Why Components of Open Sets Are Open
A component is defined by maximality among connected subsets, which by itself says nothing about whether that subset is open. Local connectedness supplies the missing local information: near each point, there are connected open neighborhoods. If the point lies in an open set \(U\), one can choose such a neighborhood inside \(U\). It then lies in the same component as the point, giving an open neighborhood contained in that component.
Throughout, a subset of a space is called connected when it is connected in its subspace topology. A space is connected if it cannot be expressed as the union of two disjoint, nonempty open subsets. A component of a space is a maximal connected subset: it is connected and is not properly contained in any other connected subset of that space.
The containment \(V\subseteq U\) is essential. A connected open neighborhood in the whole space that extends outside \(U\) cannot be used as a connected subset of \(U\). The theorem will apply the definition with \(U\) chosen to be the open set whose components we are studying.
Components and Connected Sets That Meet Them
Before proving the openness result, we establish the facts about components it uses. The construction below also ensures that every point belongs to a component, without assuming that components are already known to exist.
Proof. Let \(C_y\) be the union of all connected subsets of \(Y\) containing \(y\). The family being unioned is nonempty because the singleton \(\{y\}\) is connected: it cannot be split into two disjoint, nonempty subsets. We show that \(C_y\) is connected. Suppose instead that \(C_y=P\cup Q\), where \(P\) and \(Q\) are disjoint, nonempty, relatively open subsets of \(C_y\). Since \(y\in C_y\), it belongs to one of them; suppose \(y\in P\). For any connected set \(A\) in the union defining \(C_y\), the sets \(A\cap P\) and \(A\cap Q\) are disjoint and relatively open in \(A\), and together cover \(A\). Also \(y\in A\cap P\), so \(A\cap P\) is nonempty. If \(A\cap Q\) were nonempty, these two sets would disconnect \(A\). Thus \(A\cap Q=\varnothing\) for every such \(A\), which implies \(Q=\varnothing\), a contradiction. Hence \(C_y\) is connected. Any connected subset of \(Y\) containing \(C_y\) contains \(y\), so it is among the sets included in the union and is contained in \(C_y\). Therefore \(C_y\) is maximal connected and is a component containing \(y\). \(\square\)
We will also need the fact that two connected sets with a point in common have connected union. Here is the argument. If \(A\) and \(B\) are connected and \(A\cap B\ne\varnothing\), suppose their union had a separation into disjoint, nonempty, relatively open sets \(P\) and \(Q\). Each of \(A\) and \(B\), being connected, must lie wholly in one side of this separation. A common point belongs to both \(A\) and \(B\), so both sets must lie in the same side. Their union would then miss the other side, contradicting that it is nonempty. Thus \(A\cup B\) is connected.
Proof. The sets \(C\) and \(E\) are connected and meet, so the argument above shows that \(C\cup E\) is connected. It contains \(C\), and \(C\) is maximal among connected subsets of \(Y\). Consequently \(C\cup E=C\), which gives \(E\subseteq C\). \(\square\)
The lemma is the key use of maximality: to place a whole connected neighborhood inside a component, it is enough to know that the neighborhood meets that component. The lemma does not say that every connected set meeting a component is open. Openness will come from the local connectedness hypothesis.
The Openness Theorem
Proof. If \(U=\varnothing\), it has no components, so the assertion holds. Otherwise, let \(C\) be a component of \(U\), and take any \(x\in C\). Since \(U\) is open in \(X\), it is an open neighborhood of \(x\). By local connectedness, there is a connected open set \(V\) in \(X\) with \(x\in V\subseteq U\). Thus \(V\) is a connected subset of \(U\) and meets \(C\), since \(x\in V\cap C\). Applying the lemma in the space \(U\) gives \(V\subseteq C\). We have therefore found, for every \(x\in C\), an open neighborhood \(V\) in \(X\) contained in \(C\). The union of these neighborhoods over all \(x\in C\) is \(C\), so \(C\) is open in \(X\). Since \(C\subseteq U\), it is also open in \(U\) with the subspace topology. \(\square\)
The proof uses the openness of \(U\) at exactly one point: it makes \(U\) an open neighborhood to which the local connectedness condition applies. The conclusion is stronger than relative openness because the connected neighborhoods produced are open in \(X\) itself. The result does not assert that components of arbitrary subsets of a locally connected space are open.
Worked Applications
Worked Example: Two Components in the Real Line
Let \(X=\mathbb{R}\) with its standard topology and let \(U=(-2,0)\cup(1,3)\). We identify the components of \(U\) and check that they are open.
The sets \(A=(-2,0)\) and \(B=(1,3)\) are disjoint, nonempty, and open in \(U\); each is the complement of the other in \(U\), so each is also closed in \(U\). A connected subset \(E\subseteq U\) cannot meet both \(A\) and \(B\): if it did, \(E\cap A\) and \(E\cap B\) would be disjoint, nonempty, relatively open subsets covering \(E\), a separation. Thus every connected subset of \(U\) lies entirely in one of these two intervals.
Each open interval in \(\mathbb{R}\) is connected, by the standard connectedness theorem for intervals. Therefore neither \(A\) nor \(B\) can be enlarged to a connected subset of \(U\): any such enlargement meeting the interval cannot include points from the other interval. Hence the components of \(U\) are exactly \(A\) and \(B\).
Check: \(A\cup B=U\), and both intervals are open in \(\mathbb{R}\), so each component is open in \(X\), as the theorem predicts. The gap \([0,1]\) is not part of \(U\); the proof of maximality uses the separation of \(U\) into its two relatively open pieces, not a claim that the gap belongs to either component.
Worked Example: An Open Set with Infinitely Many Components
Let \(X=\mathbb{R}\) with its standard topology, and set \(U=\bigcup_{n\in\mathbb{N}}(2n,2n+1)\). Determine its components.
For each \(n\in\mathbb{N}\), write \(I_n=(2n,2n+1)\). Each \(I_n\) is open in \(\mathbb{R}\), so \(U\), their union, is open. Each \(I_n\) is connected by the interval theorem. If \(m\ne n\), the intervals are disjoint and separated by points not in \(U\). More explicitly, \(I_n\) and \(U\setminus I_n=\bigcup_{m\ne n}I_m\) are both open in \(U\). A connected subset of \(U\) meeting \(I_n\) and \(U\setminus I_n\) would therefore be separated by its intersections with those two sets.
It follows that a connected subset meeting \(I_n\) cannot meet any other interval. Since \(I_n\) itself is connected, it is a maximal connected subset of \(U\). Thus the components are precisely the sets \(I_n\), one for each \(n\in\mathbb{N}\). Each is open in \(X\) and in \(U\). This example shows that the theorem allows infinitely many components; it does not require a finite component list.
Check: for every \(n\), \(I_n\) is connected, and the displayed separation prevents any connected subset of \(U\) containing \(I_n\) from adding a point in a different interval. These are the two requirements—connectedness and maximality—for being a component.
Worked Example: Open Sets in a Discrete Space
Let \(X\) be a set with the discrete topology, in which every subset is open, and take any \(U\subseteq X\). Find the components of \(U\).
Every singleton \(\{x\}\) with \(x\in U\) is connected, since it has no two nonempty subsets that could form a separation. Now let \(E\subseteq U\) contain at least two points. Choose \(x\in E\). The subsets \(\{x\}\) and \(E\setminus\{x\}\) are nonempty, disjoint, and cover \(E\). They are relatively open in \(E\), because every subset of \(X\) is open. Thus they separate \(E\), so \(E\) is not connected.
Consequently the components of \(U\) are exactly its singletons. Each \(\{x\}\) is open in \(X\), so every component is open even when \(U\) is not open. This example is consistent with the theorem but also highlights that the theorem's hypotheses need not be necessary in every individual case.
Check: if \(U=\varnothing\), there are no points \(x\in U\), hence no singleton components. If \(U\ne\varnothing\), each point belongs to exactly one singleton component. The conclusion includes both cases.
A Finite Non-Hausdorff Example
The theorem does not require the space to be Hausdorff. The following finite example checks the neighborhood argument directly in a space where distinct points need not have disjoint neighborhoods.
Worked Example: Two Sierpinski Pieces
Let \(X=\{a,b,c,d\}\). Give \(X\) the topology consisting of all unions of sets from \(\{\varnothing,\{a\},\{a,b\}\}\) and \(\{\varnothing,\{c\},\{c,d\}\}\). In particular, the open sets in each of the two pieces \(\{a,b\}\) and \(\{c,d\}\) are respectively \(\varnothing,\{a\},\{a,b\}\) and \(\varnothing,\{c\},\{c,d\}\).
This space is locally connected. At \(a\), the open singleton \(\{a\}\) is a connected neighborhood; at \(b\), the open neighborhood \(\{a,b\}\) is connected. For the latter claim, the only nonempty proper open subset of \(\{a,b\}\) is \(\{a\}\), whose complement \(\{b\}\) is not open in that subspace, so there is no separation. The same checks apply to \(c\) and \(d\): \(\{c\}\) is a connected open neighborhood of \(c\), and \(\{c,d\}\) is a connected open neighborhood of \(d\). These neighborhoods give a connected open neighborhood inside any open neighborhood of the point.
Take \(U=\{a,c,d\}\). It is open, since it is the union of the open sets \(\{a\}\) and \(\{c,d\}\). In the subspace \(U\), the sets \(\{a\}\) and \(\{c,d\}\) are disjoint, open, and cover \(U\). Each is connected: the first is a singleton, and the second has the two-point topology \(\varnothing,\{c\},\{c,d\}\), with no nonempty proper subset both open and closed. Thus \(U\) is disconnected. A connected subset meeting both pieces would be separated by its intersections with them, so these two connected pieces are exactly the components of \(U\). Both are open in \(X\), in agreement with the theorem.
The space is not Hausdorff: every open neighborhood of \(b\) contains \(a\), and every open neighborhood of \(a\) contains \(a\), so any such two neighborhoods intersect. Nevertheless the component conclusion holds. Hausdorff separation is not among the hypotheses or the tools used in the proof.
A Common Mistake: Choosing a Neighborhood Too Large
A tempting but invalid proof is: “For \(x\in C\), choose a connected open neighborhood \(V\) of \(x\). Since it meets \(C\), maximality gives \(V\subseteq C\).” The problem is that the neighborhood may not lie in \(U\), so it need not be a connected subset of the space \(U\) whose component is \(C\). Maximality of \(C\) in \(U\) cannot be applied to a set extending outside \(U\).
For a concrete illustration, let \(X=\mathbb{R}\), \(U=(0,1)\cup(2,3)\), and \(x=1/2\). The interval \(V=(-1,1)\) is a connected open neighborhood of \(x\) in \(X\), but \(V\not\subseteq U\). It is therefore not a candidate for the lemma in the space \(U\). Instead, local connectedness lets us choose \(W=(1/4,3/4)\), which is connected and open in \(X\), and satisfies \(x\in W\subseteq U\). Now \(W\) is a connected subset of \(U\) meeting the component containing \(x\), so the lemma puts \(W\) inside that component.
The argument has two distinct ingredients. Maximality says that a connected set meeting a component lies within it; local connectedness, applied inside the open set, supplies an open neighborhood of every point of the component. Together they prove openness. Neither ingredient alone gives the conclusion.
Check Your Understanding
Use the definition of a component and the neighborhood argument to answer the following.
- In the theorem, where is the assumption that \(U\) is open used?
- Explain why a connected set \(E\) that meets a component \(C\) of \(U\) must be contained in \(C\).
- Let \(U=(-3,-1)\cup(1,2)\) in the standard real line. What are its components, and what prevents a connected subset from meeting both?
- In a discrete space, why can no subset containing at least two points be connected?
- In the finite example, which connected open neighborhood should be used at \(b\), and why is \(\{b\}\) not a suitable open neighborhood there?