What Counterexamples Reveal
Connectedness is preserved by several important operations, but it is not preserved by every operation involving sets and functions. The Connectedness of a Product theorem from the previous tutorial depended on specific hypotheses and a specific proof structure. Here we test some tempting claims: that unions and intersections of connected sets are connected, that subsets of connected sets are connected, or that a continuous image can be connected only when its domain is connected.
A counterexample must satisfy the hypotheses of the claim while making its conclusion fail. In particular, showing that two sets are connected is not enough to show their union is connected: we must also check how they fit together. We will use the theorem that a positive distance between two nonempty sets gives a separation, and the theorem that a connected subset lies on one side of a separation. Both were established earlier in this course.
A Union Can Fail to Be Connected
If two connected sets overlap, their union is connected by the Union of Pairwise-Intersecting Connected Sets theorem. But connected pieces need not overlap. A gap between them can leave a separation of their union.
Worked Example: Two Disjoint Intervals
Let \(A=[-3,-1]\) and \(B=[2,5]\). Each is an interval, so each is connected. They are nonempty and disjoint, and for any \(a\in A\) and \(b\in B\), \(|a-b|=b-a\geq 2-(-1)=3\). Thus \(\operatorname{dist}(A,B)\geq3>0\). The Positive Distance Gives a Separation theorem implies that \(A\) and \(B\) form a separation of \(E=A\cup B\). Therefore \(E\) is disconnected.
This does not conflict with the union theorem: \(A\cap B=\varnothing\), so its intersection hypothesis is not met. The example also shows why connectedness of each piece alone gives no conclusion about their union. The way the pieces intersect, or fail to intersect, matters.
A useful proof-checking question is therefore: what links the connected sets in the proposed union? A common point, or pairwise intersections as in the cited theorem, can provide such a link. Merely knowing that every individual set is connected does not.
An Intersection Can Fail to Be Connected
Intersections require a different caution. The Intersections of Connected Sets theorem established earlier has a common-intersection condition. Without the relevant condition, connected sets can have a disconnected intersection. In the real line, however, this particular failure is impossible: the Characterization of Connected Subsets of \(\mathbb{R}\) says that connected subsets are intervals, and an intersection of intervals is an interval or is empty. The following example therefore uses the plane.
Proof. Let \(A=\{(x,0):-1\leq x\leq1\}\) be a horizontal line segment, and let \(B=\{(x,x^2-\tfrac14):-1\leq x\leq1\}\) be a parabolic arc. The map \(x\mapsto(x,0)\) is continuous on the interval \([-1,1]\), as is the map \(x\mapsto(x,x^2-\tfrac14)\). The interval is connected, so the Continuous Images of Connected Sets theorem shows that both \(A\) and \(B\) are connected.
A point belongs to \(A\cap B\) exactly when its second coordinate is zero and \(x^2-\tfrac14=0\). Solving gives \(x^2=\tfrac14\), so \(x=-\tfrac12\) or \(x=\tfrac12\). Substitution verifies both: \((-\tfrac12)^2-\tfrac14=\tfrac14-\tfrac14=0\), and \((\tfrac12)^2-\tfrac14=\tfrac14-\tfrac14=0\). Hence \(A\cap B=\{(-\tfrac12,0),(\tfrac12,0)\}\).
These two points are separated: the singleton containing the first and the singleton containing the second are nonempty, disjoint, and have distance \(1>0\). Thus the intersection is disconnected, although each of the original sets is connected. \(\square\)
The location in \(\mathbb{R}^2\) is essential. If \(C\) and \(D\) are connected subsets of \(\mathbb{R}\), then they are intervals or empty. If their intersection is nonempty and contains \(u<v\), every \(w\) with \(u<w<v\) belongs to both \(C\) and \(D\), by the interval property. Thus the intersection is an interval, and is connected. The example warns against transferring a conclusion about the real line to general metric spaces—or a conclusion about the plane back to the real line.
A Continuous Image Need Not Preserve Disconnectedness
The Continuous Images of Connected Sets theorem gives a one-way implication: the continuous image of a connected set is connected. Its converse is false. A function can map two disconnected pieces of its domain onto the same connected range.
Proof. Take \(E=[0,1]\cup[3,4]\), and define \(f:E\to\mathbb{R}\) by \(f(x)=x\) on \([0,1]\) and \(f(x)=x-3\) on \([3,4]\). The two intervals are nonempty and have distance \(2\), so they form a separation of \(E\). Hence \(E\) is disconnected.
We check continuity on the domain \(E\), using its relative metric. Given \(x\in E\) and \(\varepsilon>0\), set \(\delta=\min\{1,\varepsilon\}\). If \(y\in E\) and \(|y-x|<\delta\), then \(x\) and \(y\) must belong to the same interval: points in different intervals are at least \(2\) apart. On either interval, the defining formula has slope \(1\), so \(|f(y)-f(x)|=|y-x|<\varepsilon\). This proves that \(f\) is continuous at every \(x\in E\).
On the first interval, \(f([0,1])=[0,1]\). On the second, \(f([3,4])=[0,1]\), since \(x-3\) ranges from \(0\) to \(1\). Therefore \(f(E)=[0,1]\), which is connected. The domain is disconnected, but its continuous image is connected. \(\square\)
Continuity does not require a function to keep distinct parts of its domain distinct in the range. Here both components are sent onto the same interval. The conclusion that the image is connected therefore cannot be used to infer that the domain is connected.
Worked Example: A Disconnected Dense Set Can Have Connected Closure
Let \(E=\mathbb{R}\setminus\{0\}\). The sets \(A=(-\infty,0)\) and \(B=(0,\infty)\) are nonempty and partition \(E\). Each is relatively open in \(E\), because each is the intersection of \(E\) with an open ray; since they are complements of one another in \(E\), each is also relatively closed. Thus they form a separation, and \(E\) is disconnected.
Every open interval around \(0\) contains nonzero real numbers, and every point other than \(0\) already belongs to \(E\). Hence \(\overline{E}=\mathbb{R}\). The real line is an interval and is connected. So the closure of a disconnected set can be connected. This does not contradict the Closure of a Connected Set theorem: that theorem says a connected set has connected closure, not that closure preserves disconnectedness.
A Connected Set Can Have Disconnected Subsets
Connectedness is a property of a set as a whole; it need not pass to every subset. Removing a point can split an interval into two pieces. The pieces may approach the removed point arbitrarily closely, so this example does not rely on positive distance between them.
Worked Example: Removing the Middle Point
Let \(J=[-1,1]\), which is connected because it is an interval, and let \(F=J\setminus\{0\}=[-1,0)\cup(0,1]\). Define \(A=[-1,0)\) and \(B=(0,1]\). They are nonempty, disjoint, and cover \(F\). Also, \(A=F\cap(-\infty,0)\) and \(B=F\cap(0,\infty)\), so both are relatively open in \(F\). Because they are complements of each other in \(F\), they are relatively closed as well. Therefore \(A\) and \(B\) form a separation of \(F\), which is disconnected.
The distance between the two pieces is \(0\), not positive: for every \(\varepsilon>0\), the points \(-\varepsilon/2\) and \(\varepsilon/2\) belong to \(A\) and \(B\), respectively, when \(0<\varepsilon\leq2\), and their distance is \(\varepsilon\). The separation comes from the relative topology, not from a gap of positive width. This is a useful reminder that positive distance is a sufficient way to prove separation, not a necessary condition.
How to Use These Counterexamples
The examples distinguish several claims that can sound similar but have different hypotheses. A union may be connected when its connected pieces intersect in the required way, but not merely because each piece is connected. An intersection of connected sets can be disconnected in a metric space, while the interval characterization prevents this for subsets of \(\mathbb{R}\). A continuous image of a connected domain is connected, but a connected image does not force a connected domain. Nor does connectedness pass automatically to subsets or to closures in the reverse direction.
When assessing a proposed rule, identify the space and the operation first. Then check the hypotheses of the relevant theorem and, if a claim seems too broad, look for a simple candidate: intervals with a gap for unions, crossing curves for intersections, or a function that identifies distinct pieces for images. Finally, verify the failure using the definition—usually by displaying a separation or by calculating the image. The next tutorial studies the structure of connected subsets of \(\mathbb{R}\), which explains why some counterexamples work on the plane but cannot work on the line.
Check Your Understanding
Use the examples and the earlier connectedness theorems to answer the following questions.
- Why does the Union of Pairwise-Intersecting Connected Sets theorem not imply that \([-3,-1]\cup[2,5]\) is connected?
- What are the two points in the intersection of the horizontal segment and parabolic arc, and what calculation verifies that they belong to both sets?
- Why can two connected subsets of \(\mathbb{R}\) not have a nonempty disconnected intersection?
- In the function example, how can continuity be checked at points in different components of the domain?
- Why does the connectedness of the closure of \(\mathbb{R}\setminus\{0\}\) not contradict the Closure of a Connected Set theorem?
- In the middle-point example, why does a separation exist even though the two pieces have distance zero?