Components as a Local Test
Connectedness asks whether a space can be separated into two disjoint, nonempty open pieces. Local connectedness asks a different question: around each point, can we find connected open neighborhoods as small as desired? The connection between these questions is captured by the components of open subsets. A component is a maximal connected piece; the characterization proved here says that local connectedness is exactly the condition that these pieces are open whenever the subset under consideration is open.
All subsets below have the subspace topology inherited from the space in which they lie. A subset is connected if it is connected with this topology: it cannot be written as the union of two disjoint, nonempty sets open in the subset.
For any \(x\in Y\), there is a unique component of \(Y\) containing \(x\). To see why, take the union of all connected subsets of \(Y\) that contain \(x\). This union is connected: in any proposed separation of the union, the side containing \(x\) must contain each of the connected subsets in its entirety. The union is therefore connected and maximal. Any two components that meet must coincide, since their union is connected and maximality forces each to contain the other.
The Characterization
Proof. First suppose \(X\) is locally connected. Let \(U\) be open in \(X\), and let \(C\) be a component of \(U\). For any \(x\in C\), local connectedness gives a connected open set \(V\) with \(x\in V\subseteq U\). Since \(V\) meets \(C\), the union \(C\cup V\) is connected: a separation of the union would separate one of the connected sets \(C\) or \(V\), because they meet. As \(C\) is maximal among connected subsets of \(U\), it follows that \(V\subseteq C\). Thus every \(x\in C\) has an open neighborhood \(V\) contained in \(C\), so \(C\) is open in \(X\).
Conversely, suppose every component of every open subset of \(X\) is open. Let \(x\in X\), and let \(U\) be any open set containing \(x\). The component \(C\) of \(U\) containing \(x\) is connected by definition, is contained in \(U\), and is open in \(X\) by the assumption. Therefore \(C\) is a connected open neighborhood of \(x\) contained in \(U\). This works for every such \(x\) and \(U\), so \(X\) is locally connected. \(\square\)
The two directions use the same objects but in opposite ways. Local connectedness supplies small connected open neighborhoods, which must fit inside a component. In the reverse direction, the component itself supplies the required connected open neighborhood. The theorem concerns components of open subsets; it does not say that components of every subset must be open.
Worked Example: Components of an Open Set in the Real Line
Let \(\mathbb{R}\) have its standard topology, and take \(U=(-2,1)\cup(2,4)\). The components of \(U\) are \((-2,1)\) and \((2,4)\).
Each interval is connected, by the connectedness of real intervals. No connected subset of \(U\) can meet both intervals: if it did, its image under the inclusion into \(\mathbb{R}\) would be a connected subset containing a point below \(1\) and a point above \(2\), and would therefore have to contain \(3/2\). But \(3/2\notin U\). Here we use the elementary fact that a connected subset of \(\mathbb{R}\) containing \(a<b\) contains every point between them: omitting an intermediate point would separate the subset into its portions below and above that point. Thus the two intervals are indeed maximal connected subsets of \(U\), and both are open in \(\mathbb{R}\).
Check: the intervals are disjoint, and \(U\) is their union. Every point of either component has an open interval around it contained in that component. This example illustrates the theorem; the real line is locally connected because every open neighborhood of a point contains a smaller open interval around that point.
Connectedness and Local Connectedness Are Different
Connectedness is a global property; local connectedness requires suitable neighborhoods at every point. Neither property implies the other. The next examples check the distinction using the theorem.
Worked Example: A Disconnected but Locally Connected Space
Let \(X=\{a,b\}\) have the discrete topology \(\mathcal{T}=\mathcal{P}(X)\), so every subset is open. The space is disconnected because \(\{a\}\) and \(\{b\}\) are disjoint nonempty open sets whose union is \(X\).
Every subset of \(X\) is also connected or a singleton: a singleton cannot be separated into two nonempty subsets. For any open \(U\subseteq X\), its components are therefore its singleton subsets when \(U\ne\varnothing\); when \(U=\varnothing\), it has no components. Every such component is open in \(X\). By the open-components characterization, \(X\) is locally connected despite being disconnected.
Check: for either point, its singleton is a connected open neighborhood contained in any open set containing that point.
Worked Example: A Connected but Not Locally Connected Space
Consider the topologist’s sine curve with its subspace topology from the standard topology on \(\mathbb{R}^2\):
This space is connected, as established in the earlier material on connectedness of the topologist’s sine curve. We show that it is not locally connected by finding a component of an open subset that is not open. Let \(U=X\cap\{(x,y)\in\mathbb{R}^2:y>0\}\). Since the upper half-plane is open in \(\mathbb{R}^2\), \(U\) is open in \(X\). Set \(V=\{0\}\times(0,1]\), and consider \(p=(0,1/2)\in V\).
The set \(V\) is connected: it is the image of the interval \((0,1]\) under the continuous map \(y\mapsto(0,y)\). We claim that \(V\) is the component of \(U\) containing \(p\). Suppose a connected subset \(S\subseteq U\) contains \(p\) and also a point \(q\) with positive first coordinate \(a\). The coordinate projection \(\pi_1(x,y)=x\) is continuous, so \(\pi_1(S)\) is connected. A continuous image of a connected space is connected, since a separation of the image would pull back to a separation of the original space. Because \(\pi_1(S)\) contains \(0\) and \(a\), it must contain every real number between them. But there is \(t\in(0,a)\) with \(\sin(1/t)<0\): choose \(n\in\mathbb{N}\) large enough that \(t=1/(3\pi/2+2\pi n)<a\). No point of \(U\) has first coordinate \(t\), a contradiction. Thus every connected subset of \(U\) containing \(p\) lies on the vertical axis. The vertical part of \(U\) is exactly \(V\), so \(V\) is the component containing \(p\).
Although \(V\) is a component of the open set \(U\), it is not open in \(X\). For each sufficiently large \(n\in\mathbb{N}\), define \(x_n=1/(2\pi n+\pi/6)\). Then \(0<x_n\leq1\) and \(\sin(1/x_n)=1/2\), so \(q_n=(x_n,1/2)\in U\setminus V\). Also \(q_n\to p\) in \(\mathbb{R}^2\). Every open neighborhood of \(p\) in \(X\) therefore contains some \(q_n\), and hence cannot be contained in \(V\). The component \(V\) is not open, so the characterization shows that \(X\) is not locally connected.
Check: the points \(q_n\) really lie in \(U\setminus V\): their second coordinate is \(1/2>0\), and their first coordinate is positive rather than zero. Their convergence to \(p\) follows from \(x_n\to0\) and their constant second coordinate.
A Common Mistake
Pitfall. Connectedness of \(X\) does not imply that every open subset \(U\subseteq X\) is connected. In the sine-curve example, one might incorrectly reason, “\(X\) is connected, so the component of \(U\) containing \(p\) is all of \(U\).” That conclusion confuses \(X\) with its open subset \(U\). For example, \(U\) contains \(p=(0,1/2)\) and also \((1/(\pi/2),1)\), since \(\sin(\pi/2)=1\); the latter point is not in \(V\). The component of \(U\) containing \(p\) is \(V\), not \(U\). The theorem detects the failure of local connectedness because this component is not open.
It is also important not to reverse the theorem’s quantifiers: it requires that components of every open subset be open. Checking only the components of \(X\) itself is not enough. A connected space has just one component, namely \(X\), which is open; that fact alone says nothing about components of smaller open subsets.
Check Your Understanding
Use the definition of components and the open-components characterization to justify your answers.
- In the discrete topology on a set \(X\), describe the components of an arbitrary open subset \(U\), including \(U=\varnothing\). Are they open in \(X\)?
- Explain why the proof from local connectedness to open components uses a connected open neighborhood contained in \(U\), rather than merely a connected neighborhood.
- In the sine-curve example, why does a connected subset containing \(p\) and a point with positive first coordinate lead to a contradiction?
- Give an example of a connected space that is locally connected, and explain how its connected open neighborhoods can be chosen.
- Why does knowing that the component of \(X\) is open fail to establish that \(X\) is locally connected?