Connecting Theorems

Intuitively, intervals of the form $(a, b)$ are connected—on the real line, you could draw a line from the number $a$ to the number $b$ without lifting up your pen. However, what do we mean by connected?

Lemma 1. Suppose there exist open sets $U, V$ such that $U \cup V = (a, b)$ . Then $U \cap V \neq \emptyset$ .

Proof. Fix $x \in U$ and $y \in V$ and suppose $x < y$ without loss of generality. Define the set

$K := \{z \in U : x < z < y\} \subseteq [x, y].$

Since $U$ is open, there exists $\epsilon_1 > 0$ such that $x \in (x-\epsilon_1, x+\epsilon_1) \subseteq U$ , so that $U$ is nonempty and bounded. Thus, $\alpha := \sup(K)$ exists. If $\alpha \in U$ , then $\alpha +\epsilon_1/2 \in U$ , a contradiction. Therefore, $\alpha \notin U$ . Since

$\alpha \in (x,y) \subseteq (a,b) = U \cup V,$

we must have $\alpha \in V$ . Since $V$ is open, there exists $\epsilon_2 > 0$ such that

$\alpha \in (\alpha - \epsilon_2, \alpha +\epsilon_2) \subseteq V.$

By the definition of $\sup(K)$ , there exists $z \in K$ such that

$z \in (\alpha - \epsilon_2, \alpha] \subseteq V.$

Therefore, $z \in K \cap V \subseteq U \cap V$ , so that $U \cap V \neq \emptyset$ , as required.

What we have shown is that $(a, b)$ cannot be separated by open sets.

Let $K$ be a topological space.

Definition 1. We say that $K$ is disconnected if there exist disjoint open subsets $U, V \subseteq K$ such that $U \cup V = K$ . We say that $K$ is connected if it is not disconnected. Whenever we consider subsets $L \subseteq K$ , we equip $L$ with the subspace topology. In particular, open intervals $(a, b) \subseteq \mathbb R$ are connected.

Theorem 1. A bounded open set $I \subseteq \mathbb R$ is of the form $(a, b)$ if and only if it is connected.

Proof. By Lemma 1, we have proven $(\Rightarrow)$ . To prove $(\Leftarrow)$ , suppose $I$ is connected. Assuming $I$ is bounded and open, we recall that if for any $x, y \in I$ such that $x < y$ and for any $z \in \mathbb R$ , $x < z < y \Rightarrow z \in K$ , then $K = (a, b)$ for some $a < b$ . That is, if $I$ is an interval that is open and bounded, then $I = (a, b)$ for some $a < b$ . Thus, we prove this interval property.

Suppose instead there exists $x, y \in K$ such that $x < y$ and there exists $z \in \mathbb R$ such that $x < z < y$ yet $z \notin K$ . Then define the sets $U := (-\infty, z) \cap K$ and $V := (z, \infty) \cap K$ , which are disjoint open subsets of $K$ such that $U \cap V = \emptyset$ . Thus, $K$ is disconnected, a contradiction.

Having done our necessary sanity checks that the connected sets in $\mathbb R$ really are the intervals (we can laboriously verify this result for all other kinds of intervals, not just open and bounded ones), we are ready to discuss the generalised intermediate value theorem, and even prove the extreme value theorem.

Theorem 2 (Intermediate Value Theorem). Let $L$ be a topological space and $f : K \to L$ be a continuous map and $M \subseteq K$ be connected.

$f(M)$ is connected.
If $L = \mathbb R$ , $f(M)$ is an interval.
For any $a, b \in M$ , if $f(a) < f(b)$ , then for any $m \in (f(a), f(b))$ , there exists $c \in (a, b)$ such that $f(c) = m$ .

Proof. We prove by contrapositive. Suppose $f(M)$ is disconnected. Find disjoint open sets $U, V \subseteq L$ such that $U \cup V \supseteq f(M)$ . Since $f$ is continuous, $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint open subsets of $K$ such that

$f^{-1}(U) \cup f^{-1}(V) = f^{-1}(U \cup V) \supseteq f^{-1}(f(M)) = M.$

Hence, $M$ is disconnected, as required.

If $L = \mathbb R$ , then $f(M) \subseteq \mathbb R$ , so that $f(M)$ being connected implies that it is an interval. Now fix $a, b \in M$ and suppose $f(a) < f(b)$ . Fix $m \in (f(a), f(b))$ . Since $f(M)$ is an interval, $(f(a), f(b)) \subseteq f(M)$ . Hence, there exists $c \in M$ such that $m = f(c)$ .

In particular, setting $M := [a, b] \subseteq \mathbb R =: K$ , we recover the vanilla intermediate value theorem:

Corollary 1. If $f : \mathbb R \to \mathbb R$ is continuous, then assuming $f(a) < f(b)$ , for any $f(a) < m < f(b)$ , there exists $c \in (a, b)$ such that $f(c) = m$ .

Furthermore, we obtain a more complete picture of the extreme value theorem:

Corollary 2 (Extreme Value Theorem). For any nonempty compact set $L \subseteq K$ , if $f : L \to \mathbb R$ is continuous, then there exists $c_1, c_2 \in L$ such that $f(L) \subseteq [f(c_1), f(c_2)]$ , where equality holds if $L$ is connected.

Proof. Since $L$ is compact, so is $f(L) \subseteq \mathbb R$ , and thus closed and bounded. Since $f(L)$ is nonempty and bounded, $m := \inf f(L)$ and $M := \sup f(L)$ exist. Since $f(L)$ is closed, $m, M \in f(L)$ . By definition, there exists $c_1,c_2 \in L$ such that $f(c_1) = m$ and $f(c_2) = M$ . Therefore, $f(L) \subseteq [m, M] = [f(c_1), f(c_2)]$ , as required.

Furthermore, if $L$ is connected, so is $f(L)$ . Thus, $f(L)$ is an interval, so that by the intermediate value theorem, $[f(c_1), f(c_2)] \subseteq f(L)$ .

What we have thus far shown is how our generalised topological notions of compactness and connectedness not only agree with our usual understanding in the context of the topology on $\mathbb R$ , but are plausibly more versatile since they don’t inherently depend on the special properties that $\mathbb R$ possesses.

A slight but still very useful generalisation of $\mathbb R$ would be the notion of metric spaces. We’ll use metric spaces to discuss the notion of completeness and even develop useful fixed-point theorems in applied fields like solving differential equations. Eventually, we’ll return to the purer aspects of point-set topology, and try to build most, if not all, of our developed ideas on compactness.

—Joel Kindiak, 1 Apr 25, 1506H

KindiakMath

Connecting Theorems

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Connecting Theorems

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