The Proof Idea: A Missing Point Creates a Cut
The Characterization of Connected Subsets of \(\mathbb{R}\), established earlier in this course, says that a set is connected exactly when it is an interval. This tutorial focuses on the proof idea behind the direction named in the title: why a connected subset of the real line cannot leave a gap between two of its points. We will use the Gap Criterion for Separation rather than re-prove it.
Suppose \(E\subseteq\mathbb{R}\) contains \(x\) and \(y\) with \(x<y\), but omits some point \(c\) between them. The cut at \(c\) divides \(E\) into points to the left of \(c\) and points to the right of \(c\). Both pieces are nonempty, since they contain \(x\) and \(y\). The Gap Criterion for Separation says precisely that this kind of gap witnesses that \(E\) is disconnected. Thus a connected set cannot have such a missing intermediate point.
The essential feature is the order on \(\mathbb{R}\). A missing point \(c\) determines two sides, and every point of \(E\) lies on one side or the other. The separation is not obtained by guessing complicated subsets: it is dictated by the location of the gap. This makes the proof both short and reusable.
This principle is a proof strategy, not a new characterization theorem: the full characterization was already established earlier. The goal here is to make the cut argument easy to recognize and to show how it leads to further conclusions without repeating that theorem's proof.
Reading the Cut Argument Carefully
When using a cut, check three details. First, the proposed cut point \(c\) must not belong to \(E\); otherwise the two displayed pieces do not cover \(E\). Second, the cut must lie strictly between two points of \(E\), so that both pieces are nonempty. Third, each piece must be open relative to \(E\). For the left piece, for example, \(E\cap(-\infty,c)\) is relatively open in \(E\), since \((-\infty,c)\) is open in \(\mathbb{R}\). The right piece is relatively open for the same reason. The Gap Criterion for Separation packages these checks into a result already available in the course.
Choose \(x,y\in E\) with \(x<y\), and suppose \(c\notin E\) with \(x<c<y\).
Set \(A=E\cap(-\infty,c)\) and \(B=E\cap(c,\infty)\). Since \(c\notin E\), every point of \(E\) lies in exactly one of these sets.
The point \(x\) belongs to \(A\), and \(y\) belongs to \(B\), so neither piece is empty.
The Gap Criterion for Separation applies, so \(E\) is disconnected. This contradicts connectedness.
The contradiction depends on the strict inequalities \(x<c<y\). If \(c=x\) or \(c=y\), the cut does not separate the two chosen points into opposite sides. A proof that says merely “split at a missing point” is incomplete unless it also verifies that the missing point lies between two members of the set.
Worked Example: Finding the Cut in a Disconnected Set
Let \(E=(-2,1)\cup(3,6]\). The points \(x=0\) and \(y=5\) belong to \(E\). Choose \(c=2\). Then \(0<2<5\) and \(2\notin E\). The two pieces determined by the cut are \(A=E\cap(-\infty,2)=(-2,1)\) and \(B=E\cap(2,\infty)=(3,6]\). They are disjoint, nonempty, and their union is \(E\). Since each is the intersection of \(E\) with an open subset of \(\mathbb{R}\), each is relatively open in \(E\). This is the cut-based separation: \(E\) is disconnected.
The endpoints verify the construction explicitly. The point \(0\) lies in \(A\) because \(0<2\), and \(5\) lies in \(B\) because \(5>2\). No point of \(E\) is omitted from \(A\cup B\), because the only real number that lies on neither side is \(2\), and \(2\notin E\).
Worked Example: Why an Endpoint Is Not a Gap Between Points
Consider \(E=[1,4)\). The number \(4\) is not in \(E\), but this alone does not show that \(E\) is disconnected. To use the cut argument, one would need points \(x,y\in E\) with \(x<4<y\). No such \(y\) exists: every \(y\in E\) satisfies \(y<4\). The missing endpoint is not a point between two members of \(E\), so it does not create the required two-sided cut.
In fact, \(E\) is an interval, and hence connected by the theorem established earlier. This example is a useful check against a common error: a set can omit points from its closure and still be connected. What matters for this argument is an omitted point strictly between two points of the set.
A New Consequence for Continuous Functions
The cut proof also clarifies why connectedness is useful when studying function values. A continuous function carries connected sets to connected sets, by the theorem on Continuous Images of Connected Sets. A connected subset of \(\mathbb{R}\) is an interval, by the Characterization of Connected Subsets of \(\mathbb{R}\). Therefore, if a continuous function takes two values, it must take every value between them. This is an intermediate-value conclusion even when the domain is not presented as a closed interval.
Proof. Since \(E\) is connected and \(f\) is continuous, the Continuous Images of Connected Sets theorem implies that \(f(E)\) is connected in \(\mathbb{R}\). By the Characterization of Connected Subsets of \(\mathbb{R}\), \(f(E)\) is an interval. The values \(f(u)\) and \(f(v)\) belong to \(f(E)\), and \(f(u)<t<f(v)\). Because \(f(E)\) is an interval, it contains every real number between any two of its elements. Hence \(t\in f(E)\). By the definition of the image \(f(E)=\{f(z):z\in E\}\), there is \(w\in E\) with \(f(w)=t\). \(\square\)
Worked Example: A Continuous Function on a Half-Open Interval
Let \(E=[-1,2)\) and \(f(x)=x^2\). The set \(E\) is an interval, so it is connected. The function \(f\) is continuous on \(E\). Take \(u=-1\) and \(v=1\); then \(f(u)=1=f(v)\), so these choices do not give distinct endpoint values for the theorem. Instead, take \(u=0\) and \(v=1\), giving \(f(u)=0\) and \(f(v)=1\). For any \(t\) with \(0<t<1\), the theorem guarantees a \(w\in[-1,2)\) with \(w^2=t\).
The value can also be checked directly: choose \(w=\sqrt{t}\). Since \(0<t<1\), we have \(0<\sqrt{t}<1\), so \(w\in E\), and \(f(w)=(\sqrt{t})^2=t\). The general theorem obtains the existence from connectedness and continuity; the formula verifies it in this example.
A Sign Consequence and Its Proof
A useful special case concerns whether a continuous function can change sign without vanishing. On a connected domain, it cannot. This fact is often applied to prove that a function is positive or negative throughout a set after checking its value at just one point and showing it has no zeros.
Proof. If \(E=\varnothing\), both universal statements hold vacuously, so the conclusion holds. Suppose \(E\ne\varnothing\). If the conclusion were false, then \(f\) would take at least one positive value and at least one negative value. Thus there would be \(u,v\in E\) with \(f(u)<0<f(v)\), after interchanging the names of the points if necessary. The Intermediate-Value Property on a Connected Set, applied with \(t=0\), gives \(w\in E\) such that \(f(w)=0\). This contradicts the assumption that \(f\) is nonzero everywhere on \(E\). Therefore \(f\) cannot take both positive and negative values. Since it takes no zero values and \(E\) is nonempty, all its values must be positive or all must be negative. \(\square\)
Worked Example: A Sign Check on a Connected Domain
Let \(E=(-3,2]\) and \(g(x)=x^2+1\). The domain \(E\) is an interval and hence connected. For every \(x\in E\), \(x^2\geq0\), so \(g(x)=x^2+1\geq1>0\). Thus \(g\) is continuous, never vanishes, and is positive throughout \(E\), in agreement with the theorem.
The theorem becomes more useful when the sign is not immediate from a formula. For example, if \(h:E\to\mathbb{R}\) is continuous, \(h(0)>0\), and one has independently shown \(h(x)\ne0\) for every \(x\in E\), then the theorem forces \(h(x)>0\) for every \(x\in E\). The connectedness of the domain prevents a change to negative values without an intervening zero.
What the Proof Does—and Does Not—Say
The cut argument depends on the order structure of the real line: every proposed gap has a left side and a right side. For the theorem that connected subsets of \(\mathbb{R}\) are intervals, this is exactly the right structure. The same argument should not be transferred mechanically to arbitrary spaces, where there may be no meaningful left and right sides or no intermediate points to test.
It is also important to keep the two directions of the characterization distinct. The cut argument explains why a connected subset of \(\mathbb{R}\) cannot omit a point between two of its elements. The converse—that every interval is connected—requires a different argument and was established separately earlier in the course. Together, the two directions give the characterization, but one direction should not be mistaken for a proof of the other.
The practical routine is simple: identify two points of the set, locate a missing point strictly between them, and use the cut to produce two nonempty relatively open pieces. If no such missing point exists for any pair, the set has the defining order property of an interval. This approach turns an abstract question about separations into a concrete check on the real line.
Check Your Understanding
Use the cut argument and the stated consequences of connectedness to answer the following questions.
- Why must a cut point lie strictly between two points of \(E\) for the Gap Criterion for Separation to apply?
- For \(E=(-5,-2]\cup[0,3)\), choose two points and a missing intermediate point that exhibit a separation.
- Why does the missing endpoint \(4\) not by itself show that \([1,4)\) is disconnected?
- State the earlier results used to prove the Intermediate-Value Property on a Connected Set.
- Suppose \(E\) is connected, \(f:E\to\mathbb{R}\) is continuous, and \(f\) never vanishes. What additional information about \(f\) at one point determines its sign everywhere?