# Equivalence of definitions of connected component

This article gives a proof/explanation of the equivalence of multiple definitions for the term Connected component (?)

View a complete list of pages giving proofs of equivalence of definitions

## Statement

We would like to show that the following definitions of connected component make sense and are equivalent.

### Definition as a subset

A connected component of a topological space is defined as a subset satisfying the following two conditions:

1. It is a connected subset, i.e., it is a connected space with the subspace topology.
2. It is not properly contained in any bigger subset that is connected.

### Definition in terms of equivalence relation

For a topological space $X$, consider the following relation: $a \sim b$ if there exists a subset of $X$ containing both $a$ and $b$ that is a connected space under the subspace topology. Then, it turns out that $\! \sim$ is an equivalence relation on $X$. The equivalence classes under $\! \sim$ are termed the connected components of $X$.

## Facts used

1. Connected union of connected subsets is connected

## Proof

### Proof that the specified relation is an equivalence relation

We prove all three aspects of an equivalence relation:

Condition Statement Proof
reflexivity $\! a \sim a$, i.e., every point is contained in a connected subset the singleton subset comprising that point is connected
symmetry $\! a \sim b \iff b \sim a$ the definition is symmetric, so this is direct
transitivity $\! a \sim b, b \sim c \implies a \sim c$ From the given, there exists a connected subset containing $a$ and $b$. Call it $D_1$. There also exists a connected subset containing $b$ and $c$. Call it $D_2$. By Fact (1), $D_1 \cup D_2$ is connected, and it contains both $a$ and $c$.

### Proof from equivalence relation to subset definition

Since $\! \sim$ is an equivalence relation, its equivalence classes are subsets. We now want to argue that:

1. Each equivalence class is a connected subset: This follows essentially from Fact (1). Let $C$ be an equivalence class. Pick any point $a$ in an equivalence class. Then, for any point $b$ in the class, let $D(a,b)$ be a connected subset containing both $a$ and $b$. The whole equivalence class is a union $\bigcup_{b \in C} D(a,b)$ and hence, by Fact (1), is connected.
2. No equivalence class is contained in a bigger connected subset: If it were, then elements outside the subset would be related to elements inside the subset, contradicting its definition as an equivalence class.

### Proof from subset definition to equivalence relation

For this, suppose $C$ is any maximal connected subset and suppose it is not an equivalence class. Let $a \in C$ and let $D$ be the equivalence class of $a$. Then, we just noted that $D$ is connected, so $C \cup D$ is connected by Fact (1), hence contradicting the maximality of $C$.