Connected Neighborhood Bases

Local Connectedness as a Neighborhood-Basis Condition

Connectedness is a global property: it asks whether a space can be separated into two disjoint nonempty open pieces. Local connectedness asks a different question at each point: inside every open neighborhood, can we find a smaller connected open neighborhood? The neighborhood-basis formulation makes this local requirement precise and gives a practical test, especially for subspaces of Euclidean space.

Throughout, a neighborhood means an open set containing the point in question. A subset \(C\) of a space is connected if it cannot be written as the union of two disjoint, nonempty sets that are open in the subspace topology on \(C\). In particular, when testing connectedness or openness in a subspace, we use that subspace's topology.

The connected open subsets in this definition must be open in \(X\), not merely in a larger space containing \(X\). The pointwise criterion below says exactly how to find them: given \(x\) and an open neighborhood \(U\), find a connected open set between \(x\) and \(U\).

Proof. Suppose first that \(X\) is locally connected. Fix \(x\in X\), and let \(U\) be any open neighborhood of \(x\). By the definition of local connectedness, \(U\) is a union of connected open subsets of \(X\). Since \(x\in U\), at least one of those subsets, say \(V\), contains \(x\). It is connected and open, and \(x\in V\subseteq U\). Thus the family of all connected open neighborhoods of \(x\) is a neighborhood basis at \(x\).

Conversely, suppose every \(x\in X\) has a neighborhood basis of connected open sets. Let \(U\subseteq X\) be open. For each \(x\in U\), the basis condition gives a connected open set \(V_x\) with \(x\in V_x\subseteq U\). Consequently \(U=\bigcup_{x\in U}V_x\): every \(V_x\) lies in \(U\), and every point \(x\in U\) lies in its corresponding \(V_x\). If \(U=\varnothing\), it is the union of the empty family. Hence every open set is a union of connected open subsets, and \(X\) is locally connected. \(\square\)

This proof also explains the strength of the criterion. It is not enough that a point lie in some connected neighborhood; there must be a connected open neighborhood inside each prescribed open neighborhood. That containment is what makes the sets a basis.

Intervals with the Subspace Topology

An interval \(J\subseteq\mathbb{R}\) is given the subspace topology inherited from the standard topology on \(\mathbb{R}\). Thus its open sets have the form \(J\cap O\), where \(O\) is open in \(\mathbb{R}\). An interval is locally connected: around any point, intersect a sufficiently small open interval in \(\mathbb{R}\) with \(J\). Such intersections are open in \(J\), and the interval property ensures they are intervals, hence connected.

The same reasoning works at every point of any interval, including its endpoints. Given a relatively open neighborhood \(U=J\cap O\) of \(x\), choose \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq O\). Then \(J\cap(x-\varepsilon,x+\varepsilon)\) is a relatively open interval containing \(x\) and contained in \(U\). The connectedness of intervals in \(\mathbb{R}\) is the standard connectedness result from the earlier material on connectedness.

Convex Subspaces of Euclidean Space

A subset \(A\subseteq\mathbb{R}^n\) is convex if, whenever \(a,b\in A\) and \(t\in[0,1]\), the point \((1-t)a+tb\) belongs to \(A\). Give \(A\) the subspace topology from the Euclidean topology on \(\mathbb{R}^n\). Convexity lets us build connected neighborhoods by intersecting \(A\) with small Euclidean balls.

Proof. If \(A=\varnothing\), its only open subset is empty, which is the union of the empty family of connected open sets. Now suppose \(A\ne\varnothing\). Fix \(x\in A\), and let \(U\) be an open neighborhood of \(x\) in \(A\). By the definition of the subspace topology, \(U=A\cap O\) for some open \(O\subseteq\mathbb{R}^n\). Since \(x\in O\), there is \(r>0\) with \(B(x,r)\subseteq O\), where \(B(x,r)=\{y\in\mathbb{R}^n:\|y-x\|<r\}\). Set \(V=A\cap B(x,r)\). Then \(V\) is open in \(A\), \(x\in V\), and \(V\subseteq U\).

It remains to prove that \(V\) is connected. If \(a,b\in V\) and \(t\in[0,1]\), convexity of \(A\) gives \((1-t)a+tb\in A\). Also, the triangle inequality gives \(\|(1-t)a+tb-x\|\leq(1-t)\|a-x\|+t\|b-x\|<(1-t)r+tr=r\). Thus the whole line segment from \(a\) to \(b\) lies in \(V\). The map \(\gamma:[0,1]\to V\) defined by \(\gamma(t)=(1-t)a+tb\) is continuous and joins \(a\) to \(b\); if \(a=b\), the constant map does so. Hence \(V\) is path-connected. The standard result that path-connected spaces are connected, established in the earlier material on path-connectedness, now gives that \(V\) is connected. We have found a connected open neighborhood inside every \(U\), so the criterion proves that \(A\) is locally connected. \(\square\)

Why the Subspace Topology Matters

Not every subspace of Euclidean space is locally connected. The criterion detects this by asking whether connected open neighborhoods can be made arbitrarily small. In \(\mathbb{Q}\subseteq\mathbb{R}\), the topology is the subspace topology, so a basic open neighborhood of \(q\in\mathbb{Q}\) is \(\mathbb{Q}\cap(a,b)\), where \(a<q<b\).

Common pitfall. One might try to conclude that a set is open in a subspace because it looks open in the ambient space, or reject it because it is not ambient-open. Neither test is correct. For instance, \([0,\tfrac14)\) is not open in \(\mathbb{R}\), but it is open in \([0,1]\) because \([0,\tfrac14)=[0,1]\cap(-1,\tfrac14)\). The relevant question is always whether the set is the intersection of the subspace with an ambient open set.

The basis criterion is useful because it turns local connectedness into a neighborhood-by-neighborhood task. In intervals and convex subsets, intersections with small open intervals or balls give the required connected open sets. In \(\mathbb{Q}\), the connected subsets are too small to be open, so the same test proves failure.