Connectedness at Small Scales
Connectedness describes a space as a whole: a space is connected if it cannot be split into two disjoint, nonempty open parts. Local connectedness asks a different question. Near each point, and inside every prescribed open neighborhood, can we find a smaller connected open neighborhood? The scale matters: one connected neighborhood around a point is not enough if a smaller prescribed neighborhood contains no such set.
All neighborhoods below are taken in the topology of the space being discussed. In particular, when we use a subset such as \(\mathbb{Q}\) or \([0,1]\), it has the subspace topology from \(\mathbb{R}\). A subset is connected if it is connected with its subspace topology: it cannot be written as a union of two disjoint, nonempty sets that are open in that subspace.
This definition says that connected open neighborhoods can be found as small as a given open neighborhood requires. It does not say that \(X\) itself is connected, or that every open subset of \(X\) is connected. The word “open” is important: a connected set containing \(x\) need not be a neighborhood of \(x\), and a connected neighborhood need not be open unless that is required.
Intervals Give Connected Neighborhoods
We will use the standard real-line fact that every interval is connected. Here is the separation argument, which also explains why intervals are useful for this local question.
Proof. Suppose, for a contradiction, that \(J=A\cup B\), where \(A\) and \(B\) are disjoint, nonempty, relatively open subsets of \(J\). Choose one point from each set. By interchanging their names if needed, take \(a\in A\), \(b\in B\) with \(a<b\). The set \(S=A\cap[a,b]\) is nonempty because it contains \(a\), and it is bounded above by \(b\). Let \(c=\sup S\). Since \(a\leq c\leq b\) and \(J\) is an interval containing \(a\) and \(b\), we have \(c\in J\). Thus \(c\in A\) or \(c\in B\).
If \(c\in A\), then \(c\ne b\), so \(c<b\). Relative openness of \(A\) gives a \(\delta>0\) such that \(J\cap(c-\delta,c+\delta)\subseteq A\). There is a point \(t\in J\) with \(c<t<\min(c+\delta,b)\), since \(J\) contains every point between \(c\) and \(b\). Then \(t\in A\cap[a,b]=S\) and \(t>c\), contradicting that \(c\) is an upper bound of \(S\).
If \(c\in B\), then \(c\ne a\), so \(c>a\). Relative openness of \(B\) gives a \(\delta>0\) such that \(J\cap(c-\delta,c+\delta)\subseteq B\). Because \(c=\sup S\), there is an \(s\in S\) with \(c-\delta<s\leq c\). This point belongs to \(A\), and it lies in \(J\cap(c-\delta,c+\delta)\), which is contained in \(B\). That contradicts \(A\cap B=\varnothing\). Both cases are impossible, so no such separation exists and \(J\) is connected. \(\square\)
Worked Example: The Real Line
Give \(\mathbb{R}\) its standard topology and fix \(x\in\mathbb{R}\). We check the local condition for an arbitrary open neighborhood \(U\) of \(x\), not just for one convenient choice of \(U\).
Since \(U\) is open and contains \(x\), there is an \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq U\). Put \(V=(x-\varepsilon,x+\varepsilon)\). This set is open in \(\mathbb{R}\), contains \(x\), and is an interval. By the interval theorem, it is connected. Therefore \(x\in V\subseteq U\), as required. Since the argument works for every \(x\) and every open neighborhood \(U\), \(\mathbb{R}\) is locally connected.
Check: the containment used is exactly \((x-\varepsilon,x+\varepsilon)\subseteq U\), which follows from openness at \(x\); no claim is made that \(U\) itself is connected.
Worked Example: The Unit Interval
Let \(I=[0,1]\) have the subspace topology inherited from the standard topology on \(\mathbb{R}\). Fix \(x\in I\) and an open neighborhood \(U\) of \(x\) in \(I\). By the definition of the subspace topology, there is an open set \(O\subseteq\mathbb{R}\) such that \(U=O\cap I\). Since \(x\in O\), choose \(\varepsilon>0\) with \((x-\varepsilon,x+\varepsilon)\subseteq O\). Then set \(V=(x-\varepsilon,x+\varepsilon)\cap I\).
The set \(V\) is open in \(I\), contains \(x\), and is contained in \(U\). It is an interval in \(\mathbb{R}\): it consists of the points between its endpoints, including an endpoint of \(I\) when that endpoint lies in \(V\). The interval theorem makes \(V\) connected. For example, if \(x=0\) and \(\varepsilon=1/4\), then \(V=[0,1/4)\), which is open in \(I\) and connected. Thus \(I\) is locally connected, including at its endpoints.
Check: \(V\subseteq U\) because \(V\subseteq O\) and \(V\subseteq I\), so \(V\subseteq O\cap I=U\). The example at \(0\) also shows why neighborhoods in \(I\) need not be open intervals in \(\mathbb{R}\).
Discrete Spaces and the Need for Openness
Worked Example: A Discrete Space
Let \(X\) be any set with the discrete topology \(\mathcal{T}=\mathcal{P}(X)\), so every subset is open. If \(x\in X\) and \(U\) is any open neighborhood of \(x\), take \(V=\{x\}\). Then \(V\) is open, contains \(x\), and is contained in \(U\).
A singleton is connected: it cannot be partitioned into two nonempty subsets, since it has only one point. Hence \(V\) is a connected open neighborhood and \(X\) is locally connected. If \(X=\varnothing\), the same conclusion holds by the vacuous case in the definition.
Check: the topology is discrete, so \(\{x\}\in\mathcal{T}\); this step would not hold for an arbitrary topology.
Connectedness of a set and being a neighborhood are separate requirements. A singleton is always connected, but it need not be open. The rational subspace makes this distinction decisive.
Worked Example: The Rational Subspace
Give \(\mathbb{Q}\) the subspace topology inherited from the standard topology on \(\mathbb{R}\). First, every connected subset of \(\mathbb{Q}\) has at most one point. To see this, suppose a subset \(C\subseteq\mathbb{Q}\) contains \(p<q\). Choose an irrational \(\alpha\) with \(p<\alpha<q\), using the density of the irrationals in \(\mathbb{R}\). Then
These two sets are disjoint, nonempty (they contain \(p\) and \(q\), respectively), and open in \(C\), because each is the intersection of \(C\) with an open subset of \(\mathbb{R}\). They form a separation of \(C\). Thus \(C\) is not connected, proving the claim.
Now fix \(x\in\mathbb{Q}\) and use \(U=\mathbb{Q}\), an open neighborhood of \(x\) in itself. If there were a connected open neighborhood \(V\) of \(x\) contained in \(U\), the claim just proved would force \(V\) to have at most one point. Since \(x\in V\), this means \(V=\{x\}\). But \(\{x\}\) is not open in \(\mathbb{Q}\): every relative open neighborhood of \(x\) contains \(\mathbb{Q}\cap(x-\varepsilon,x+\varepsilon)\) for some \(\varepsilon>0\), and density of \(\mathbb{Q}\) gives a rational in that interval different from \(x\). This contradiction shows that \(\mathbb{Q}\) is not locally connected.
Check: \(\{x\}\) is connected, but it is not a connected open neighborhood of \(x\) in \(\mathbb{Q}\). Connectedness alone does not satisfy the definition.
Local Connectedness Is Preserved by Homeomorphisms
Local connectedness is a property of the topology, not of the names or coordinates used for points. A homeomorphism is a bijection that is continuous and whose inverse is continuous. We verify directly that it preserves the local condition.
Proof. First suppose \(X\) is locally connected. Let \(y\in Y\) and let \(W\) be an open neighborhood of \(y\). Write \(x=f^{-1}(y)\). Continuity of \(f\) gives that \(f^{-1}(W)\) is open in \(X\) and contains \(x\). Local connectedness of \(X\) supplies a connected open set \(V\) with \(x\in V\subseteq f^{-1}(W)\). The continuous image \(f(V)\) is connected: if it had a separation into disjoint nonempty relatively open sets, their preimages under \(f|_V\) would give a separation of \(V\). Also, \(f(V)\) is open in \(Y\), since \(f(V)=(f^{-1})^{-1}(V)\) and \(f^{-1}\) is continuous. Finally, \(y\in f(V)\subseteq W\). This proves local connectedness of \(Y\). Applying the same argument to the homeomorphism \(f^{-1}:Y\to X\) proves the converse. \(\square\)
A Common Pitfall
A tempting but incorrect argument is: “Every connected subset of \(\mathbb{Q}\) has at most one point, and singletons are connected, so \(\mathbb{Q}\) is locally connected.” The conclusion does not follow. For local connectedness, the connected set must also be open and must fit inside each prescribed open neighborhood. In the rational example, a connected open neighborhood \(V\) of \(x\) would have to satisfy \(V=\{x\}\), but \(\{x\}\) is not open in \(\mathbb{Q}\). The correction is to check all parts of the condition \(x\in V\subseteq U\), including that \(V\) is open.
The contrast among the examples is useful: intervals supply connected open neighborhoods at every scale; discrete spaces use open singletons; and \(\mathbb{Q}\) has connected singletons but lacks the required open neighborhoods. Thus the local condition cannot be decided by checking only whether some connected sets exist near a point.
Check Your Understanding
For each question, specify the topology being used and check both connectedness and openness where they are relevant.
- State the definition of local connectedness at a point \(x\), including the quantifier over open neighborhoods \(U\).
- Why does the proof for \(\mathbb{R}\) require an interval contained in an arbitrary open neighborhood, rather than just noting that \(\mathbb{R}\) is connected?
- In the discrete topology on a nonempty set \(X\), identify a connected open neighborhood inside a given open set \(U\ni x\).
- Explain why the conclusion that connected subsets of \(\mathbb{Q}\) have at most one point does not establish that \(\mathbb{Q}\) is locally connected.
- Use the homeomorphism theorem to explain why a space homeomorphic to \([0,1]\) is locally connected.