Open Subspaces of Locally Connected Spaces

Local Connectedness and Relative Neighborhoods

Local connectedness is a condition about finding connected neighborhoods as small as needed. When a space is replaced by a subspace, its open sets are defined relative to the larger space, so a neighborhood that is open in the subspace need not be open in the original space. The distinction matters: if the subspace is open, its relative neighborhoods are also open in the ambient space, and local connectedness transfers. Without that hypothesis, the conclusion can fail.

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 \(C\). A neighborhood in this tutorial means an open set containing the point. For \(Y\subseteq X\), the subspace topology on \(Y\) consists of the sets \(Y\cap U\), where \(U\) is open in \(X\).

Definition: A space \(X\) is locally connected if, for every \(x\in X\) and every open neighborhood \(U\) of \(x\), there is a connected open set \(V\) such that \(x\in V\subseteq U\). Equivalently, each point has a basis consisting of connected open neighborhoods.

The quantifiers are important: it is not enough to find one connected neighborhood of each point. The connected neighborhoods must be available inside every prescribed open neighborhood. The open-subspace result preserves precisely this basis condition.

A Fact About Subspace Topologies

Lemma: If \(V\subseteq Y\subseteq X\), then the subspace topology on \(V\) inherited from \(Y\) is the same as the subspace topology on \(V\) inherited directly from \(X\). Consequently, a subset \(V\) is connected in one of these topologies if and only if it is connected in the other.

Proof. Every open set in \(Y\) has the form \(Y\cap U\) for some open \(U\subseteq X\). Its intersection with \(V\) is \(V\cap(Y\cap U)=V\cap U\), since \(V\subseteq Y\). Conversely, every set \(V\cap U\), for \(U\) open in \(X\), is also \(V\cap(Y\cap U)\), with \(Y\cap U\) open in \(Y\). Thus the two collections of open subsets of \(V\) agree. Connectedness is defined entirely by which subsets are open, so it agrees as well. \(\square\)

This lemma lets us check connectedness of a set inside the ambient space even when we are discussing it as a subset of a subspace. It does not say that every open subset of \(Y\) is open in \(X\); that additional fact requires \(Y\) itself to be open.

The Open-Subspace Theorem

Theorem: If \(X\) is locally connected and \(Y\subseteq X\) is open, then \(Y\), with its subspace topology, is locally connected.

Proof. If \(Y=\varnothing\), the local connectedness condition holds vacuously because \(Y\) has no points. Now suppose \(Y\ne\varnothing\). Let \(y\in Y\), and let \(W\) be any open neighborhood of \(y\) in \(Y\). By the definition of the subspace topology, \(W=Y\cap U\) for some open set \(U\) in \(X\). Both \(Y\) and \(U\) are open in \(X\), so their intersection \(W\) is open in \(X\), not merely in \(Y\). It contains \(y\). Since \(X\) is locally connected, there is a connected open set \(V\) in \(X\) with \(y\in V\subseteq W\). In particular \(V\subseteq Y\), and \(V\) is open in \(Y\) because \(V=Y\cap V\) and \(V\) is open in \(X\). By the lemma, \(V\) is connected with the topology inherited from \(Y\). We have therefore found a connected open neighborhood \(V\) in \(Y\) with \(y\in V\subseteq W\). Since \(y\) and \(W\) were arbitrary, \(Y\) is locally connected. \(\square\)

The crucial step is the sentence that makes \(W\) open in \(X\): it uses that \(Y\) is open. Only then can local connectedness of \(X\) be applied directly to \(W\). The connected neighborhood supplied by \(X\) is contained in \(W\), so it is automatically a subset of \(Y\) too.

Worked Example: An Open Interval in the Real Line

Consider \(X=\mathbb{R}\) with its standard topology and \(Y=(0,1)\) with the subspace topology. To verify local connectedness at a particular point, take \(y=1/2\) and the relative open neighborhood \(W=(2/5,3/5)\) in \(Y\). The interval \(V=(9/20,11/20)\) contains \(y\), since \(9/20<10/20=1/2<11/20\). It is contained in \(W\), since \(2/5=8/20<9/20\) and \(11/20<12/20=3/5\). It is open in \(Y\), and it is connected because intervals in \(\mathbb{R}\) are connected. Thus this neighborhood has a connected open neighborhood inside it.

The same construction works at every point and at every relative open neighborhood: since \(Y\) is open in \(\mathbb{R}\), that relative neighborhood is open in \(\mathbb{R}\), and it contains a sufficiently small open interval around its chosen point. The interval is connected. The theorem explains why this local construction works throughout \(Y\).

Worked Example: A Subspace of a Finite Discrete Space

Let \(X=\{a,b,c\}\) have the discrete topology, in which every subset is open, and let \(Y=\{a,c\}\). The set \(Y\) is open in \(X\). In the subspace topology on \(Y\), every subset is open: for example, \(\{a\}=Y\cap\{a\}\) and \(\{c\}=Y\cap\{c\}\).

For the point \(a\), any open neighborhood \(W\) in \(Y\) contains \(\{a\}\), and \(\{a\}\) is itself open and connected. It is connected because a one-point set cannot be divided into two nonempty subsets. The same argument applies at \(c\). Hence every relative open neighborhood of either point contains a connected open neighborhood of that point, so \(Y\) is locally connected.

This example also shows why connectedness alone is not the issue: the available connected neighborhoods can be very small. The theorem ensures that when \(Y\) is open, the ambient connected neighborhoods can be chosen inside any prescribed relative neighborhood.

Why Openness Cannot Be Dropped

A connected subset of the real line must contain every real number between any two of its points. We record the argument because it makes the failure for the rationals explicit.

Lemma: If \(J\subseteq\mathbb{R}\) is an interval and \(J\) is connected, then whenever \(a,b\in J\) with \(a<b\), every \(c\) satisfying \(a<c<b\) belongs to \(J\). More generally, any connected subset \(E\subseteq\mathbb{R}\) that contains \(a\) and \(b\) contains every point between them.

Proof. The interval assertion follows from the definition of interval. For the general assertion, suppose \(E\) is connected, \(a,b\in E\), \(a<b\), and \(c\in(a,b)\) is not in \(E\). Then \(E\cap(-\infty,c)\) and \(E\cap(c,\infty)\) are disjoint, nonempty subsets of \(E\): the first contains \(a\), and the second contains \(b\). They are open in \(E\), as they are intersections with open subsets of \(\mathbb{R}\), and together they cover \(E\) because \(c\notin E\). They give a separation of \(E\), contradicting connectedness. Therefore every such \(c\) belongs to \(E\). \(\square\)

Worked Example: The Rationals Are Not Locally Connected

Give \(\mathbb{R}\) its standard topology and let \(Y=\mathbb{Q}\) have the subspace topology. First, \(Y\) is not open in \(\mathbb{R}\): for any \(q\in\mathbb{Q}\) and any \(\varepsilon>0\), choose \(n\in\mathbb{N}\) large enough that \(0<\sqrt{2}/n<\varepsilon\). Then \(q+\sqrt{2}/n\) lies in \((q-\varepsilon,q+\varepsilon)\), but is irrational, so that interval is not contained in \(\mathbb{Q}\).

Now consider the relative open neighborhood \(W=\mathbb{Q}\cap(-1,1)\) of \(0\). Suppose there were a connected open neighborhood \(V\) of \(0\) in \(\mathbb{Q}\) with \(V\subseteq W\). If \(V\) contained a point \(r\ne0\), then \(0\) and \(r\) would be distinct points of the connected subset \(V\subseteq\mathbb{R}\). The lemma implies that \(V\) contains every real number between \(0\) and \(r\). There is an irrational number strictly between them, contradicting \(V\subseteq\mathbb{Q}\). Thus \(V\) would have to be \(\{0\}\).

But \(\{0\}\) is not open in \(\mathbb{Q}\). Every relative open neighborhood of \(0\) contains \(\mathbb{Q}\cap(-\delta,\delta)\) for some \(\delta>0\), and this set contains a nonzero rational: choose \(n\in\mathbb{N}\) with \(1/n<\delta\), so \(1/n\in\mathbb{Q}\cap(-\delta,\delta)\) and \(1/n\ne0\). Therefore no connected open neighborhood \(V\) as supposed exists. The space \(\mathbb{Q}\) is not locally connected.

For comparison, \(\mathbb{R}\) is locally connected because every open neighborhood of a point contains an open interval around that point, and intervals are connected. Thus this is a non-open subspace of a locally connected space for which local connectedness fails.

The Neighborhood Argument: A Common Pitfall

Common mistake. Given a subspace-open neighborhood \(W\) of \(y\in Y\), one might write, “\(W\) is open in \(X\), so choose a connected open neighborhood in \(X\) inside \(W\).” This confuses relative openness with ambient openness. For \(Y=\mathbb{Q}\subseteq\mathbb{R}\), the set \(W=\mathbb{Q}\cap(-1,1)\) is open in \(\mathbb{Q}\), but it is not open in \(\mathbb{R}\): every real interval around \(0\) contains irrational numbers. The attempted application of local connectedness of \(\mathbb{R}\) to \(W\) is therefore not justified. The correction is to check the hypothesis \(Y\) open first. If it holds, write \(W=Y\cap U\) with \(U\) open in \(X\); then \(W\) is open in \(X\), and the theorem’s proof applies. If it does not hold, the theorem provides no such conclusion, as the rational subspace demonstrates.

The openness assumption is sufficient, not a claim that every non-open subspace fails to be locally connected. Its role in the proof is specific: it lets each relative open neighborhood serve as an ambient open neighborhood. Once a connected ambient neighborhood has been found inside it, the inherited topology preserves connectedness.

Check Your Understanding

Use the definition of local connectedness and track which topology makes each set open.

In the theorem’s proof, where exactly is the assumption that \(Y\) is open used?
Show that every subspace of a discrete space is locally connected.
Why does the connected-subset lemma rule out a connected subset of \(\mathbb{Q}\) containing two distinct points?
Give an example of a non-open subspace of a locally connected space that is nevertheless locally connected.
In the rational example, why is it not enough to note that \(\mathbb{Q}\) has many connected subsets consisting of one point?