What Local Connectedness Tests
A space can be connected as a whole and still fail to look connected near some of its points. Local connectedness tests that local behavior: around each point, can we find connected open sets as small as we need? The real line gives a positive example; the rationals, the Cantor set, and the topologist’s sine curve show different ways the test can fail.
Throughout, subsets inherit the subspace topology unless another topology is specified. A space is connected if it cannot be written as the union of two disjoint, nonempty open subsets. A subset is connected when it is connected with its subspace topology.
Thus the test concerns connected open neighborhoods, not just connected subsets that happen to contain the point. We will use the standard fact that the continuous image of a connected space is connected. To see why, suppose the image had a separation into disjoint nonempty relatively open sets. Their preimages under the continuous map would be disjoint nonempty relatively open sets partitioning the original space, contradicting its connectedness.
The Real Line: Small Connected Neighborhoods
In \(\mathbb{R}\) with its standard topology, every point of every open set lies in a sufficiently small open interval contained in that set. Open intervals are connected, so these intervals provide the required connected open neighborhoods. Here is the interval fact, including the order argument behind it.
Proof. Suppose, to the contrary, that \(J=A\cup B\) is a separation: \(A\) and \(B\) are disjoint, nonempty, relatively open subsets of \(J\). Choose one point from each set; after interchanging their names if necessary, take \(a\in A\), \(b\in B\), and \(a<b\). Let \(c=\sup(A\cap[a,b])\), which exists because \(A\cap[a,b]\) is nonempty and bounded above. Since \(J\) is an interval, \(c\in J\). If \(c\in A\), then \(c\ne b\), and relative openness of \(A\) gives points of \(A\cap[a,b]\) greater than \(c\), contradicting the definition of \(c\). If \(c\in B\), then \(c>a\), and relative openness of \(B\) gives a left-hand interval around \(c\) in \(J\) contained in \(B\). The definition of supremum gives a point of \(A\cap[a,b]\) in that interval, again a contradiction. Since \(A\) and \(B\) partition \(J\), these cases exhaust the possibilities. Therefore \(J\) is connected. \(\square\)
Worked Example: A Neighborhood in the Real Line
Take \(x=2\) and the open neighborhood \(U=(1,3)\) in standard \(\mathbb{R}\). The open interval \(V=(3/2,5/2)\) contains \(2\), and its endpoints satisfy \(1<3/2<2<5/2<3\). Thus \(2\in V\subseteq U\). The interval proposition shows that \(V\) is connected.
The same construction works at every point \(x\) of every open set \(U\subseteq\mathbb{R}\): openness gives some \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq U\), and this interval is connected. Hence \(\mathbb{R}\) is locally connected at every point.
Check: the chosen neighborhood is open in \(\mathbb{R}\), not merely connected. Both requirements in the definition are met.
The Rationals: Connected Subsets Are Too Small
Give \(\mathbb{Q}\) the subspace topology inherited from standard \(\mathbb{R}\). If a subset of \(\mathbb{Q}\) contains distinct points, an irrational number between them divides it into two nonempty relatively open pieces. Consequently, every connected subset of \(\mathbb{Q}\) has at most one point. But no point of \(\mathbb{Q}\) is isolated, so a singleton cannot be an open neighborhood. This prevents local connectedness.
Proof. Let \(D\subseteq\mathbb{Q}\) contain distinct points \(a<b\). Choose an irrational \(r\) with \(a<r<b\), using the density of the irrationals in \(\mathbb{R}\). Then \(D\) is the union of \(D\cap(-\infty,r)\) and \(D\cap(r,\infty)\): no point of \(D\) equals \(r\), since \(D\subseteq\mathbb{Q}\). Both pieces are nonempty, as they contain \(a\) and \(b\), respectively. They are disjoint and open in \(D\), so \(D\) is disconnected. Thus connected subsets of \(\mathbb{Q}\) have at most one point.
Every open neighborhood of any \(q\in\mathbb{Q}\) contains another rational: it contains \((q-\varepsilon,q+\varepsilon)\cap\mathbb{Q}\) for some \(\varepsilon>0\), and the density of \(\mathbb{Q}\) supplies a rational different from \(q\) in that interval. Hence no singleton is open in \(\mathbb{Q}\). Since every connected subset is empty or a singleton, there is no connected open neighborhood of \(q\). This holds for every \(q\), so \(\mathbb{Q}\) is not locally connected. \(\square\)
Worked Example: An Irrational Cut in a Rational Neighborhood
Consider \(U=(-1,1)\cap\mathbb{Q}\), an open neighborhood of \(0\) in \(\mathbb{Q}\). A tempting but incorrect argument calls \(U\) connected because \((-1,1)\) is an interval in \(\mathbb{R}\). The interval proposition applies to \((-1,1)\), not to its rational subspace.
For a direct check, \(r=\sqrt{2}/2\) is irrational and satisfies \(0<r<1\), since \(0<\sqrt{2}<2\). The sets \(U\cap(-\infty,r)\) and \(U\cap(r,\infty)\) are disjoint and relatively open in \(U\); they cover \(U\), because no rational equals \(r\). Both are nonempty: \(-1/2\) lies in the first and \(1/2\) lies in the second, since \(1/2<\sqrt{2}/2\). Thus \(U\) is disconnected.
Correction: connectedness of an interval in \(\mathbb{R}\) does not transfer automatically to a subspace obtained by deleting its irrational points. The irrational cut gives an explicit separation.
The Cantor Set: Clopen Gaps Separate Points
Let \(C\) be the standard middle-thirds Cantor set, with the subspace topology from \(\mathbb{R}\). At stage \(n\) of its construction, \(C\) lies in a union of closed intervals of length \(3^{-n}\); the open middle thirds removed during construction are gaps in \(C\). Any two distinct points of \(C\) eventually lie in different stage intervals. A removed gap between those intervals separates the two points within \(C\).
Worked Example: Separating Two Cantor-Set Points
Take \(x=0\) and \(y=1\), both in \(C\). The first removed gap is \((1/3,2/3)\). The two sets \(C\cap(-\infty,1/2)\) and \(C\cap(1/2,\infty)\) are disjoint and relatively open in \(C\), cover \(C\), and contain \(0\) and \(1\), respectively. The cut point \(1/2\) is not in \(C\), because it lies in the removed gap. This illustrates how a gap separates points while leaving a partition of the Cantor set.
More generally, if \(x<y\) lie in \(C\), choose \(n\) with \(3^{-n}<y-x\). They cannot lie in the same stage-\(n\) interval, whose length is \(3^{-n}\). The construction has a removed open gap between the stage intervals containing them. A point \(r\) in that gap is not in \(C\), so \(C\cap(-\infty,r)\) and \(C\cap(r,\infty)\) separate any subset of \(C\) containing both \(x\) and \(y\). Thus every connected subset of \(C\) has at most one point.
There are no isolated points of \(C\). Indeed, let \(x\in C\) and let \(U\) be a relative open neighborhood of \(x\). For sufficiently large \(n\), the stage-\(n\) interval containing \(x\) has length \(3^{-n}\) smaller than the radius of an ambient interval around \(x\) contained in \(U\). Both endpoints of that stage interval belong to \(C\); at least one is different from \(x\), and both lie in \(U\). Therefore \(U\) contains another point of \(C\).
Every connected subset of \(C\) is empty or a singleton, but no singleton is open in \(C\). Therefore no point has a connected open neighborhood, and \(C\) is not locally connected.
The Topologist’s Sine Curve: Connectedness Does Not Suffice
Define the topologist’s sine curve as the subspace \(T\) of standard \(\mathbb{R}^2\) given by \(T=G\cup L\), where \(G=\{(x,\sin(1/x)):x>0\}\) and \(L=\{(0,y):-1\le y\le1\}\). Its vertical segment \(L\) is approached by the oscillating graph. We show that no point \(p=(0,y)\) on this segment has a connected open neighborhood small enough to fit inside the open ball of radius \(1/2\) about \(p\).
First, every open neighborhood of \(p\) in \(T\) meets \(G\). Choose \(\theta\) with \(\sin\theta=y\). For sufficiently large \(n\), \(x_n=1/(\theta+2\pi n)>0\), and \(x_n\to0\) while \(\sin(1/x_n)=y\). Thus \((x_n,y)\in G\) and \((x_n,y)\to p\).
Proof. Fix \(p=(0,y)\in L\), and let \(U=T\cap B_{\|\cdot\|}(p,1/2)\), where \(B_{\|\cdot\|}\) is the open Euclidean ball. Suppose there were a connected open set \(W\) in \(T\) with \(p\in W\subseteq U\). Since every neighborhood of \(p\) meets \(G\), choose \(q=(x_0,\sin(1/x_0))\in W\cap G\), so \(x_0>0\).
The coordinate projection \(\pi_x:T\to\mathbb{R}\), \(\pi_x(x',y')=x'\), is continuous, so \(\pi_x(W)\) is connected. It contains \(0\) and \(x_0\); a connected subset of \(\mathbb{R}\) containing two points contains the interval between them, by the interval proposition (or its separation argument). Hence \([0,x_0]\subseteq\pi_x(W)\). For every \(t\in(0,x_0)\), the only point of \(T\) with first coordinate \(t\) is \((t,\sin(1/t))\). Since \(t\in\pi_x(W)\), that point belongs to \(W\).
Choose \(n\) large enough that \(t_+=1/(\pi/2+2\pi n)\) and \(t_-=1/(3\pi/2+2\pi n)\) are both less than \(x_0\). The corresponding graph points have second coordinates \(\sin(\pi/2+2\pi n)=1\) and \(\sin(3\pi/2+2\pi n)=-1\), so both belong to \(W\). For \(y\in[-1,1]\), at least one of \(|1-y|\) and \(|-1-y|\) is at least \(1\): their sum is \(2\). The corresponding graph point is therefore at Euclidean distance at least \(1\) from \(p\), so it does not belong to \(U\). This contradicts \(W\subseteq U\). No such connected open \(W\) exists, proving the claim for every \(p\in L\). \(\square\)
Worked Example: Why the Projection Test Forces Oscillation
Take the representative point \(p=(0,0)\). If a connected open neighborhood \(W\) of \(p\) were contained in \(T\cap B_{\|\cdot\|}(p,1/2)\), then \(W\) would meet the graph at some \((x_0,\sin(1/x_0))\) with \(x_0>0\). Connectedness of \(\pi_x(W)\) forces every \(t\) with \(0<t<x_0\) to occur as a first coordinate in \(W\). In particular, for large \(n\), both \(t_+=1/(\pi/2+2\pi n)\) and \(t_-=1/(3\pi/2+2\pi n)\) lie in that range. The unique graph points above them are \((t_+,1)\) and \((t_-,-1)\). Their distances from \(p\) are at least \(1\), contradicting containment in the radius-\(1/2\) ball.
Check: the contradiction uses that the neighborhood is connected, so its projection contains the whole interval of first coordinates; it does not follow just from the existence of graph points near \(p\).
A Useful Comparison and a Common Pitfall
The real line succeeds because small open intervals are connected. In \(\mathbb{Q}\) and \(C\), connected subsets are too small to be open neighborhoods. On the sine curve, oscillation prevents any connected open neighborhood of a point on the vertical segment from staying inside a sufficiently small ball. These arguments test local connectedness directly; they do not infer it merely from whether a whole space is connected.
Common pitfall. A connected space need not be locally connected. The topologist’s sine curve is connected, but the proposition above rules out connected open neighborhoods at points of its vertical segment. A second error is to treat a subspace of an interval as though it were itself an interval: the rational neighborhood \((-1,1)\cap\mathbb{Q}\) is separated by the irrational cut \(\sqrt{2}/2\), despite the connectedness of \((-1,1)\) in \(\mathbb{R}\).
Check Your Understanding
For each question, use the connected-open-neighborhood test and identify the topology being used.
- Why does every open neighborhood of \(x\in\mathbb{R}\) contain a connected open neighborhood of \(x\)?
- Use an irrational cut to show that \((0,1)\cap\mathbb{Q}\) is disconnected.
- Why does the absence of isolated points matter when concluding that \(C\) is not locally connected?
- In the sine-curve argument, why does connectedness of \(\pi_x(W)\) force it to contain every \(t\) between \(0\) and \(x_0\)?
- Does the connectedness of an entire space imply that it is locally connected? Give an example from this tutorial.