3.2Connected Subspaces of the Real Line

Proof.

Recall that a subset $Y \subseteq L$ is said to be convex if, for any $a < b \in Y$, the interval $[a, b]$ of $Y$ is a subset of $Y$ (that is, all elements of $L$ in between $a$ and $b$ lie in $L$). Note $L$ is a convex subset of itself, and the intervals and rays in $L$ are exactly the convex subsets of $L$, so let $Y$ be an arbitrary convex subset of $L$—we prove it is connected.

For the sake of contradiction, suppose $Y = A \cup B$ is a separation. Pick some $a \in A$ and $b \in B$ (without loss of generality, let $a < b$), then note $[a, b] \subseteq Y$ by convexity of $Y$. Since $A, B$ form a cover for $Y$, note that

forms a cover for $[a, b]$. Trivially, it follows that $A_0$ and $B_0$ are disjoint and open in the subspace topology $[a, b] \subseteq Y$. Thus $[a, b] = A_0 \cup B_0$ is a separation. Now let $c = \sup(A_0)$, which is well-defined since $A_0$ is a nonempty, bounded subset of a linear continuum. Since $[a, b]$ contains its maximum element $b$, it follows that $c \le b$ and so $c \in [a, b]$. We show $c \notin A_0 \cup B_0 = [a, b]$ to yield a contradiction.

First, suppose $c \in B_0$. Then $c \ne a$, since $A_0 \cap B_0 = \emptyset$. By virtue of $B_0$ being open in $[a, b]$, there is some $a < d < b$ such that $(d, c] \subseteq B_0$. Then note we have have $(d, b] = (d, c] \cup (c, b]$. Indeed, $(c, b] \cap A_0 = \emptyset$ since $c$ is greater than every element in $A_0$. Moreover, $(d, c] \subseteq B_0$, and so $(d, c] \cap A_0 = \emptyset$ since $B_0 \cap A_0 = \emptyset$. Thus $d$ is an upper bound for $A$ that is less than $c$, a contradiction.

Now suppose $c \in A_0$. Then note $c \ne b \in B_0$, so $a \le c < b$. Since $A_0 \subseteq [a, b]$ is open, there is some $e \in A_0$ such that $[c, e) \in A_0$. Then there is some $c < z < e$ such that $z \in A_0$, a contradiction since $c$ is meant to be an upper bound.

Thus $Y$ cannot be separated. ❦

Proof.

Let $A = f(X) \cap (-\infty, r)$ and $B = f(X) \cap (r, \infty)$. Note that $A, B$ are disjoint and non-empty, since $f(a) \in A$ and $f(b) \in B$. Each is open in $f(X)$, being the intersection of an open set in $Y$ (a ray) with $f(X)$ itself. Thus, if there is no $c \in X$ such that $f(c) = r$, then $f(X) = A \cup B$ would constitute a separation, which is a contradiction since the continuous image of a connected set is connected. Thus such a $c$ exists. ❦

Proof.

Suppose $X$ is path-connected, and for the sake of contradiction, let $X = A \cup B$ constitute a separation. For some $x \in A$ and $y \in B$, let $f: [a, b] \to X$ be a path from $x$ to $y$. Then $f([a, b])$ is a connected subset of $X$, so it belong entirely in either $A$ or $B$, a contradiction since its image has elements in both sets. ❦

Solution.

1. Suppose $f: (0, 1) \to (0, 1]$ is a homeomorphism. Since $f$ is surjective, there is some $c \in (0, 1)$ such that $f(c) = 1$. Then define $\tilde f$ by a domain restriction $\tilde{f}: (0, 1) \setminus \{c\} \to (0, 1)$, which is also a homeomorphism. But then note that even though the domain of $\tilde f^{-1}$ is connected, its image is not, a contradiction. An analogous argument works for all other pairs.

2. Consider $X = (0, 1]$ and $Y = [0, 1]$. Then let $f$ be the identity and let $g(x) = 0.5x + 0.5$.

Proof.

Fix $a < b \in X$. If there were no $c \in X$ such that $a < c < b$, then $b$ is the immediate successor to $a$, and so $X = (-\infty, b) \cup (a, \infty)$ constitutes a separation for $X$, which is impossible—thus $c \in X$.

Now suppose $O \subseteq X$ is nonempty and bounded above—we show there exists a supremum for $O$. Let $A$ be the set of all upper bounds on $O$, and for the sake of contradiction, suppose $A$ has no least element. Thus, for any $a \in A$, there is some $a' \in A$ such that $a' < a$. Note then that $A$ contains no limit points of $A^c$ and vice versa, so $A \cup A^c = X$ is a separation, a contradiction. ❦

Solution.

Immediately, $f$ is injective, since if $a \ne b$, it follows that $f(a) \ne f(b)$ by hypothesis—thus $f$ is a bijection. Let $B = (a, b)$ be a basic open set in $X$ (that is, either an interval or ray). For any $x < y \in B$, it follows that $f(x) < f(y)$ by hypothesis. If there is some $c \in Y$ such that $f(x) < c < f(y)$, then by surjectivity there is some $r \in X$ such that $f(r) = c$. By order preserving properties of $f$, $r$ must be such that $x < r < y$, and so $r \in B$. Thus $f$ sends an interval/ray in $X$ to an interval/ray in $Y$. Note that $x \in B$ if and only if $a < x < b$; coupling this with the convexity of $f(B)$, we have that $f(B) = (f(a), f(b))$, an open interval in $Y$. An analogous argument can be used for $f^{-1}$ as well, which proves $f$ is a homeomorphism.

Solution.

1. Let $\{X_\lambda\}_{\lambda \in \Lambda}$ be a set of path-connected topological spaces, and let $X = \prod_{\lambda \in \Lambda} X_\lambda$. Let $\vb{x} = (x_\lambda)_{\lambda \in \Lambda}, \vb{y} = (y_\lambda)_{\lambda \in \Lambda} \in X$ be points. Note that for each $\lambda$, there is a continuous path $\gamma_\lambda: [a, b] \to X_\lambda$ such that $\gamma_\lambda(a) = x_\lambda$ and $\gamma_\lambda(b) = y_\lambda$. Now we may define

   $$\gamma: [a, b] \to X \qquad \gamma(t) = (\gamma_\lambda(t))_{\lambda \in \Lambda},$$

   then we see $\gamma(a) = \vb{x}$ and $\gamma(b) = \vb{y}$. Moreover, a map into the product topology is continuous so long as the components are continuous. Thus the product of path-connected spaces is path-connected.

2. No. Let $A$ be the image of the topologist's sine curve.

3. Let $x, y \in X$. Then $x, y \in f^{-1}(f(X)) \subseteq X$. By path-connectedness of $X$, there is some $\gamma: [a, b] \to f^{-1}(f(X))$ such that $\gamma(a) = x$ and $\gamma(b) = y$. Then consider $f \circ \gamma: [a, b] \to f(X)$. $f \circ \gamma$ is the composition of continuous functions, so it is continuous. Moreover, $(f \circ \gamma)(a) = f(x)$ and $(f \circ \gamma)(b) = f(y)$. Thus this is a continuous path from $f(x)$ to $f(y)$, and since these are arbitrary points of $f(X)$, we conclude $f(X)$ is path-connected.

4. Suppose $x, y \in \bigcup_{\lambda \in \Lambda} X_\lambda$: more specifically, let $\lambda_x$ and $\lambda_y$ be such that $x \in X_{\lambda_x}$ and $y \in X_{\lambda_y}$. By hypothesis there is some $z \in \bigcap_{\lambda \in \Lambda} X_\lambda$, so $z$ is in both subspaces. By path-connectedness of each space, note there exists paths from $x$ to $z$ and $y$ to $z$ in the respective spaces. Merging these paths gives a new path from $x$ to $y$ that is contained in the union of all the path-connected spaces, completing the proof.

Proof.

Fix $x_0 \in U$, and let $O \subseteq U$ be the set of all points $x \in U$ such that there is a path from $x$ to $x_0$. Let $x \in O$ be arbitrary: since $x \in U$ as well, there is some open ball $B$ such that $x \in B \subseteq U$. For every $y \in B$, note there is a path from $y$ to $x$ (since balls in $\bR^2$ are path-connected), and there is a path from $x$ to $x_0$, so there is a path from $y$ to $x_0$. Thus $y \in O$, and since $y$ is arbitrary, this implies $B \subseteq O$. Since there is a open set about each point of $O$ contained in $O$, we have that $O$ is open.

Now consider $O^c$. Once again, let $x \in O^c$ be arbitrary, then there is some open ball $B$ such that $x \in B \subseteq U$. For the sake of contradiction, suppose there is some $y \in B \cap O$. Then there is a path connection $x_0 \to y \to x$, which is a contradiction, since $x \in O$ in this case. Thus $B \subseteq O^c$ and so $O^c$ is open.

Finally, note that $O^c$ are disjoint, open sets such that $U = O \cup O^c$. Since $U$ is connected, no separation exists, so one of these sets must be empty. Certainly, $x_0 \in O$, so $O^c$ is empty and thus $O = U$, meaning $U$ is path-connected. ❦