Start by Checking the Definition
Many errors in connectedness arguments are not failures of advanced theory; they are failures to check the sets in a proposed separation. The two pieces must be nonempty, disjoint, relatively open, and together equal the entire space. Other common mistakes reverse a valid theorem about continuous images, or apply a union theorem without its shared-point hypothesis. We will check each issue directly from the definition.
For a subset \(A\) of a topological space \(X\), the subspace topology on \(A\) consists of all sets \(A\cap U\), where \(U\) is open in \(X\). A set open in \(A\) need not be open in \(X\).
Thus, when checking a proposed separation of \(A\), openness means relative openness in \(A\), and the union must equal \(A\), not merely most of it. Missing even one point invalidates the proposed separation.
Relative Openness Is Not the Same as Covering
Worked Example: A Missing Midpoint
Consider \(X=[0,1]\) with the subspace topology inherited from \(\mathbb{R}\). A proposed argument says that \(X\) is disconnected because it is the union of \(U=[0,\frac12)\) and \(V=(\frac12,1]\).
Both sets are indeed open in \(X\). Specifically, \(U=X\cap(-1,\frac12)\) and \(V=X\cap(\frac12,2)\), and the intervals on the right are open in \(\mathbb{R}\). The sets are nonempty and disjoint, but they do not cover \(X\): \(\frac12\in X\), while \(\frac12\notin U\) and \(\frac12\notin V\). So they are not a separation.
There is a second possible error here: calling \(U\) open in \(\mathbb{R}\). It is not, since it contains \(0\) but no open interval around \(0\) is contained in \(U\). It is open in \(X\), which is exactly what a separation of \(X\) requires. The corrected conclusion is not that these two sets separate \(X\), but that this particular attempt fails. To establish connectedness we need to rule out every possible separation.
Proof. The empty set and a one-point set have no pair of disjoint nonempty subsets, so neither has a separation. Now let \(J\) be an interval containing at least two points. Suppose, for contradiction, that \(J=U\cup V\) is a separation. Choose one point from each set. After exchanging the names \(U,V\) if necessary, there are \(u\in U\) and \(v\in V\) with \(u<v\). Since \(J\) is an interval, \([u,v]\subseteq J\). Let \(S=U\cap[u,v]\). This set is nonempty, contains \(u\), and is bounded above by \(v\); let \(c=\sup S\). Then \(c\in[u,v]\subseteq J\), so \(c\) belongs to exactly one of \(U,V\).
If \(c\in U\), then \(c\ne v\), because \(v\in V\). Since \(U\) is open in \(J\), some \(\delta>0\) satisfies \((c-\delta,c+\delta)\cap J\subseteq U\). Choose \(t\) with \(c<t<\min(c+\delta,v)\). Then \(t\in U\cap[u,v]=S\), contradicting that \(c\) is an upper bound for \(S\). If \(c\in V\), then \(c\ne u\), because \(u\in U\). Relative openness gives \(\delta>0\) such that \((c-\delta,c+\delta)\cap J\subseteq V\). By the definition of supremum, there is \(s\in S\) with \(c-\delta<s\le c\). Since \(c\notin U\), \(s\ne c\), so \(s\in V\), contradicting \(s\in S\subseteq U\). Both possibilities are impossible; therefore no separation exists. \(\square\)
Do Not Omit the Intermediate Point
The midpoint example is a useful warning: two relatively open pieces can be disjoint without being a separation. A valid disconnection proof must verify the covering condition explicitly. Conversely, to prove an interval connected, it is not enough to refute one proposed split; the theorem above rules out all splits by using the order structure of the real line.
Worked Example: Correcting a Separation Claim
Let \(E=[0,1]\cup[2,3]\subseteq\mathbb{R}\), with the subspace topology. The two sets \(C=[0,1]\) and \(D=[2,3]\) are intervals, hence connected by the theorem. A mistaken argument might say that a union of connected sets must be connected, and therefore \(E\) is connected. That conclusion does not follow: connectedness of the two pieces alone is not enough.
In fact, \(C=E\cap(-1,\frac32)\) and \(D=E\cap(\frac32,4)\). Thus both are open in \(E\). They are nonempty, disjoint, and \(E=C\cup D\). They form a separation, so \(E\) is disconnected. Notice the distinction from the earlier midpoint attempt: here there is no point of \(E\) between the two pieces left out of the union.
Check: \(C\cap D=\varnothing\), \(C\cup D=[0,1]\cup[2,3]=E\), and each set is the intersection of \(E\) with an open real interval. All four requirements for a separation are met.
Continuous Images and Continuous Preimages
A continuous map sends a connected space to a connected image. The direction matters: the theorem concerns the image of a connected domain, not the preimage of a connected target.
Proof. Suppose \(f(X)\) has a separation \(P,Q\). Continuity means that the preimage of every open set is open; applied to the relatively open sets \(P,Q\) in \(f(X)\), this gives relatively open sets \(f^{-1}(P),f^{-1}(Q)\) in \(X\). Both are nonempty because \(P,Q\subseteq f(X)\). They are disjoint, and they cover \(X\), since \(P,Q\) partition \(f(X)\). They would therefore separate \(X\), contradicting its connectedness. Hence \(f(X)\) is connected. \(\square\)
Worked Example: A Connected Target with a Disconnected Preimage
Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^2\), and let \(K=[1,4]\). The function is continuous, and \(K\) is an interval, hence connected. But the preimage is not connected:
Indeed, \(1\le x^2\le4\) is equivalent to \(1\le |x|\le2\), which gives exactly the two displayed intervals. Set \(P=[-2,-1]\) and \(Q=[1,2]\). In the subspace \(f^{-1}(K)\), these sets are relatively open because \(P=f^{-1}(K)\cap(-3,0)\) and \(Q=f^{-1}(K)\cap(0,3)\). They are nonempty, disjoint, and their union is \(f^{-1}(K)\), so they form a separation.
Check: \(f^{-1}(K)\) contains \(-2,-1,1,2\), since their squares are \(4,1,1,4\), respectively; it excludes \(0\), since \(0^2=0\notin[1,4]\). This confirms why the preimage has two separated pieces. The valid continuous-image theorem does not assert that this preimage is connected.
What a Shared Point Does—and Does Not—Guarantee
Proof. Suppose \(C\cup D\) had a separation \(P,Q\). The sets \(C\cap P\) and \(C\cap Q\) are relatively open in \(C\), are disjoint, and cover \(C\). Since \(C\) is connected, at least one must be empty; because \(C\) is nonempty, all of \(C\) lies in one side, say \(P\). The same argument shows that all of \(D\) lies in \(P\) or all of \(D\) lies in \(Q\). Since \(C\cap D\ne\varnothing\) and \(C\subseteq P\), \(D\) cannot lie in \(Q\). Thus \(D\subseteq P\), so \(C\cup D\subseteq P\), contradicting that \(Q\) is nonempty. Therefore \(C\cup D\) is connected. \(\square\)
The shared point is a sufficient condition for connectedness of the union, not a necessary one. Without it, either outcome is possible.
Worked Example: Disjoint Connected Sets with Connected Union
In \(\mathbb{R}\), take \(C=[0,\frac12)\) and \(D=[\frac12,1]\). These are intervals, so each is connected. They are disjoint: \(\frac12\notin C\), while \(\frac12\in D\). Their union is \([0,1]\), which is an interval and therefore connected.
This example does not contradict the common-point theorem: that theorem provides a sufficient hypothesis, and here its shared-point hypothesis is absent. It does show why “the sets must share a point for their union to be connected” is an invalid converse. A disjoint union of connected sets can be connected, as it is here, or disconnected, as in the preceding example.
Check: The union includes every real number from \(0\) to \(1\), including \(\frac12\) through \(D\), and includes no number outside that interval. Thus \(C\cup D=[0,1]\), not \([0,1]\setminus\{\frac12\}\).
A Reliable Checklist
When a connectedness argument seems plausible, check its logical direction and its sets before relying on intuition. For a proposed separation, verify nonemptiness, disjointness, relative openness, and equality of the union with the whole space. For a continuity argument, check whether the claim concerns an image or a preimage. For a union, check whether the hypotheses actually include a shared point; if not, inspect the union itself rather than assuming either outcome.
Common mistake. “\(C\) and \(D\) are connected, so \(C\cup D\) is connected.” This claim is false: for \(C=[0,1]\) and \(D=[2,3]\), both sets are connected intervals, but the relatively open sets \(C=E\cap(-1,\frac32)\) and \(D=E\cap(\frac32,4)\), where \(E=C\cup D\), form a separation of \(E\). The correction is to use the common-point theorem only when \(C\cap D\ne\varnothing\), or to prove connectedness or disconnectedness of a particular union directly.
Check Your Understanding
For each question, check all separation conditions rather than relying on the names of the sets.
- In \(X=[0,1]\), are \([0,\frac13)\) and \((\frac13,1]\) a separation? Which condition fails?
- Give a continuous function and a connected subset of its codomain whose preimage is disconnected.
- Why does the common-point theorem apply to two connected sets with a shared point? Identify where the shared point is used in the proof.
- Can two disjoint connected subsets of \(\mathbb{R}\) have a connected union? Give an example and verify the union.
- For \(E=[0,1]\cup[2,3]\), write each piece as \(E\cap U\) for an open set \(U\subseteq\mathbb{R}\), and use this to verify relative openness.