[Munkres Topology] 23. Connected Spaces

 This article is one of the posts in the Textbook Commentary Project.

The definition of connectedness for a topological space is a quite natural one. One says that a space can be 'seoarated' if it can be broken up into two 'globs' - disjoint open sets. Otherwise, one says that it is connected.

Definition. Let $X$ be a topological space. A separation of $X$ is a pair, $U$, $V$ of disjoint nonempty open subsets of $X$ whose union is $X$. The space $X$ is said to be connected if there does not exists a separation of $X$.

Also, a space $X$ is connected if and only is the only subsets of $X$ that are both open and closed in $X$ are the empty set and $X$ itself.

Proof. [For seperation. $\Leftarrow$] For if $A$ is a nonempty proper subset of $X$ that is both open and closed in $X$, then the sets $U=A$ and $V=X-A$ constitute a separation of $X$, for they are open, disjoint, and nonempty, and their union is $X$. [For seperation. $\Rightarrow$] Conversely, if $U$ and $V$ form a separation of $X$, then $U$ is nonempty and different from $X$, and it is both open and closed in $X$.

Lemma 23.1. If $Y$ is a subspace of $X$, a separation of $Y$ is a pair of disjoint nonempty sets $A$ and $B$ whose union is $Y$, neither of which contains a limit point of the other. The space $Y$ is connected if there exists no separation of $Y$.

Proof. [$\Rightarrow$] Suppose first that $A$ and $B$ form a separation of $Y$. Then $A$ is both open and closed in $Y$. The closure of $A$ in $Y$ is the set $\bar{A}\cap Y$ (where $\bar{A}$ as usual denotes the closure of $A$ in $X$). Since $A$ is closed in $Y$, $A=\bar{A}\cap Y$; or to say the same thing, $\bar{A}\cap B=\varnothing$(Since $B\subset Y$). Since $\bar{A}$ is the union of $A$ and its limit points, $B$ contains no limit points of $A$. A similar argument shows that $A$ contians no limit point of $B$.

[$\Leftarrow$] Conversely, suppose that $A$ and $B$ are disjoint nonempty sets whose union is $Y$, neither of which contains a limit point of the other. Then $\bar{A}\cap B=\varnothing$ and $A\cap \bar{B}=\varnothing$; therefore, we conclude that $\bar{A}\cap Y=A$ and $\bar{B}\cap Y=B$. Thus both $A$ and $B$ are closed in $Y$, and since $A=Y-B$ and $B=Y-A$, they are open in $Y$ as well.

Lemma 23.2. If the sets $C$ and $D$ form a separation of $X$, and if $Y$ is a connected subspace of $X$, then $Y$ lies entirely withing either $C$ or $D$.

Proof. Since $C$ and $D$ are both open in $X$, the sets $C\cap Y$ and $D\cap Y$ are open in $Y$. These two sets are disjoint and thier union is $Y$; if they were both nonempty, they would constitute a separation of $Y$. Therefore, one of then is empty. Hence $Y$ must lie entirely in $C$ or in $D$.

Theorem 23.3. The union of a collection of connected subspaces of $X$ that have a point in common is connected.

Proof. Let $\{A_\alpha\}$ be a collection of connected subspaces of a space $X$; let $p$ be a point of $\bigcap A_\alpha$. We prove that the space $Y=\bigcup A_\alpha$ is connected. Suppose that $Y=C\cup D$ is separation of $Y$. The point $p$ is in one of the sets $C$ or $D$; suppose $p\in C$. Since $A_\alpha$ is connected, it must lie entirely in either $C$ or $D$, and it cannot lie in $D$ because if contains the point $p$ of $C$. Hence $A_\alpha \subset C$ for every $\alpha$, so that $\bigcup A_\alpha \subset C$, contradicting the fact that $D$ is nonempty.

Theorem 23.4. Let $A$ be a connected subspace of $X$. If $A\subset B \subset \bar{A}$, then $B$ is also connected.

Proof. Let $A$ be connected and let $A\subset B \subset \bar{A}$. Suppose that $B=C\cup D$ is a separation of $B$. By Lemma 23.2, the set $A$ must lie entirely in $C$ or in $D$; suppose that $A\subset C$. Then $\bar{A}\subset \bar{C}$; since $\bar{C}$ and $D$ are disjoint, $B$ cannot intersect $D$ (since $B\subset \bar{C}$). This contradicts the fact that $D$ is a nonempty subset of $B$.

Theorem 23.5. The iamge of a connected space under a continuous map is connected.

Proof. Let $f:X\rightarrow Y$ be a continuous map; let $X$ be connected. We with to prove the image space $Z=f(X)$ is connected. Since the map obtained from $f$ by restricting its range to the space $Z$ is also continuous, it suffices to consider the case of a continuous surjective map $$g:X\rightarrow Z.$$ Suppose that $Z=A\cup B$ is a separation of $Z$ into two disjoint nonempty sets open in $Z$. Then $g^{-1}(A)$ and $g^{-1}(B)$ are disjoint sets whose union is $X$; they are open in $X$ because $g$ is continuous, and nonempty because $g$ is surjective. Therefore, they form a separation of $X$, contradicting the assumption that $X$ is connected.

Theorem 23.6. A finite cartesian product of connected spaces is connected.

Proof. We prove tha theorem first for the product of two connected space $X$ and $Y$. This proof is easy to visualize. Choose a 'base point' $a\times b$ in the product $X\times Y$. Note that the 'horizontal slice' $X\times b$ is connected, being homeomorphic with $X$, and each 'vertical slicve' $x\times Y$ is connected, being homeomorphic with $Y$. As a result, each $T-shaped$ spac $$T_x=(X\times b)\cup (x\times Y)$$ is connected, being the union of two connected spaces that have the point $x\times b$ in common. See Figure 23.2. 

Now form the union $\bigcup_{x\in X} T_x$ of all these T-shaped spaces. This union is connected because it is the union of a collection of connected spaces that have the point $a\times b$ in common. Since this union equals $X\times Y$, the space $X\times Y$ is connected.

The proof for any finite product of connected spaces follows by induction. using the fact that $X_1\times \cdots\times X_n$ is homeomorphic with $(X_1\times \cdots\times X_{n-1})\times X_n$.