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. $\square$
Remark 5.7.4. Let $X$ be a topological space and $x \in X$. Let $Z \subset X$ be the connected component of $X$ passing through $x$. Consider the intersection $E$ of all open and closed subsets of $X$ containing $x$. It is clear that $Z \subset E$. In general $Z \not= E$. For example, let $X = \{ x, y, z_1, z_2, \ldots \} $ with the topology with the following basis of opens, $\{ z_ n\} $, $\{ x, z_ n, z_{n + 1}, \ldots \} $, and $\{ y, z_ n, z_{n + 1}, \ldots \} $ for all $n$. Then $Z = \{ x\} $ and $E = \{ x, y\} $. We omit the details.
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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