> [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]]. $X$ is **not connected** if there are disjoint [[Open Set|open sets]] that cover it > $ > \exists U, V \in \topo: X = U \cup V, U \cap V = \emptyset, U, V \ne \emptyset > $ > [!definition] > > Let $(X, \topo)$ be a topological space. A [[Set|set]] of points $A \subseteq X$ is **not connected** if > $ > \exists U, V \in \topo: A \subseteq U \cup V, U \cap V \ne \emptyset, U, V \cap A \ne \emptyset > $ > [!theorem] > > Let $(X, \topo)$ be a topological space, then $X$ is connected if and only if the only subsets of $X$ that are both open and [[Closed Set|closed]] are $\emptyset$ and $X$. > > *Proof*. Suppose that $X$ is not connected, then there exists non-empty open sets $U, V$ such that $U \cup V = X$ and $U \cap V = \emptyset$. Therefore $V = U^c$, $U = V^c$, and both sets are both closed and open. > > Suppose that there exists a non-empty closed and open set $E \subset X$, then $E^c$ is also open and $E \cup E^c$ is a partition of $X$ with open sets. > [!theorem] > > Let $(X, \topo)$ be a topological space, $\seqi{E}$ be a family of connected sets with $\bigcap_{i \in I}E_i$ non-empty, then $\bigcup_{i \in I}E_i$ is connected. > > *Proof*. Let $U, V \subset X$ such that $U \cup V \supset \bigcup_{i \in I}E_i$ and $U \cap V = \emptyset$. Let $x \in \bigcap_{i \in I}E_i$ and assume without loss of generality that $x \in U$. For each $i \in I$, since $E_i$ is connected, $U \cap E_i \ne \emptyset$ implies that $E_i \subset U$. Thus $V \cap \bigcup_{i \in I}E_i = \emptyset$. > [!theorem] > > Let $(X, \topo)$ be a topological space and $E \subset X$ be a connected set, then $\ol{E}$ is also connected. > > *Proof*. Let $U, V$ be open sets such that $U, V$ intersect $\ol{E}$. Let $x \in U \cap \ol{E}$, then $U$ is a [[Neighbourhood|neighbourhood]] of $x$ and $U \cap E \ne \emptyset$. Similarly, $V \cap E \ne \emptyset$. As $E$ is connected, $U \cap V \cap E$ is non-empty. Therefore $U \cap V \cap \ol{E}$ is non-empty and $\ol{E}$ is connected. > [!definition] > > Let $(X, \topo)$ be a topological space and $x \in X$, then there exists a unique maximal connected subset of $X$, which is closed, known as the [[Connected Component|connected component]] of $x$. > > *Proof*. Let $\mathcal{U} = \bracs{E \subset X: x \in E, E \text{ connected}}$, we claim that $F = \bigcup_{E \in \mathcal{U}}E$ is the desired set. > - All sets in $\mathcal{U}$ contain $x$, so $F$ is connected. > - $F$ contains every connected set that contains $x$, so $F$ is maximal. > - $F \supset \ol{F}$, so $F$ is closed. > [!theorem] > > Let $(X, \topo_X)$ and $(Y, \topo_Y)$ be topological spaces and $f: X \to Y$ be a [[Continuity|continuous]] function. Then $f(U)$ is connected if $U$ is connected. > > *Proof*. Suppose that $f(U)$ is not connected, then there exists disjoint non-empty open sets $A, B \in \topo_Y$ that cover $f(U)$. Their preimage $f^{-1}(A), f^{-1}(B)$ are non-empty open sets. Since $A, B$ are disjoint, $f^{-1}(A)$ and $f^{-1}(B)$ are also disjoint, which contradicts the fact that $U$ is connected. Therefore $f(U)$ must also be connected. > [!theorem] > > Let $X, Y$ be connected topological spaces, then $X \times Y$ is also connected. > > *Proof*. Fix $x \in X$, then the cross $x \times Y \cup X \times y$ is connected for each $y \in Y$. Since > $ > X \times Y = \bigcap_{y \in Y}(x \times Y \cup X \times y) \supset x \times Y > $ > the product $X \times Y$ is also connected. > [!theorem] > > Let $\seqi{X}$ be a family of connected topological spaces, then $\prod_{i \in I}X_i$ is connected. > > *Proof*. Let $x \in \prod_{i \in I}X_i$. For each $K \subset I$ finite, let $A_K = \prod_{i \in K}X_i$, then the set > $ > Y_K = \prod_{i \in I}Y_i \quad Y_i = \begin{cases} > X_i &i \in K \\ > \bracs{x_i} &i \not\in K > \end{cases} > $ > is connected. Since $\bigcap_{K \subset I, K\text{ finite}}Y_K \supset \bracs{x}$, $A = \bigcup_{K \subset I, K\text{ finite}}Y_K$ is connected. Therefore $\prod_{i \in I}X_i = \overline{A}$ is also connected.