A Missing Point Creates a Separation
In the standard topology on \(\mathbb{R}\), points are ordered, so between any two distinct points there are many possible cut points. If a subset contains points on both sides of a cut but omits the cut itself, the subset falls into a left part and a right part. These parts are disjoint, nonempty, and open in the subset. Thus they form a separation, which connectedness forbids.
For \(A\subseteq\mathbb{R}\), we always give \(A\) the subspace topology: its open sets are the sets \(A\cap U\), where \(U\) is open in the standard topology on \(\mathbb{R}\). A separation of \(A\) is a pair of disjoint, nonempty, relatively open subsets whose union is \(A\). The space \(A\) is connected if it has no separation. The empty set and a one-point set are connected under this definition, since neither can be split into two nonempty pieces.
The order matters in the definition only to state which point lies between the other two. If the two given points occur in the opposite order, we can switch their names. The essential case is therefore \(x<t<y\): the two points are in the subset and the intermediate point is not.
Separating at a Gap
Proof. Define \(A_- = A\cap(-\infty,c)\) and \(A_+ = A\cap(c,\infty)\). The rays \((-\infty,c)\) and \((c,\infty)\) are open in the standard topology on \(\mathbb{R}\), so \(A_-\) and \(A_+\) are open in the subspace topology on \(A\). They are disjoint because no real number is both less than and greater than \(c\). Since \(c\notin A\), every point of \(A\) is either less than \(c\) or greater than \(c\), so \(A_-\cup A_+=A\). Finally, \(x\in A_-\) and \(y\in A_+\), so both pieces are nonempty. They form a separation of \(A\); hence \(A\) is disconnected. \(\square\)
The hypothesis that the cut point is missing is what makes the two pieces cover \(A\). The hypotheses \(x<c<y\) ensure that neither piece is empty. Without those conditions, the displayed sets need not give a separation.
Worked Example: A Two-Point Set
Consider \(A=\{-2,4\}\) with the subspace topology from \(\mathbb{R}\). The point \(1\) lies between the points of \(A\), since \(-2<1<4\), and \(1\notin A\). Cutting at \(c=1\) gives
Both sets are relatively open because they are intersections of \(A\) with open rays. They are disjoint, neither is empty, and their union is \(\{-2,4\}=A\). Thus this is a separation and \(A\) is disconnected.
Check: \(-2<1\) puts \(-2\) in \(A_-\), while \(4>1\) puts \(4\) in \(A_+\). The only elements of \(A\) are \(-2\) and \(4\), so the two pieces account for every element and omit no point of \(A\).
Worked Example: Two Closed Pieces with a Gap
Let \(A=[-3,0]\cup[2,5]\), again with the subspace topology from \(\mathbb{R}\). The cut point \(c=1\) is not in \(A\), and the points \(0\) and \(2\) of \(A\) satisfy \(0<1<2\). The two sides of the cut are
For the first equality, the portion \([-3,0]\) lies below \(1\), while every point of \([2,5]\) lies above \(1\). For the second, the portion \([2,5]\) lies above \(1\), while every point of \([-3,0]\) lies below \(1\). Each piece is relatively open in \(A\), since each is the intersection of \(A\) with the indicated open ray. They are nonempty, disjoint, and their union is \(A\). Therefore \(A\) is disconnected.
Check: the left piece contains \(0\), and the right piece contains \(2\), so both are nonempty. The endpoints \(-3,0,2,5\) are all assigned to the indicated side, and there are no points of \(A\) in the gap \((0,2)\). In particular, the cut point \(1\) contributes no missing element to the union.
Connected Subsets Must Contain Every Intermediate Point
Proof. Let \(C\) be connected, and take \(x,y\in C\) with \(x\leq t\leq y\). If \(x=y\), then \(x\leq t\leq x\) forces \(t=x\), so \(t\in C\). Now suppose \(x<y\). If \(t=x\) or \(t=y\), then \(t\in C\) because \(x,y\in C\). It remains to consider \(x<t<y\). Assume, seeking a contradiction, that \(t\notin C\). The lemma applied to \(A=C\) and \(c=t\) gives a separation of \(C\), contradicting that \(C\) is connected. Thus \(t\in C\), as required. \(\square\)
This argument also covers the degenerate cases. If \(C\) is empty or has only one point, there cannot be two distinct points of \(C\) with a missing point strictly between them. The proof does not require \(C\) to be open or closed in \(\mathbb{R}\), nor does it require \(C\) to be bounded.
Worked Example: Applying the Theorem to an Interval
The open interval \(C=(-5,4)\) is connected by the earlier result that every interval is connected. Its points \(-4\) and \(3\) belong to \(C\), because \(-5<-4<4\) and \(-5<3<4\). The theorem therefore puts every \(t\) satisfying \(-4\leq t\leq3\) in \(C\). For instance, the intermediate values \(-3\), \(0\), and \(2\) all satisfy the required inequalities:
Consequently \(-3,0,2\in C\). The statement concerns points between the chosen points \(-4\) and \(3\); it does not assert that the endpoints \(-5\) and \(4\) belong to \(C\).
Check: each of \(-3\), \(0\), and \(2\) is also strictly between \(-5\) and \(4\), so each is indeed an element of the specified open interval. The conclusion is consistent with direct membership in \(C\).
Why Relative Openness Matters
The cut pieces in the lemma are open in \(A\), not necessarily open as subsets of the whole real line. That distinction is essential: connectedness of \(A\) is defined using its subspace topology. To show a subset \(E\subseteq A\) is relatively open, it is enough to exhibit an open set \(U\subseteq\mathbb{R}\) with \(E=A\cap U\). The lemma does exactly this with the open rays on either side of the missing point.
Common pitfall. For \(A=[-1,0]\cup[2,3]\) and \(c=1\), one might incorrectly claim that \(A\cap(-\infty,1)=[-1,0]\) is open in \(\mathbb{R}\). It is not: every open neighborhood of \(0\) in \(\mathbb{R}\) contains points greater than \(0\), so no such neighborhood is contained in \([-1,0]\). The correct claim is that \([-1,0]\) is open in \(A\), because \(A\cap(-\infty,1)=[-1,0]\), and \((-\infty,1)\) is open in \(\mathbb{R}\). The same reasoning applies to \([2,3]=A\cap(1,\infty)\). These relative-open pieces form a separation of \(A\).
The theorem turns a topological hypothesis into a concrete order condition: a connected subset cannot have a gap between two of its points. The proof works because a missing real number provides a cut, and the ordinary open rays on either side of that cut induce a separation. The converse direction—that a subset with the between-points property is connected—uses the connectedness of intervals established earlier; the argument here needs only the direction stated in the theorem.
Check Your Understanding
For each question, identify the cut point and check relative openness in the specified subspace topology.
- Let \(A=\{-5,2\}\). Use \(c=-1\) to write down a separation of \(A\).
- For \(A=[-4,-2]\cup[1,6]\), cut at \(c=0\). What are the two pieces of the separation?
- Why does the proof of the theorem not need a separate argument when \(t=x\) or \(t=y\)?
- Give the precise reason that \(A\cap(-\infty,c)\) is relatively open in \(A\), even if it is not open in \(\mathbb{R}\).
- Suppose \(C\subseteq\mathbb{R}\) is connected and contains \(-2\) and \(5\). Which points \(t\) are guaranteed to belong to \(C\)?