Prove a Connectedness Theorem

Connectedness Through Separations

The contraction mapping theorem concerns distances and limits. Connectedness asks a different question: can a space be split into two nonempty pieces that are open relative to the space and do not meet? For subsets of the real line, the order structure gives a precise answer. A set is connected exactly when it contains every real number between any two of its points—that is, exactly when it is an interval.

The word “relative” matters. To decide whether a subset is connected, we use its own topology: a set open in the subset is the intersection of that subset with an open set in the ambient space. In particular, the pieces in a separation need not be open in the whole real line.

The interval definition includes open, closed, half-open, unbounded, and singleton intervals. The empty set also satisfies the stated between-points condition. Both the empty set and every singleton are connected: neither can be written as the union of two disjoint nonempty sets.

Every Interval Is Connected

The key step is to rule out a separation of an interval. If a separation existed, choose one point from each side and consider the points on the first side between them. Their supremum would have to belong to one side or the other. Relative openness on either side then contradicts the supremum property.

Proof. The empty set and singleton cases are connected as noted above. Now suppose \(I\) contains at least two points, and assume for contradiction that \(I=U\cup V\) is a separation. Choose one point from each side. After exchanging the names of the sides if necessary, there are \(a\in U\) and \(b\in V\) with \(a<b\). Since \(I\) is an interval, \([a,b]\subseteq I\).

Let \(S=U\cap[a,b]\). This set is nonempty because \(a\in S\), and it is bounded above by \(b\). Let \(c=\sup S\), which exists by the least upper bound property of the real numbers. Since \(a\leq c\leq b\) and \([a,b]\subseteq I\), we have \(c\in I=U\cup V\).

First suppose \(c\in U\). Since \(b\in V\) and \(U\cap V=\varnothing\), we have \(c<b\). The set \(U\) is open relative to \(I\), so there is an \(r>0\) such that \(I\cap(c-r,c+r)\subseteq U\). Choose \(h>0\) with \(h<r\) and \(h<b-c\). Then \(c+h\in[a,b]\subseteq I\), and \(c+h\in(c-r,c+r)\), so \(c+h\in U\cap[a,b]=S\). This contradicts that \(c\) is an upper bound of \(S\), because \(c+h>c\).

Now suppose \(c\in V\). Since \(a\in U\), we have \(c>a\). Relative openness of \(V\) gives an \(r>0\) such that \(I\cap(c-r,c+r)\subseteq V\). Choose \(\varepsilon>0\) with \(\varepsilon<r\) and \(\varepsilon<c-a\). Because \(c=\sup S\), the number \(c-\varepsilon\) is not an upper bound of \(S\); hence there is an \(s\in S\) with \(s>c-\varepsilon\). Also \(s\leq c\). Since \(s\in U\) but \(c\in V\), the disjointness of \(U\) and \(V\) implies \(s\neq c\), so \(s<c\). Thus \(s\in I\cap(c-r,c+r)\subseteq V\), contradicting \(s\in S\subseteq U\).

Both possible locations of \(c\) lead to a contradiction. Therefore \(I\) has no separation and is connected. \(\square\)

The proof uses the order completeness of \(\mathbb{R}\), specifically the existence of the supremum of a nonempty bounded set. Its central technique is to take a boundary point between the two proposed pieces and show that neither relatively open side can contain that point consistently with the supremum property.

Connected Subsets of the Real Line Are Intervals

The converse has a shorter proof. If a set contains two points but omits a point strictly between them, the omitted point divides the set into the portion to its left and the portion to its right. Those portions form a separation.

Proof. The implication from interval to connected was proved in the theorem Every Interval in \(\mathbb{R}\) Is Connected. For the reverse implication, suppose \(E\) is connected. We show that \(E\) is an interval. If it were not, there would be \(a,b\in E\) and \(c\notin E\) such that \(a<c<b\), after ordering the two endpoints if needed.

Define \(U=E\cap(-\infty,c)\) and \(V=E\cap(c,\infty)\). These sets are open in the subspace topology on \(E\), since each is the intersection of \(E\) with an open subset of \(\mathbb{R}\). They are disjoint. They are both nonempty because \(a\in U\) and \(b\in V\). Finally, \(U\cup V=E\), since \(c\notin E\). Thus \(U,V\) form a separation of \(E\), contradicting the assumption that \(E\) is connected. Hence \(E\) is an interval. \(\square\)

The two directions have different proof structures. To prove an interval connected, the supremum argument excludes every possible separation. To prove connected implies interval, one missing intermediate point explicitly constructs a separation. Together, these arguments make the interval characterization both a theorem and a practical test.

Worked Applications

Unions with a Common Point

The interval characterization is specific to subsets of the real line. A useful connectedness result also applies in any topological space: a union of connected sets that all share a point is connected. Each set must lie wholly on one side of any proposed separation, and the common point forces all the sets to lie on the same side.

Proof. If \(J=\varnothing\), then \(W=\varnothing\), which is connected because it cannot be the union of two nonempty sets. Now suppose \(J\neq\varnothing\). Assume for contradiction that \(W=P\cup Q\) is a separation. The sets \(P,Q\) are disjoint, nonempty, and relatively open in \(W\).

Fix \(j\in J\). The sets \(C_j\cap P\) and \(C_j\cap Q\) are relatively open in \(C_j\), are disjoint, and have union \(C_j\). Since \(C_j\) is connected, they cannot both be nonempty. Therefore \(C_j\) lies wholly in \(P\) or wholly in \(Q\). But \(p\in C_j\) for every \(j\), so each \(C_j\) must lie in the same side as \(p\). It follows that their union \(W\) lies wholly in that side, contradicting that the other side of the separation is nonempty. Thus \(W\) is connected. \(\square\)

Why Relative Openness and the Hypotheses Matter

A separation is not merely a division into two disjoint nonempty pieces. The pieces must also be open relative to the set being tested. For example, \([0,1]\) is connected even though it is not open in \(\mathbb{R}\). Its connectedness follows from being an interval, not from its openness or closedness in the ambient line.

Likewise, a set may be connected even when the ambient space is disconnected; the singleton example shows this directly. There is no conflict: the pieces in a separation need not be open in the whole ambient space, but their intersections with the subset determine whether they are open in the subset. A separation of a subset by relatively open pieces is a separation of that subset itself.

When applying the interval characterization, check for a missing point strictly between two points of the set. Finding one proves disconnectedness immediately. If there is no such point, the set is an interval, and the supremum theorem proves connectedness. For unions in a general space, verify both that each set is connected and that there is a shared point; without the common point, the conclusion can fail, as the separated set \(\{0,1\}\) demonstrates.