This blog post was originally supposed to be about my one of my favorite characterizations of connectedness called chain-connectedness. After a few drafts, I realized that chain-connectedness isn’t rich enough for an entire post. I also realized that, while visually appealing, chain-connectedness isn’t the most suitable notion of connectedness for some applications. My perspective on connectedness has recently grown, and I now have a trifecta of favorite characterizations of connectedness.

Before discussing chain-connectedness, I want to first introduce the idea of connectedness as an induction principle.

Connectedness as an Induction Principle

Our initial characterization of connectedness will be as follows:

Definition (Connected)

A topological space $X$ is said to be disconnected provided that there exists two non-empty, disjoint, open subsets of $X$ which union to $X$. The space $X$ is said to be connected provided that it is not disconnected.

Intuitively, this notion of connectedness says that the non-empty open subsets of $X$ cannot separate $X$.

Over at Pseudonium’s Blog, connectedness as an induction Principle is motivated by and generalizes real induction. Roughly, the induction principle characterizes a connected space by its clopen subsets, of which there can only be two — the whole space and the empty space.

Theorem (Induction Principle for Connectedness)

A topological space $X$ is connected if and only if for every subset $S$ of $X$ such that

• $S$ is non-empty
• $S$ is open
• $S$ is closed

then $S = X$.

Proof

Suppose that $X$ is connected. Let $S$ be a subset of $X$ satisfying the three properties above, i.e., $S$ is a non-empty clopen subset. If $S$ is a proper subset of $X$, then $X - S$ is also a non-empty, open subset of $X$; $X$ is thus the union of two non-empty, disjoint, open subsets, a contradiction.

Conversely, suppose $X$ is a space satisfying the inductive principle — that the only non-empty clopen subset of $X$ is $X$. Let $U$ and $V$ be two non-empty, disjoint, open subsets which union to $X$. The set $U = X - V$ is a non-empty, clopen subset of $X$. By the inductive principle on $X$, $U = X$, contradicting the fact that $V$ is non-empty. Since $X$ cannot be disconnected, $X$ is connected.

This characterization tells us necessary and sufficient conditions that a subset of a connected space must satisfy to be the entire space. If a subset $S$ of a connected space $X$ is defined by a predicate $\phi$ on $X$ as $S = \{x \in X : \phi(x)\}$, the induction principle for connectedness tells us that $\phi$ is true at every point of $X$ provided that

• $\phi(x)$ is true at some point $x \in X$,
• on $S$, $\phi$ is locally true, meaning that for each inhabitant $x \in S$, there is an open set $U$ such that $x \in U \subseteq S$ and $\phi(u)$ is true for all $u \in U$, and
• on $X - S$, $\phi$ is locally false.

We have met the first of my three favorite perspectives on connectedness.

It is hard not to mention that this last observation leads naturally to the characterization of connectedness by continuous functions into discrete spaces; however, this isn’t one of the three perspectives I will be discussing.

Chain-Connectedness

After learning about chain-connectedness during my undergrad (my original favorite characterization), I found it to be extremely intuitive, easy to visualize, and useful for topological proofs that extend certain local properties on connected spaces to global ones.

Preliminaries

The preliminary notion of a chain is useful for defining chain-connectedness:

Definition (Chain and Chain-Connected Open Covers)

Let $X$ be a topological space and $\mathcal{U}$ be an open cover of $X$ (a collection of open subsets of $X$ which union to $X$).

Given two open subsets $U,V \in \mathcal{U}$, a chain from $U$ to $V$ in $\mathcal{U}$ is a finite sequnece $\{U_0,\dots,U_{n}\}$ in $\mathcal{U}$ such that

• $U_0 = U$,
• $U_n = V$, and
• $U_{i} \cap U_{i + 1} \neq \emptyset$ for each $0 \leq i < n$.

For $x \in U$ and $y \in V$, the set $\{U_{0},\dots,U_{n}\}$ is called a chain from $x$ to $y$ in $\mathcal{U}$.

The open cover $\mathcal{U}$ is a chain-connected open cover if there is a chain between any two non-empty open sets in $\mathcal{U}$, or equivalently, between any two points of $X$.

It is worth pointing out again that a chain in an open cover $\mathcal{U}$ must consist of open sets from $\mathcal{U}$.

Definition (Chain-Connected)

A topological space $X$ is said to be chain-connected if every open cover of $X$ is a chain-connected open cover of $X$.

Explicitly, $X$ is chain-connected if for every open cover $\mathcal{U}$ of $X$ and any two non-empty open sets $U,V \in \mathcal{U}$, there is a finite sequence $\{U_{0},\dots,U_{n}\}$ in $\mathcal{U}$ such that

• $U_{0} = U$,
• $U_{n} = V$, and
• $U_{i} \cap U_{i + 1} \neq \emptyset$ for each $0 \leq i < n$.

Before moving on, let’s contrast this with the notion of path-connectedness:

Definition (Path and Path-Connected)

A path in a topological space $X$ is a continuous function $\gamma : I \to X$, where $I = [0,1]$ is the unit interval (with the usual topology).

The space $X$ is said to be path-connected provided that there is a path between any two points of $X$.

The notion of chain-connectedness may be seen as a discretization of path-connectedness; instead of any two points being joined by a path of points in $X$, roughly, any two open sets (or equivalently, points) are joined by a path of open sets in $X$. This observation suggests a natural visualization of chains and chain-connectedness.

The main reason I like chain-connectedness so much is because of its visual and geometric nature which I feel other characterizations lack; it becomes more intuitive after trying to apply it, as we will now.

How Chain-Connectedness is Linked to Connectedness

It is a familiar theorem that every path-connected space is connected. The discrete analogue of this is also true; every chain-connected space is connected. Moreover, these notions of connectedness are equivalent.

Theorem (Chain-Connectedness and Connectedness are Equivalent)

A topological space $X$ is connected if and only it is chain-connected.

Proof

Apply the induction principle for connectedness!

Let $\mathcal{U}$ be an open cover of $X$. Declare two points $x$ and $y$ of $X$ to be equivalent if and only if there is a chain in $\mathcal{U}$ from $x$ to $y$ and write $x \sim y$; indeed, this defines an equivalence relation on $X$.

Say that $x_0$ is a point of $X$; define $S = \{x \in X : x \sim x_0\}$.

• $S$ is non-empty: $x_0 \in S$.
• $S$ is open: for $x \in S$, there is a chain $\{U_0,\dots,U_n\} \subseteq \mathcal{U}$ from $x_0$ to $x$. For each $y \in U_n$, the same chain is a chain from $x_0$ to $y$, thus $U_n$ is an open neighborhood of $x$ contained in $S$.
• $S$ is closed: for $x \in X - S$, there is an open set $U \in \mathcal{U}$ containing $x$. $U$ cannot intersect $S$, since otherwise there would be a chain from $x_0$ to $x$.

By the inductive principle for connectedness, $S = X$, thus $X$ is chain-connected.

Conversely, suppose $X$ is chain-connected. Let $S$ be a non-empty clopen subset of $X$. The set $\mathcal{U} = \{S, X - S\}$ is an open cover of $X$, so there is a chain in $\mathcal{U}$ between any two of its non-empty open subsets. No subset of $\mathcal{U}$ forms a chain from $S$ to $X - S$, therefore $X - S$ must be empty, and thus $S = X$. Having demonstrated that $X$ satisfies the induction principle for connectedness, $X$ is connected.

The above proof shows us how properties that hold locally can be patched together via chains of open sets.

The equivalence relation above will be referred to as the equivalence relation determined by chains.

Having shown that these two notions of connectedness are equivalent, we have met the second of my three favorite perspectives on connectedness.

The Final Equivalence

Any subset $S$ of a set $X$ induces an equivalence relation $x\sim y$ if and only if $x,y \in S$ or $x,y \in X - S$. If $X$ is a topological space, and $S$ is clopen, then the equivalence classes of $\sim$ are both open. This means that the distinction of points of $X$ by belonging to $S$ or not belonging to $S$ vanishes locally. In a connected space, that is enough for the distinction to vanish globally.

This observation leads us to my final favorite perspective on connectedness. At least one other source calls this the local-to-global lemma. First, it is useful to make precise what I meant by a distinction vanishing locally and globally.

Definition ((Locally) Trivial Equivalence Relation)

An equivalence relation $\sim$ on a set $X$ is a trivial relation if $\sim \, = X \times X$, that is, if every element of $X$ is related to every other element of $X$, or, if $\sim$ has a single equivalence class.

If $\sim$ is a relation on a topological space $X$, $\sim$ is called locally trivial if there is an open neighborhood $U$ around each point of $X$ on which $x \sim y$ for all $x,y \in U$ (equivalently, if the equivalence classes of $\sim$ are open).

Locally trivial equivalence relations are ways to distinguish of points of a space where the distinction vanishes locally, i.e., each point is contained in an open neighborhood where all points are “equal”.

Theorem (The Local-to-Global Lemma)

A topological space $X$ is connected if and only if every locally trivial equivalence relation on $X$ is (globally) trivial.

Proof

Suppose $X$ is connected. Let $\sim$ be a locally trivial equivalence relation on $X$. Each equivalence class of $\sim$ is non-empty and clopen. By the induction principle for connectedness, each equivalence class is equal to $X$. The equivalence relation $\sim$ has a unique equivalence class and is therefore trivial.

Conversely, let $S$ be a non-empty clopen subset of $X$. The equivalence relation $x \sim y$ if and only if $x,y \in S$ or $x,y \in X - S$ is a locally trivial equivalence relation, and is thus promoted to a globally trivial one. Let $x_0 \in S$. Every element $x \in X$ is related to $x_0$, and so $S = X$. By the induction principle for connectedness, $X$ is connected.

The local-to-global lemma frames connectedness as the principle that any distinction which vanishes locally must vanish globally.

It is interesting to specialize the above proof of the converse direction to the case when $S$ is defined by a predicate on $X$. It is also interesting to specialize the above proof of the forwards direction when $\sim$ is the equivalence relation determined by chains.

Applications

Here are a few applications of my favorite notions of connectedness.

Proposition

If $X$ is connected and locally path-connected, then $X$ is path-connected.

Proof

Path-connectedness is an equivalence relation on any space; being locally path-connected means that it is a locally trivial equivalence relation (locally, any two points are equivalent). By the local-to-global lemma, $X$ is path-connected.

Proposition

Every locally constant function on a connected space $X$ is constant.

Proof

Let $f : X \to Y$ be a locally constant function on a connected space $X$.

Say that $x \sim y$ if and only if $f(x) = f(y)$. Any distinctions among points with respect to $\sim$ vanishes locally ($f$ is locally constant); by the local-to-global lemma, $f$ is constant.

Proposition

The product of finitely many connected spaces is connected.

Proof

By induction, it suffices to show that the product of two connected spaces is connected. We use chain-connectedness.

Let $X$ and $Y$ be connected and $\mathcal{U}$ an open cover of $X \times Y$. The projection $\pi_X : X \times Y \to X$ is an open map, and so $\pi_X(\mathcal{U}) = \{\pi_X(U) : U \in \mathcal{U}\}$ is an open cover of $X$, likewise for $Y$.

Given $(a,b),(c,d) \in X \times Y$, there is a finite chain $\{\pi_X(U_0),\dots,\pi_X(U_n)\}$ in $\pi_X(\mathcal{U})$ from $a$ to $c$. Likewise, there is a finite chain $\{\pi_Y(V_0),\dots,\pi_Y(V_m)\}$ in $\pi_Y(\mathcal{U})$ from $b$ to $d$. The set $\{U_0,\dots,U_n,V_0,\dots,V_m\}$ is a chain in $\mathcal{U}$ from $(a,b)$ to $(c,d)$ since $(c,b) \in U_n \cap V_0$.

Proposition

If $A$ is a connected subspace of $X$, if $A \subseteq B \subseteq \mathtt{cl}\,A$, then $B$ is connected.

Proof

We use chain-connectedness.

Let $\mathcal{U}$ be an open cover of $B$ and fix $x,y \in B$. There are four cases:

1. $(x \in A, y \in A)$ There is a chain in $\mathcal{U}$ from $x$ to $y$ since $A$ is connected.

2. $(x \in A, y \in B - A)$ In this case, $y$ is a limit point of $A$; there is some neighborhood $U \in \mathcal{U}$ of $y$ and some $z \in U \cap (A - \{y\})$. By (chain-)connectedness of $A$, there is a chain in $\mathcal{U}$ from $x$ to $z$. Appending $U$ to this chain produces a chain in $\mathcal{U}$ from $x$ to $y$.

3. $(x \in B - A, y \in A)$ This case is symmetric.

4. $(x \in B - A, y \in B - A)$ Just as in case 2., there is an open set $U_y \in \mathcal{U}$ containing $y$ and some $z_y \in U_y \cap (A - \{y\})$; likewise for $x$. By (chain-)connectedness of $A$, is a chain in $\mathcal{U}$ from $z_x$ to $z_y$. Prepending $U_x$ and appending $U_y$ yields a chain from $x$ to $y$ in $\mathcal{U}$.

This next batch of applications uses a bit more machinery, so I will only give proof sketches.

Closing Thoughts