## 5.7 Connected components

Definition 5.7.1. Let $X$ be a topological space.

We say $X$ is *connected* if $X$ is not empty and whenever $X = T_1 \amalg T_2$ with $T_ i \subset X$ open and closed, then either $T_1 = \emptyset $ or $T_2 = \emptyset $.

We say $T \subset X$ is a *connected component* of $X$ if $T$ is a maximal connected subset of $X$.

The empty space is not connected.

Lemma 5.7.2. Let $f : X \to Y$ be a continuous map of topological spaces. If $E \subset X$ is a connected subset, then $f(E) \subset Y$ is connected as well.

**Proof.**
Omitted.
$\square$

Lemma 5.7.3. Let $X$ be a topological space.

If $T \subset X$ is connected, then so is its closure.

Any connected component of $X$ is closed (but not necessarily open).

Every connected subset of $X$ is contained in a unique connected component of $X$.

Every point of $X$ is contained in a unique connected component, in other words, $X$ is the union of its connected components.

**Proof.**
Let $\overline{T}$ be the closure of the connected subset $T$. Suppose $\overline{T} = T_1 \amalg T_2$ with $T_ i \subset \overline{T}$ open and closed. Then $T = (T\cap T_1) \amalg (T \cap T_2)$. Hence $T$ equals one of the two, say $T = T_1 \cap T$. Thus clearly $\overline{T} \subset T_1$ as desired.

Pick a point $x\in X$. Consider the set $A$ of connected subsets $x \in T_\alpha \subset X$. Note that $A$ is nonempty since $\{ x\} \in A$. There is a partial ordering on $A$ coming from inclusion: $\alpha \leq \alpha ' \Leftrightarrow T_\alpha \subset T_{\alpha '}$. Choose a maximal totally ordered subset $A' \subset A$, and let $T = \bigcup _{\alpha \in A'} T_\alpha $. We claim that $T$ is connected. Namely, suppose that $T = T_1 \amalg T_2$ is a disjoint union of two open and closed subsets of $T$. For each $\alpha \in A'$ we have either $T_\alpha \subset T_1$ or $T_\alpha \subset T_2$, by connectedness of $T_\alpha $. Suppose that for some $\alpha _0 \in A'$ we have $T_{\alpha _0} \not\subset T_1$ (say, if not we're done anyway). Then, since $A'$ is totally ordered we see immediately that $T_\alpha \subset T_2$ for all $\alpha \in A'$. Hence $T = T_2$.

To get an example where connected components are not open, just take an infinite product $\prod _{n \in \mathbf{N}} \{ 0, 1\} $ with the product topology. Its connected components are singletons, which are not open.
$\square$

Lemma 5.7.5. Let $f : X \to Y$ be a continuous map of topological spaces. Assume that

all fibres of $f$ are connected, and

a set $T \subset Y$ is closed if and only if $f^{-1}(T)$ is closed.

Then $f$ induces a bijection between the sets of connected components of $X$ and $Y$.

**Proof.**
Let $T \subset Y$ be a connected component. Note that $T$ is closed, see Lemma 5.7.3. The lemma follows if we show that $f^{-1}(T)$ is connected because any connected subset of $X$ maps into a connected component of $Y$ by Lemma 5.7.2. Suppose that $f^{-1}(T) = Z_1 \amalg Z_2$ with $Z_1$, $Z_2$ closed. For any $t \in T$ we see that $f^{-1}(\{ t\} ) = Z_1 \cap f^{-1}(\{ t\} ) \amalg Z_2 \cap f^{-1}(\{ t\} )$. By (1) we see $f^{-1}(\{ t\} )$ is connected we conclude that either $f^{-1}(\{ t\} ) \subset Z_1$ or $f^{-1}(\{ t\} ) \subset Z_2$. In other words $T = T_1 \amalg T_2$ with $f^{-1}(T_ i) = Z_ i$. By (2) we conclude that $T_ i$ is closed in $Y$. Hence either $T_1 = \emptyset $ or $T_2 = \emptyset $ as desired.
$\square$

Lemma 5.7.6. Let $f : X \to Y$ be a continuous map of topological spaces. Assume that (a) $f$ is open, (b) all fibres of $f$ are connected. Then $f$ induces a bijection between the sets of connected components of $X$ and $Y$.

**Proof.**
This is a special case of Lemma 5.7.5.
$\square$

Lemma 5.7.7. Let $f : X \to Y$ be a continuous map of nonempty topological spaces. Assume that (a) $Y$ is connected, (b) $f$ is open and closed, and (c) there is a point $y\in Y$ such that the fiber $f^{-1}(y)$ is a finite set. Then $X$ has at most $|f^{-1}(y)|$ connected components. Hence any connected component $T$ of $X$ is open and closed, and $f(T)$ is a nonempty open and closed subset of $Y$, which is therefore equal to $Y$.

**Proof.**
If the topological space $X$ has at least $N$ connected components for some $N \in \mathbf{N}$, we find by induction a decomposition $X = X_1 \amalg \ldots \amalg X_ N$ of $X$ as a disjoint union of $N$ nonempty open and closed subsets $X_1, \ldots , X_ N$ of $X$. As $f$ is open and closed, each $f(X_ i)$ is a nonempty open and closed subset of $Y$ and is hence equal to $Y$. In particular the intersection $X_ i \cap f^{-1}(y)$ is nonempty for each $1 \leq i \leq N$. Hence $f^{-1}(y)$ has at least $N$ elements.
$\square$

Definition 5.7.8. A topological space is *totally disconnected* if the connected components are all singletons.

A discrete space is totally disconnected. A totally disconnected space need not be discrete, for example $\mathbf{Q} \subset \mathbf{R}$ is totally disconnected but not discrete.

Lemma 5.7.9. Let $X$ be a topological space. Let $\pi _0(X)$ be the set of connected components of $X$. Let $X \to \pi _0(X)$ be the map which sends $x \in X$ to the connected component of $X$ passing through $x$. Endow $\pi _0(X)$ with the quotient topology. Then $\pi _0(X)$ is a totally disconnected space and any continuous map $X \to Y$ from $X$ to a totally disconnected space $Y$ factors through $\pi _0(X)$.

**Proof.**
By Lemma 5.7.5 the connected components of $\pi _0(X)$ are the singletons. We omit the proof of the second statement.
$\square$

Definition 5.7.10. A topological space $X$ is called *locally connected* if every point $x \in X$ has a fundamental system of connected neighbourhoods.

Lemma 5.7.11. Let $X$ be a topological space. If $X$ is locally connected, then

any open subset of $X$ is locally connected, and

the connected components of $X$ are open.

So also the connected components of open subsets of $X$ are open. In particular, every point has a fundamental system of open connected neighbourhoods.

**Proof.**
Omitted.
$\square$

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