Two Useful Moves in a Connectedness Proof
The previous tutorial used the Intermediate Value Theorem to turn endpoint information into an existence proof. Connectedness proofs often have a different shape: assume a separation exists, then show that some structural feature of the set makes the separation impossible. Two useful features are density and contact through a closure. They let us prove connectedness even when a set is larger than a familiar connected subset, or when two connected pieces do not actually meet.
Recall that a separation of a set \(F\) is a representation \(F=U\cup V\) by disjoint, nonempty sets \(U\) and \(V\) that are relatively open in \(F\). The Clopen Criterion for Connectedness and the Relative-Closure Criterion for a Separation, established earlier in this course, provide equivalent ways to work with this definition. Here we will use separations directly and make each contradiction explicit.
Write the set as the union of disjoint, nonempty relatively open pieces.
Connectedness forces that set to lie entirely in one piece of the separation.
Show that the other piece would have to contain points arbitrarily close to the first, contradicting relative openness.
Enlarging a Connected Set Without Losing Connectedness
A connected set may be enlarged by adding some or all of its limit points. The key restriction is that every added point must still be a limit of points from the original connected set. In that situation, a proposed separation cannot isolate the added points from the original set.
Proof. If \(E=\varnothing\), then \(\overline{E}=\varnothing\), so the inclusions force \(F=\varnothing\), which is connected. Suppose now that \(E\ne\varnothing\). Assume, for contradiction, that \(F=U\cup V\) is a separation. The sets \(E\cap U\) and \(E\cap V\) are disjoint, relatively open in \(E\), and have union \(E\). Since \(E\) is connected, they cannot both be nonempty. As \(E\ne\varnothing\), it follows that \(E\) lies entirely in one side; after interchanging the names of the sides if necessary, suppose \(E\subseteq U\).
Because \(V\) is nonempty, choose \(z\in V\). The inclusion \(F\subseteq\overline{E}\) implies \(z\in\overline{E}\). Since \(V\) is relatively open in \(F\), there is an \(r>0\) such that the open ball \(B(z,r)\) satisfies \(B(z,r)\cap F\subseteq V\). But \(z\in\overline{E}\) means that every open ball centered at \(z\) meets \(E\). In particular, choose \(e\in B(z,r)\cap E\). Since \(E\subseteq F\), this point belongs to \(B(z,r)\cap F\), so \(e\in V\). This contradicts \(E\subseteq U\) and \(U\cap V=\varnothing\). Thus \(F\) has no separation and is connected. \(\square\)
The proof uses more than the fact that \(E\) is contained in \(F\). The inclusion \(F\subseteq\overline{E}\) ensures that every point of \(F\), including a point in a putative second side of a separation, is approached by points of \(E\). Relative openness would then force a neighborhood around that point to exclude \(E\), which is impossible.
Worked Example: Adding Any Part of a Disk’s Boundary
Let \(E=\{(x,y)\in\mathbb{R}^2:x^2+y^2<1\}\), the open unit disk. This set is convex, so the line segment between any two of its points stays in \(E\); hence it is path connected and therefore connected. Its closure is the closed unit disk. Indeed, points outside the closed disk have a neighborhood disjoint from \(E\), while each boundary point \(p\), with \(\|p\|=1\), is the limit of the points \((1-1/n)p\in E\) for integers \(n\geq2\).
Now let \(S\) be any subset of the unit circle, with no restrictions on how many points it contains or how it is arranged, and put \(F=E\cup S\). Then
The theorem gives that \(F\) is connected. In particular, adding just one boundary point, adding a closed arc, or adding a highly scattered collection of boundary points does not disconnect the disk. The added points need not form a connected set of their own; it is their being limit points of the connected disk that matters.
Joining Connected Sets Through a Closure Contact
The union of two connected sets is connected when they intersect; this is a special case of the Union of Pairwise-Intersecting Connected Sets Theorem established earlier. Actual intersection is not always necessary. If a point of one connected set lies in the closure of the other, relative openness again prevents a separation from pulling the sets apart.
Proof. Suppose, to obtain a contradiction, that \(A\cup B=U\cup V\) is a separation. Restricting this separation to \(A\) shows that \(A\cap U\) and \(A\cap V\) cannot both be nonempty: if they were, they would be a separation of \(A\), contrary to its connectedness. Thus \(A\) lies entirely in \(U\) or entirely in \(V\). By interchanging the names of \(U\) and \(V\) if needed, assume \(A\subseteq U\).
Choose \(p\in\overline{A}\cap B\). Because \(B\) is connected, its intersections with \(U\) and \(V\) cannot both be nonempty; otherwise their restrictions would separate \(B\). If \(p\in U\), then \(B\) must lie entirely in \(U\). If instead \(p\in V\), relative openness of \(V\) in \(A\cup B\) gives an \(r>0\) such that \(B(p,r)\cap(A\cup B)\subseteq V\). But \(p\in\overline{A}\) guarantees some \(a\in B(p,r)\cap A\). Since \(A\subseteq U\), this point cannot belong to \(V\), contradicting the ball inclusion. Therefore \(p\notin V\), so \(p\in U\), and connectedness of \(B\) forces \(B\subseteq U\). We have obtained \(A\cup B\subseteq U\), contradicting the nonemptiness of \(V\). Hence \(A\cup B\) is connected. \(\square\)
Worked Example: Joining Two Pieces That Do Not Intersect
In \(\mathbb{R}^2\), let
Each set is connected: \(A\) and \(B\) are line segments or rays parameterized by intervals, and intervals are connected. They are disjoint, since every point of \(A\) has negative first coordinate while every point of \(B\) has first coordinate zero. Nevertheless, \((0,0)\in B\) and \((0,0)\in\overline{A}\), because the points \((-1/n,0)\in A\) converge to \((0,0)\). The closure-contact theorem therefore shows that \(A\cup B\) is connected. The two pieces meet in the limiting sense that is needed by the proof, even though their intersection is empty.
Why the Closure Contact Must Be in the Union
It is important not to weaken the closure-contact hypothesis to \(\overline{A}\cap\overline{B}\ne\varnothing\). A shared limit point that belongs to neither set may not prevent a separation. The next example checks the distinction.
Worked Example: A Shared Limit Point That Is Not Enough
Let \(A=\{(x,0):x<0\}\) and \(B=\{(x,0):x>0\}\). Both sets are connected, and their closures meet at the origin:
However, \((0,0)\notin A\cup B\), so neither \(\overline{A}\cap B\) nor \(A\cap\overline{B}\) is nonempty. Their union is disconnected. In fact, \(A\) and \(B\) are disjoint, nonempty, and relatively open in \(A\cup B\): each is the intersection of the union with one of the open half-planes \(x<0\) and \(x>0\). Thus they form a separation. The argument in the closure-contact theorem needs the contact point to belong to one of the sets, where relative openness of the separation can be applied.
These examples illustrate a general habit for proof writing: check where the limiting point lies, not just whether two closures intersect. To contradict relative openness, one needs a point in a side of the proposed separation and points of the other set arbitrarily close to it. A point omitted from both sets provides no such starting point.
A Practical Checklist
When a connectedness problem involves a larger set or a union, begin by identifying the exact structure available. If a known connected set \(E\) is contained in the set of interest \(F\), check whether every point of \(F\) lies in \(\overline{E}\). If two connected sets are to be joined, check whether they intersect or whether one contains a point in the closure of the other. Then assume a separation and use relative openness at the point that density or closure contact supplies.
These are sufficient conditions, not characterizations. A set may be connected even if it cannot be presented as an enlargement of a particular connected subset, and two connected sets may have a connected union for reasons not covered by the closure-contact criterion. A proof should use the condition it has actually verified, without claiming that it is necessary.
Check Your Understanding
Use the definitions and proof strategies from this tutorial to answer the following questions.
- If \(E\) is connected and \(E\subseteq F\subseteq\overline{E}\), what contradiction arises if a point of a proposed second side of a separation lies in \(F\)?
- Why must the density-enlargement theorem handle the case \(E=\varnothing\) separately?
- In the closure-contact theorem, why does a point \(p\in\overline{A}\cap B\) contradict the possibility that \(p\) lies in the side opposite \(A\)?
- For the two rays on opposite sides of the origin, why does \(\overline{A}\cap\overline{B}\ne\varnothing\) fail to establish connectedness of their union?
- Give an example of a set \(F\) obtained by adding points to the open unit disk that the dense-enlargement theorem guarantees is connected.