## 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.**
Let $A \subset f(E)$ an open and closed subset of $f(E)$. Because $f$ is continuous, $f^{-1}(A)$ is an open and closed subset of $E$. As $E$ is connected, $f^{-1}(A) = \emptyset $ or $f^{-1}(A) = E$. However, $A\subset f(E)$ implies that $A = f(f^{-1}(A))$. Indeed, if $x\in f(f^{-1}(A))$ then there is $y\in f^{-1}(A)$ such that $f(y) = x$ and because $y\in f^{-1}(A)$, we have $f(y) \in A$ i.e. $x\in A$. Reciprocally, if $x\in A$, $A\subset f(E)$ implies that there is $y\in E$ such that $f(y) = x$. Therefore $y\in f^{-1}(A)$, and then $x\in f(f^{-1}(A))$. Thus $A = \emptyset $ or $A = f(E)$ proving that $f(E)$ is connected.
$\square$

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 disjoint 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 $\overline{T} \subset T_1$. This implies (1) and (2).

Let $A$ be a nonempty set of connected subsets of $X$ such that $\Omega = \bigcap _{T \in A} T$ is nonempty. We claim $E = \bigcup _{T \in A} T$ is connected. Namely, $E$ is nonempty as it contains $\Omega $. Say $E = E_1 \amalg E_2$ with $E_ i$ closed in $E$. We may assume $E_1$ meets $\Omega $ (after renumbering). Then each $T \in A$ meets $E_1$ and hence must be contained in $E_1$ as $T$ is connected. Hence $E \subset E_1$ which proves the claim.

Let $W \subset X$ be a nonempty connected subset. If we apply the result of the previous paragraph to the set of all connected subsets of $X$ containing $W$, then we see that $E$ is a connected component of $X$. This implies existence and uniqueness in (3).

Let $x \in X$. Taking $W = \{ x\} $ in the previous paragraph we see that $x$ is contained in a unique connected component of $X$. Any two distinct connected components must be disjoint (by the result of the second paragraph).

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.**
For all $x\in X$ let write $\mathcal{N}(x)$ the fundamental system of connected neighbourhoods of $x$ and let $U\subset X$ be an open subset of $X$. Then for all $x\in U$, $U$ is a neighbourhood of $x$, so the set $\{ V \in \mathcal{N}(x) | V\subset U\} $ is not empty and is a fundamental system of connected neighbourhoods of $x$ in $U$. Thus $U$ is locally connected and it proves (1).

Let $x \in \mathcal{C} \subset X$ where $\mathcal{C}$ is the connected component of $x$. Because $X$ is locally connected, there exists $\mathcal{N}$ a connected neighbourhood of $x$. Therefore by the definition of a connected component, we have $\mathcal{N} \subset \mathcal{C}$ and then $\mathcal{C}$ is a neighbourhood of $x$. It implies that $\mathcal{C}$ is a neighbourhood of each of his point, in other words $\mathcal{C}$ is open and (2) is proven.
$\square$

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