3.3Components and Local Connectedness

Proof.

1. Reflexivity. $x \sim x$ since $\{x\} \subseteq X$ is a connected subspace.

2. Symmetry. Immediate.

3. Transitivity. If $x \sim y$ and $y \sim z$, then there are connected subsets $A, B \subseteq X$ where $x, y \in A$ and $y, z \in B$. Then $A \cup B$ is a union that shares the point $y$, so it is connected, and it contains $x, z$. Thus $x \sim z$.

❦

Proof.

Since the connected components of $X$ are equivalence classes, they form a partition for $X$. If $C \subseteq X$ were connected and were such that $C \cap [x_1]_{\sim} \ne \emptyset$ and $C \cap [x_2]_{\sim} \ne \emptyset$, where $\sim$ is the equivalence relation that induces the components, then $x_1 \sim x_2$ and so $[x_1]_{\sim} = [x_2]_{\sim}$—thus $C$ intersects exactly one equivalence class and is also fully contained in one.

Fix some $x_0 \in X$, and note $[x_0]_{\sim}$ is a component. For any $x \in [x_0]_{\sim}$, there is some connected subspace $C_x$ for which $x, x_0 \in C_x$. With the previous part in mind, note $C_x \subseteq [x_0]_{\sim}$, and so $[x_0]_{\sim} = \bigcup_{x \in [x_0]_{\sim}} C_x$, which is connected since each $C_x$ has the shared point $x_0$. Thus every component is connected. ❦

Proof.

1. Reflexivity. $x \sim x$ since $\gamma: [a, a] \to X$, where $\gamma(a) = x$, is a path from $x$ to $x$.

2. Symmetry. Immediate.

3. Transitivity. Suppose $x \sim y$ and $y \sim z$. Then there are continuous maps $\gamma_1: [a, b] \to X$ and $\gamma_2: [b, c] \to X$ such that $\gamma_1$ parameterizes a path from $x$ to $y$ and $\gamma_2$ parameterizes a path from $y$ to $z$. By the pasting lemma, note that $\gamma: [a, c] \to X$ defined by “pasting” $\gamma_1, \gamma_2$ is a path from $x$ to $z$, and so $x \sim z$.

❦

Proof.

The proof is identical to the proof about connected components. The proof for showing path components are path-connected is an immediate application of transitivity. ❦

Proof.

$(\Longrightarrow)$: Suppose $X$ is locally connected. Let $U \subseteq X$ be open, and let $C \subseteq U$ be some connected component of $U$. For each $x \in C$, there is some open, connected neighborhood $V_x$ of $x$ such that $x \in V_x \subseteq U$. Since $V_x \cap C \ne \emptyset$ and $C$ is a connected component, it follows that $V_x \subseteq C$ since $V_x$ is connected too. Thus $C = \bigcup_{x \in C} V_x$, the arbitrary union of open sets, which is open.

$(\Longleftarrow)$: Let $U \subseteq X$ be an arbitrary open subset. Then note, for each $x \in U$, there is some connected component $C \subseteq U$ such that $x \in C \subseteq U$. By hypothesis, $C$ is open, completing the proof. ❦

Proof.

The proof is identical to the proof of the previous theorem. ❦

Proof.

Let $P$ be a path component of $X$. Since $P$ is path-connected, it is also connected, so it is fully contained in some connected component $C \subseteq X$, proving the first statement.

Now assume $X$ is locally path-connected. To prove the second statement, we show $C \subseteq P$. For the sake of contradiction, suppose $P \subsetneq C$. Let $Q$ be the union of all path components of $X$ that are different from $P$ and intersect $C$. Since all path components are contained in exactly one connected component, it follows that $Q \subseteq C$. Then note

Since every path component of a locally path-connected set is open, we have that $P, Q$ are disjoint, nonempty open subsets whose union is $C$, meaning $C$ is separated (a contradiction, since $C$ is connected). Thus $C \subseteq P$, and so $P = C$. ❦

Proof.

Suppose $C \subseteq \mathbb{R}_\ell$ contains at least two distinct points, $x < y$. Then $C = \big((-\infty, y) \cap C\big) \cup \big([y, \infty) \cap C\big)$ constitutes a separation of $C$, so $C$ is disconnected. Thus, the only nonempty connected subspaces of $\mathbb{R}_\ell$ are singletons, which are therefore the connected components. Since every path-connected space is connected, the path components must also be exactly the singletons.

Note that for some $f: \bR \to \bR_\ell$ to be continuous, the image of a connected set must be connected. Since $\bR$ is connected in its standard topology, it follows that $f(\bR)$ is connected, but the only connected subsets of $\bR_\ell$ are singletons, so it follows that $f(\bR) = \{*\}$, so all continuous maps are constant maps. ❦

Solution.

1. Note that the arbitrary product of connected and path-connected sets is connected and path-connected in the product topology. Thus $\bR^\omega$ is the only connected and path-connected component in $\bR^\omega$ with the product topology.

2. $(\Longrightarrow)$: Suppose $\vb{x}, \vb{y}$ are in the same connected component. Then there is some connected $C' \subseteq \bR^\omega$ such that $\vb{x}, \vb{y} \in C'$. Now consider a translation $T: \bR^\omega \to \bR^\omega$ defined by $T(\vb{x}) = \vb{x} - \vb{y}$. This is a homeomorphism, so $C := T(C')$ is also a connected component. Thus we have $\vb{x - y}, \vb{0} \in C$.

   First, we show $\bR^\omega$ can be separated by the subspaces of bounded and unbounded sequences. Let $O \subseteq \bR^\omega$ be the subspace of all bounded sequences. For each bounded sequence $x \in O$, define $B_x = B_{\bar{\rho}}(x, 1/2)$, a ball about $x$ defined by the uniform metric of radius $1/2$. For each $y \in B_x$, note that $|x_n - y_n| < 1/2$ for every $n \in \bZ_+$, and so $y$ is bounded as well, meaning $y \in O$. Thus $x \in B_x \subseteq O$, so $O = \bigcup_{x \in O} B_x$. An analogous procedure can be performed to show $O^c$ is open, since drawing a ball of radius $1/2$ about an unbounded sequence still lies in $O^c$. Thus $\bR^\omega = O \cup O^c$, a separation.

   We know $\vb{0}$ is bounded. If $\vb{x - y}$ were unbounded, then $C = (O \cap C) \cup (O^c \cap C)$ would be a valid separation for a supposedly connected set (a contradiction), so $\vb{x - y}$ is forced to be bounded.

   $(\Longleftarrow)$: Suppose $\vb{x} - \vb{y}$ is a bounded sequence. Let $O \subseteq \bR^\omega$ be the subspace of all bounded sequences. We will show $O$ is connected, which will complete the proof.

   We proceed by showing $O$ is path-connected. Note the translation $T: \bR^\omega \to \bR^\omega$ given by $T(\vb{z}) = \vb{z} - \vb{y}$ is a homeomorphism. Since $\vb{x} - \vb{y}$ and $\vb{y} - \vb{y} = \vb{0}$ are both bounded sequences, we have $T(\vb{x}), T(\vb{y}) \in O$. We can show there is a path from $\vb{h} = \vb{x} - \vb{y}$ to $\vb{0}$ in $O$. We may define a map $\gamma: [0, 1] \to O$ by $\gamma(t) = t\vb{h}$. Indeed, $\gamma(0) = \vb{0}$ and $\gamma(1) = \vb{h}$, so to show $\gamma$ is a path, we simply show $\gamma$ is continuous.

   Since $\vb{h} \in O$, there is some $M \in \bR$ such that $|h_n| < M$ for every $n \in \bZ_+$. Fix some $t_0 \in [0, 1]$ and some $1 > \epsilon > 0$. Let $\delta = \epsilon/M$, then let $t \in [0, 1]$ be such that $|t - t_0| < \delta$. Then

   $$\begin{align*} \bar{\rho}(\gamma(t), \gamma(t_0)) &= \sup_{n \in \bZ_+} \left\{ \min\left(\abs{th_n - t_0 h_n}, 1\right) \right\} \\ &= \sup_{n \in \bZ_+} \left\{ \abs{t - t_0} \abs{h_n} \right\} \\ &< \sup_{n \in \bZ_+} \left\{ (\epsilon/M)(M) \right\} \\ &= \sup_{n \in \bZ_+} \{\epsilon\} = \epsilon. \end{align*}$$

   If $\epsilon \ge 1$, then $\delta$ may be anything since the uniform metric is at most $1$. Thus $\gamma$ is continuous, and we conclude $O$ is path-connected, which implies $O$ is also connected. Because $T(\vb{x})$ and $T(\vb{y})$ both belong to the connected subspace $O$, they are in the same component. Since $T^{-1}$ is a homeomorphism, it maps components to components, meaning $\vb{x}$ and $\vb{y}$ must also belong to the same component in $\bR^\omega$, completing the proof.

Solution.

Let $O \subseteq X$ be open and connected. Fix $x \in O$ and any open neighborhood $U \subseteq O$ of $x$. Since $U$ is an open subset of the locally path-connected space $X$, all of $U$'s path components are open. Thus there is some open, path component (which is thus path-connected) $[x]_\sim$ such that $x \in [x]_\sim \subseteq U$. Thus $O$ is locally path-connected, and so its path components and connected components coincide. Since $O$ is connected, its only connected component is $O$, and so $O$ is also path-connected, completing the proof.