Real Analysis: Global Metric Topology
Connectedness
Motivation
Consider the real interval $(0, 1)$ under the standard metric. We can see that the set is geometrically "one piece," that is, there are no gaps separating any of the points. In contrast, the set $(0, 1) \cup (3, 4)$ is made up of "two pieces," where one is $(0, 1)$ and the other is $(3, 3).$ We can see geometrically that points either lie in one interval or the other, but that there is space in between the two. We would like a way to rigorously describe this notion of a set being made of a single piece or multiple disconnected pieces.
The most similar notion we have rigorously described so far is the idea of an isolated point of a set. Recall that an isolated point of a set is one for which their exists a neighborhood that does not include any other points of the set. We will not generalize isolated points to "isolated sets," although we will call them disconnected sets instead.
Connected Spaces
To formalize the notion of disconnected sets, we first formalize the notion of a disconnected space. A metric space $(X, d)$ is disconnected if $X$ can be represented as the union of two nonempty open sets. A metric space is connected if it is not disconnected. Likewise, a subset $A$ of $X$ is a disconnected subset if the subspace $(A, d)$ is disconnected, and $A$ connected if it is not disconnected.
This definition lines up with our geometric intuition. Open sets do not contain their boundaries, so even if two disjoint open sets share any boundary points, the boundary points still "take up space" between them. If the subspace does not contain the boundaries of the open sets or any other "space" in between them, then the subspace is even more clearly split in twain. The intuition expands to subsets, for if two sets are disconnected as a subspace, then the addition of more points around them in a superspace will not add any additional points to the original set, and thus it will remain as two pieces.
Connectedness is a global topological property, which means it a property of a metric space or a set as a whole, and not a property of a particular point with respect to a set, or local topological property. Thus while it makes sense to discuss the interior points of a set, it makes no sense to discuss "connected points" of a set. However, as with all concepts in metric topology, the connectedness of a space depends on the underlying metric, and sets that are connected under one metric may be disconnected under another.
Conditions for Connectedness
The following conditions are equivalent for connectedness, i.e. they hold if and only if $(X, d)$ is connected:
 Open Union Criterion: $X$ is not the union of two nonempty open subsets. (This is definition given above.)
 Closed Union Criterion: $X$ is not the union of two nonempty closed subsets.
 OpenClosed Criterion: The only two subsets of $X$ that are both open and closed are $X$ and $\varnothing.$
 Boundary Criterion: Every nonempty open subset of $X$ has a nonempty boundary.
 Separated Criterion: There are no sets $A$ and $B$ such that $A \cup B = X$ and $A \cap \overline{B} = \varnothing$ and $\overline{A} \cap B = \varnothing.$
Connected Subsets of $\mathbb{R}$
The connected subsets of $\mathbb{R}$ are what we might expect: only singleton sets, intervals, and $\mathbb{R}$ itself are connected. All other subsets of $\mathbb{R},$ including $\mathbb{Q},$ are disconnected.
Problems
Show that the empty set is connected.
A set is disconnected if it can be expressed as the union of two disjoint nonempty open subsets. The empty set does not contain any nonempty subsets, therefore it is not disconnected, and so therefore it is connected.
Equivalent definitions I: Show that each of the following is an equivalent definition for connectedness.
 Closed Union Criterion: $X$ is not the union of two disjoint nonempty closed proper subsets.
 OpenClosed Criterion: The only two subsets of $X$ that are both open and closed are $X$ and $\varnothing.$
 Boundary Criterion: Every nonempty open proper subset of $X$ has a nonempty boundary.
 Separated Criterion: There are no sets $A$ and $B$ such that $A \cup B = X$ and $A \cap \overline{B} = \varnothing$ and $\overline{A} \cap B = \varnothing.$

Closed Union Criterion: First, let $(X, d)$ be disconnected. Then there are two nonempty disjoint open proper subsets $U, V \subset X$ such that $U \cup V = X.$ Since $U$ and $V$ form a partition of $X,$ it follows that $U^c = V$ and $V^c = U,$ and so $U$ and $V$ are both closed. Therefore $X$ is the union of two disjoint nonempty closed proper subsets.
Conversely, suppose there are two nonempty disjoint closed proper subsets $U, V \subset X$ such that $U \cup V = X.$ Since $U$ and $V$ form a partition of $X,$ it follows that $U^c = V$ and $V^c = U,$ and so therefore $U$ and $V$ are open as well. Therefore $X$ is the union of two disjoint nonempty open proper subsets, so $X$ is connected. 
OpenClosed Criterion: First, let $(X, d)$ be connected. Assume there are two disjoint open subsets $U, V \subseteq X$ such that $U \cup V = X.$ Since $X$ is connected, it cannot be the case that both $U$ and $V$ are proper subsets. Therefore one of them is $X$ itself, which implies the other is $\varnothing.$
Conversely, assume $X$ and $\varnothing$ are the only two clopen subsets of $(X, d).$ Proof by contradiction. Assume there are two nonempty disjoint open proper subsets $U, V \subset X$ whose union is $X.$ Since $U$ and $V$ form a partition of $X,$ it follows that $U^c = V$ and $V^c = U,$ and so $U$ and $V$ are closed as well. But this is a contradiction, since only $X$ and $\varnothing$ can be clopen. 
Boundary Criterion: First, let $(X, d)$ be connected. Assume $U$ is a nonempty open proper subset of $X.$ Then $U^c$ is closed and $U \cup U^c = X.$ It must be that $U^c$ is not open, as otherwise $X$ would be disconnected. Therefore $\partial (U^c) \neq \varnothing,$ which in turn means that $\partial U \neq \varnothing.$
Conversely, assume that every nonempty open proper subset of $X$ has a nonempty boundary. Proof by contradiction. Assume $X$ is disconnected such that there exist two disjoint open proper subsets $U$ and $V$ of $X$ whose union is $X.$ It follows that $U^c = V$ and $V^c = U$ and so $U$ and $V$ are both closed as well. But this is a contradiction, since clopen sets have empty boundaries. Therefore $X$ was connected after all. 
Separated Criterion: First, let $(X, d)$ be disconnected. Then there are two disjoint open nonempty proper subsets $U, V \subset X$ whose union is $X.$ Since $U$ and $V$ form a partition of $X,$ it follows that $U^c = V$ and and $V^c = U,$ and so $U$ and $V$ are closed as well. Therefore $U = \overline{U}$ and $V = \overline{V}.$ Thus $U \cap \overline{V} = \varnothing = \overline{U} \cap V.$
Conversely, let $U$ and $V$ be two nonempty sets such that $U \cap \overline{V} = \varnothing$ and $\overline{U} \cap V = \varnothing$ and $U \cup V = X.$ It follows from the first two facts that $U$ and $V$ are disjoint. It follows from the third that $U^c = \overline{V}$ and $V^c = \overline{U}.$ Since the closure of a set is always closed, it follows that $\overline{U}$ and $\overline{V}$ are closed, from which it follows that $\left(\overline{U}\right)^c = V$ is open and $\left(\overline{V}\right)^c = U$ is open. Thus $X$ is a union of two disjoint open nonempty proper subsets and therefore disconnected.
Equivalent definitions II: Show that each of the following is equivalent to the Closed Union Criterion:
 OpenClosed Criterion: The only two subsets of $X$ that are both open and closed are $X$ and $\varnothing.$
 Boundary Criterion: Every nonempty open proper subset of $X$ has a nonempty boundary.
 Separated Criterion: There are no sets $A$ and $B$ such that $A \cup B = X$ and $A \cap \overline{B} = \varnothing$ and $\overline{A} \cap B = \varnothing.$

OpenClosed Criterion: First, let $(X, d)$ fulfill the closed union criterion. Assume there are two disjoint open subsets $U, V \subseteq X$ such that $U \cup V = X.$ Since $U$ and $V$ form a partition of $X,$ it follows that $U^c = V$ and $V^c = U,$ and so $U$ and $V$ are open. Since $X$ fulfills the closed union criterion, it cannot be the case that both $U$ and $V$ are proper subsets. Therefore one of them is $X$ itself, which implies the other is $\varnothing.$
Conversely, assume $X$ and $\varnothing$ are the only two clopen subsets of $(X, d).$ Proof by contradiction. Assume there are two nonempty disjoint closed proper subsets $U, V \subset X$ whose union is $X.$ Since $U$ and $V$ form a partition of $X,$ it follows that $U^c = V$ and $V^c = U,$ and so $U$ and $V$ are open as well. But this is a contradiction, since only $X$ and $\varnothing$ can be clopen. 
Boundary Criterion: First, let $(X, d)$ fulfill the closed union criterion. Assume $U$ is a nonempty closed proper subset of $X.$ Then $U^c$ is open and $U \cup U^c = X.$ It must be that $U^c$ is not closed, as otherwise $X$ would not fulfill the closed union criterion. Therefore $\partial (U^c) \neq \varnothing,$ which in turn means that $\partial U \neq \varnothing.$
Conversely, assume that every nonempty open proper subset of $X$ has a nonempty boundary. Proof by contradiction. Assume $X$ does not fulfill the closed union criterion such that there exist two nonempty disjoint closed proper subsets $U$ and $V$ of $X$ whose union is $X.$ Since $U$ and $V$ form a partition of $X,$ it follows that $U^c = V$ and $V^c = U$ and so $U$ and $V$ are both open as well. But this is a contradiction, since clopen sets have empty boundaries. Therefore $X$ fulfilled the closed union criterion after all. 
Separated Criterion: First, let $(X, d)$ not fulfill the closed union criterion. Then there are two disjoint closed nonempty proper subsets $U, V \subset X$ whose union is $X.$ Since $U$ and $V$ form a partition of $X,$ it follows that $U^c = V$ and and $V^c = U,$ and so $U$ and $V$ are open as well. Therefore $U = \overline{U}$ and $V = \overline{V}.$ Thus $U \cap \overline{V} = \varnothing = \overline{U} \cap V.$
Conversely, let $U$ and $V$ be two nonempty sets such that $U \cap \overline{V} = \varnothing$ and $\overline{U} \cap V = \varnothing$ and $U \cup V = X.$ It follows from the first two facts that $U$ and $V$ are disjoint. It follows from the third that $U^c = \overline{V}$ and $V^c = \overline{U}.$ Since the closure of a set is always closed, it follows that $\overline{U}$ and $\overline{V}$ are closed, from which it follows that $\left(\overline{U}\right)^c = V$ is open and $\left(\overline{V}\right)^c = U$ is open. Thus $X$ is a union of two nonempty disjoint closed proper subsets and therefore does not fulfill the closed union criterion.
Equivalent definitions III: Show that both of the following are equivalent to the openclosed criterion is equivalent:
 Boundary Criterion: Every nonempty open proper subset of $X$ has a nonempty boundary.
 Separated Criterion: There are no sets $A$ and $B$ such that $A \cup B = X$ and $A \cap \overline{B} = \varnothing$ and $\overline{A} \cap B = \varnothing.$

Boundary Criterion: First, let $(X, d)$ fulfill the openclosed criterion. Assume $U$ is a nonempty open proper subset of $X.$ Then $U^c$ is a nonempty proper closed subset of $X$ and $U \cup U^c = X.$ It must be that $U^c$ is not open, as otherwise $X$ would not fulfill the openclosed union criterion. Therefore $\partial (U^c) \neq \varnothing,$ which in turn means that $\partial U \neq \varnothing.$
Conversely, assume that every nonempty open proper subset of $X$ has a nonempty boundary. Proof by contradiction. Assume $X$ does not fulfill the openclosed criterion such that there exist two disjoint nonempty clopen proper subsets $U, V \subset X$ whose union is $X.$ Since $U$ and $V$ are clopen, they have empty boundaries. But this is a contradiction, since by assumption $U$ and $V$ must have nonempty boundaries. Therefore $X$ was fulfilled the openclosed criterion after all. 
Separated Criterion: First, let $(X, d)$ not fulfill the openclosed criterion. Then there exists a clopen nonempty proper subset $U \subset X.$ Since $U$ is clopen, it follows that $U^c$ is clopen. This then implies that $U = \overline{U}$ and $U^c = \overline{U^c},$ and so $U \cap \overline{U^c} = \varnothing$ and $\overline{U} \cap U^c = \varnothing.$
Conversely, let $U$ and $V$ be two nonempty sets such that $U \cap \overline{V} = \varnothing$ and $\overline{U} \cap V = \varnothing$ and $U \cup V = X.$ It follows from the first two facts that $U$ and $V$ are disjoint. It follows from the third that $U^c = \overline{V}$ and $V^c = \overline{U}.$ Since the closure of a set is always closed, it follows that $\overline{U}$ and $\overline{V}$ are closed, from which it follows that $\left(\overline{U}\right)^c = V$ is open and $\left(\overline{V}\right)^c = U$ is open. Thus $U$ and $V$ are nonempty clopen proper subsets of $X.$
Eqvuialent definitions IV: Show that the Boundary Criterion and the Separated Criterion are equivalent.
First, assume $(X, d)$ does not fulfill the Boundary Criterion. Then there is a nonempty open proper subset $U$ with an empty boundary. It follows that $U \cup \partial U = U \cup \varnothing = \overline{U} = U.$ Likewise, $U^c$ is closed. Because $\partial U = \partial (U^c),$ it follows that $U^c \cup \partial (U^c) = U^c \cup \varnothing = \overline{U^c} = U^c.$ Therefore $U \cap \overline{U^c} = \overline{U} \cap U^c = \varnothing$ and $U \cup U^c = X.$
Conversely, assume $(X, d)$ does not fulfill the Separated Criterion. Then there are two nonempty proper subsets $U, V$ such that $U \cap \overline{V} = \varnothing$ and $\overline{U} \cap V = \varnothing$ and $U \cup V = X.$ It follows from the first two facts that $U$ and $V$ are disjoint. From the third fact it then follows that $U^c = V$ and $V^c = U.$ Since $U \cap \partial V = \varnothing,$ it follows that $\partial V \subseteq V,$ and so $V$ is closed. Likewise, since $V \cap \partial U = \varnothing,$ it follows that $\partial U \subseteq U,$ and so $U$ is closed. Since $U^c = V,$ we see that $V$ is open, and likewise since $V^c = U,$ we see that $U$ is also open. Therefore $U$ and $V$ are both clopen and thus have empty boundaries.
Is $\mathbb{R}$ connected under the discrete metric?
No. Every subset of a metric space with the discrete metric is both open and closed. Therefore $(\infty, 0]$ and $(0, \infty)$ are both nonempty and open, and their union is $\mathbb{R}.$ Thus $\mathbb{R}$ is disconnected under the discrete metric.
Is every discrete metric space disconnected?
No. The singleton space $(\{x\}, d)$ is connected under any metric, including the discrete metric, since its only subsets are $\{x\}$ and $\varnothing,$ both of which are clopen.
Show that if $\overline{A} \cap \overline{B} = \varnothing,$ then $A$ and $B$ are separated. Show that the converse is false.
Assume $\overline{A} \cap \overline{B} = \varnothing.$ Since $A \cap \overline{B} \subseteq \overline{A} \cap \overline{B},$ it follows that $A \cap \overline{B} = \varnothing.$ Likewise, since $\overline{A} \cap B \subseteq \overline{A} \cap \overline{B}$, it follows that $\overline{A} \cap B = \varnothing.$ Therefore $A$ and $B$ are separated.
Conversely, consider the sets $(0,1)$ and $(1, 0).$ The two sets are disconnected, as $\overline{(0, 1)} \cap (1, 2) = [0, 1] \cap (1, 2) = \varnothing$ and $(0, 1) \cap \overline{(1, 2)} = (0, 1) \cap [1, 2] = \varnothing.$ However, $\overline{(0, 1)} \cap \overline{(1, 2)} = [0, 1] \cap [1, 2] = \{1\} \neq \varnothing.$
Show that every singleton set is connected.
Let $(X, d)$ be a metric space, and let $A = \{x\} \subseteq X.$ The only subsets of $(A, d)$ are $A$ itself and $\varnothing.$ Therefore $A$ cannot be the union of two proper nonempty open subsets. Therefore $(A, d)$ is a connected subspace, and so $A$ is a connected subset of $(X, d).$
Show that finite subsets of $\mathbb{R}$ of cardinality greater than $1$ are disconnected.
Let $R$ be a finite subset of $\mathbb{R}$ where $R > 1.$ It follows that $R$ has at least two elements. Take $a = \text{min}(R)$ and $r = \frac{1}{2}\min\{R  \{a\}\}.$ Then $B_{r}(a)$ is an open ball containing $a.$ Likewise, $\{a\}^c = R  \{a\} = (a + r, \infty) \cap R$ is open in $R.$ Since $\{a\}$ and $(R  \{a\})$ are two nonempty disjoint open proper subsets of $R$ whose union is $R,$ it follows that $R$ is disconnected.
Show that $\mathbb{R}$ is connected.
Proof by contradiction. Assume $\mathbb{R}$ is disconnected. Then there are two nonempty open proper subsets $S, T \subset \mathbb{R}$ such that $S \cap T = \varnothing$ and $S \cup T = \mathbb{R}.$ It follows that $S$ and $T$ are clopen and thus have empty boundaries. With these facts established, we show that in fact $S = \mathbb{R},$ which is a contradiction.
Since $S$ is nonempty, it contains at least one point, call it $x.$ Since $S$ is open, $x$ is an interior point, and so there exists some $r_0 > 0$ such that $B_{r_0}(x) \subseteq S.$ Consider the set of all such radii $R = \{ r \in \mathbb{R} : B_{r}(x) \subseteq S \}.$ Since $S$ is a proper subset of $\mathbb{R},$ it must be that $\sup(R)$ is finite, lest $B_{\infty}(x) = S = \mathbb{R}.$ It follows that either $x  \sup(R)$ or $x + \sup(R)$ is not contained within $B_{\sup(R)}(x)$ and thus not contained within $S.$ Without loss of generality, assume this point is $x  \sup(R)$ for now, and call it $x_l$ for brevity. It follows that for every $q > 0$ that $B_{q}(x_l)$ contains the point $x_l$ outside of $B_{r_0}(x)$ and thus outside of $S$. Likewise, $B_{q}(x_l)$ contains the point $x_l + \frac{q}{2}$ inside $B_{r_0}(x)$ and thus inside $S.$ But this means that $x_l$ is a boundary point of $S,$ which is a contradiction, since $S$ has an empty boundary. Assuming instead that $x + \sup(R)$ is outside $S$ is to no avail, since the same logic shows that $x + \sup(R)$ would then also be a boundary point. Thus $\sup(R)$ could not have been finite after all, and so $S = B_{\infty}(x) = \mathbb{R}$. But this is in turn a contradiction, since $S$ is a proper subset of $\mathbb{R}.$ Thus $S$ could not have been a nonempty clopen proper subset of $\mathbb{R}.$ Thus $\mathbb{R}$ is not disconnected after all, and so $\mathbb{R}$ is connected.
Show that every interval on $\mathbb{R}$ is connected.
Proof by contradiction. Consider some open interval $(a, b) \subseteq \mathbb{R},$ and assume $(a, b)$ is disconnected such that $(a, b) = S \cup T$ where $S$ and $T$ are both nonempty open subsets of $(a, b)$ such that $S \cap T = \varnothing.$ It follows that $S$ and $T$ are clopen and therefore have empty boundaries. We use this fact to show that $S = (a, b)$ and $T = \varnothing,$ a contradiction.
Since $S$ is open and nonempty, it contains a point $x$ for which there is a positive $r \in \mathbb{R}$ such that $B_r(x) \subseteq S.$ Consider the set of radii for such balls, $R = \{ r \in \mathbb{R} : B_{r}(x) \subseteq S \}.$ It follows that $\sup(R) < \max\{x  a, b  a\},$ lest $B_{\sup(R)}(x) = S = (a, b)$ and thus $T = \varnothing.$ Then at least one of $x  \sup(R)$ or $x + \sup(R)$ is a boundary point of $B_{\sup(R)}(x).$ But then this point cannot be a boundary point of $S,$ since $\partial S = \varnothing,$ and so $\sup(R) \geq \max\{x  a, b  a\}$ after all. But this implies $S = (a, b)$ and $T = \varnothing,$ which is a contradiction. Thus $(a, b)$ is connected after all.
The proofs for half open intervals and closed intervals are identical.
Show that every nonempty connected subset of $\mathbb{R}$ is either a singleton set, an interval, or $\mathbb{R}$ itself.
We have already shown that singleton sets, intervals, and $\mathbb{R}$ itself are all connected. It remains for us to show that any other kind of nonempty subset is not connected.
Proof by contradiction. Let $A$ be a nonempty proper subset of $\mathbb{R}$ such that $A$ is connected and is neither a singleton set nor an interval. Since $A$ is not a singleton set, it has at least two elements. Therefore $\inf(A) \neq \sup(A).$ Since $A$ is not an interval, there is some $c \in A^c$ such that $\infty \leq \inf(A) < c < \sup(A) \leq \infty.$ Set $S = (\infty, c)$ and $T = (c, \infty).$ Then $S' = S \cap A$ is open in $A$ and $T' = T \cap A$ is open in $A$ and $S' \cup T' = A.$ But this implies that $A$ is disconnected. It must be that $A$ was either empty, a point, an interval, or all of $\mathbb{R}$ after all.
Show that $\mathbb{Q}$ is disconnected.
Consider $\sqrt{2} \in \mathbb{Q}^c,$ and note that $(\infty, \sqrt{2})$ and $(\sqrt{2}, \infty)$ are open subsets of $\mathbb{R}.$ Let $S = (\infty, \sqrt{2}) \cap \mathbb{Q}$ and $T = (\sqrt{2}, \infty) \cap \mathbb{Q}.$ It follows that $S$ and $T$ are open in $\mathbb{Q}$ and that $S \cup T = \mathbb{Q}.$ Therefore $\mathbb{Q}$ is disconnected.