\} a neighbourhood. We need to show that U xU_x contains a connected neighbourhood of xx.

But, by definition, U xU_x contains an open neighbourhood of xx and by assumption this decomposes as the disjoint union of its connected components. One of these contains xx. Since in a disjoint union space all summands are open, this is the required connected open neighbourhod.

Examples

Example

Every discrete topological space is locally connected.

Example

(Euclidean space is locally connected)

For nn \in \mathbb{N} the Euclidean space n\mathbb{R}^n (with its metric topology) is locally connected.

Proof

By nature of the Euclidean metric topology, every neighbourhood U xU_x of a point xx contains an open ball containing xx. Moreover, every open ball clearly contains an open cube, hence a product space i{1,,n}(x iϵ,x i+ϵ)\underset{i \in \{1, \cdots, n\}}{\prod} (x_i-\epsilon, x_i + \epsilon) of open intervals which is still a neighbourhood of xx.

Now intervals are connected (by this example) and products of connected spaces are connected (by this example). This shows that ever open neighbourhood contains a connected neighbourhood.

Proposition

(open subspace of locally connected space is locally connected)

Every open subspace of a locally connected space is itself locally connected

Proof

This is immediate from def. .

Remark

Warning

A connected topological space need not be locally connected.

Example

The topologist's sine curve is connected but not locally connected.

Examples of locally connected spaces include topological manifolds.

Finally,

Definition

A space XX is totally disconnected topological space if its connected components are precisely the singletons of XX.

In other words, a space is totally disconnected if its coreflection into LocConnLocConn is discrete. Such spaces recur in the study of Stone spaces.

The category of totally disconnected spaces is a reflective subcategory of TopTop. The reflector sends a space XX to the space X/X/\sim whose points are the connected components of XX, endowed with the quotient topology induced by the projection q:XX/q: X \to X/\sim. Details may be found at totally disconnected space.

Properties

The category of locally connected spaces

Let i:LocConnTopi \colon LocConn \hookrightarrow Top be the full subcategory inclusion of locally connected spaces into all of Top. The following result is straightforward but useful.

Theorem

LocConnLocConn is a coreflective subcategory of TopTop, i.e., the inclusion ii has a right adjoint RR. For XX a given space, R(X)R(X) has the same underlying set as XX and the coarsest locally connected topology that is finer than the original topology on XX.

Being a coreflective category of a complete and cocomplete category, the category LocConnLocConn is also complete and cocomplete. Of course, limits and particularly infinite products in LocConnLocConn are not calculated as they are in TopTop; rather one takes the limit in TopTop and then retopologizes it according to Theorem . (For finite products of locally connected spaces, we can just take the product in TopTop – the result will be again locally connected.)

Cohesion over sets

Let Γ:LocConnSet\Gamma \colon LocConn \to Set be the underlying set functor, and let ,Δ:SetLocConn\nabla, \Delta \colon Set \to LocConn be the functors which assign to a set the same set equipped with the codiscrete and discrete topologies, respectively. Let Π 0:LocConnSet\Pi_0 \colon LocConn \to Set be the functor which assigns to a locally connected space the set of its connected components.

Theorem

There is an adjoint quadruple of adjoint functors

Π 0ΔΓ:SetLocConn \Pi_0 \dashv \Delta \dashv \Gamma \dashv \nabla \colon Set \to LocConn

and moreover, the functor Π 0\Pi_0 preserves finite products.

While LocConnLocConn is not a topos, this is the adjunction structure as on a cohesive topos.

Proof

The proof is largely straightforward; we point out that the continuity of the unit XΔΠ 0XX \to \Delta \Pi_0 X is immediate from a locally connected space’s being the coproduct of its connected components. As for Π 0\Pi_0 preserving finite products, write locally connected spaces XX, YY as coproducts of connected spaces

X= iC i;Y= jD j;X = \sum_i C_i; \qquad Y = \sum_j D_j;

then their product in LocConnLocConn coincides with their product in TopTop, and is

X×Y i,jC i×D jX \times Y \cong \sum_{i, j} C_i \times D_j

where each summand C i×D jC_i \times D_j is connected by Result . From this it is immediate that Π 0\Pi_0 preserves finite products.

Accordingly the category of sheaves on a locally connected space is a locally connected topos. For related discussions, see also cohesive topos.

Quotients of locally connected spaces

Lemma

A quotient space of a locally connected space XX is also locally connected.

Proof

Suppose q:XYq: X \to Y is a quotient map, and let VYV \subseteq Y be an open neighborhood of yYy \in Y. Let C(y)C(y) be the connected component of yy in VV; we must show C(y)C(y) is open in YY. For that it suffices that C=q 1(C(y))C = q^{-1}(C(y)) be open in XX, or that each xCx \in C is an interior point. Since XX is locally connected, the connected component U xU_x of xx in q 1(V)q^{-1}(V) is open, and the subset q(U x)Vq(U_x) \subseteq V is connected, and therefore q(U x)C(y)q(U_x) \subseteq C(y) (as C(y)C(y) is the maximal connected subset of VV containing q(x)q(x)). Hence U xq 1(C(y))=CU_x \subseteq q^{-1}(C(y)) = C, proving that xx is interior to CC, as desired.

The conclusion does not follow if q:XYq: X \to Y is merely surjective; e.g., there is a surjective (continuous) map from \mathbb{R} to (a version of) the Warsaw circle, but the latter is not locally connected.

Last revised on January 29, 2021 at 06:59:43. See the history of this page for a list of all contributions to it.