Failure Of Local Connectedness On The Sine Curve

The Sine Curve and Its Limiting Segment

Connectedness is a global property: it says that a space cannot be split into two disjoint, nonempty open pieces. Local connectedness asks for a stronger kind of control near each point. On the sine curve, the oscillating graph accumulates on a vertical segment, and those two features behave differently nearby. We will show that every point of the limiting segment has arbitrarily small neighborhoods that contain no connected neighborhood of that point.

Let \(X\) be the following subset of \(\mathbb{R}^2\), with the subspace topology inherited from the usual Euclidean topology:

The first set is the oscillating graph; the second is its limiting vertical segment. Write \(\pi:X\to\mathbb{R}\) for the horizontal-coordinate map, \(\pi(x,y)=x\). It is continuous, since it is the restriction to \(X\) of the coordinate projection on \(\mathbb{R}^2\).

Thus, to rule out local connectedness at \(p\), it is enough to find an open neighborhood \(U\) of \(p\) that contains no connected neighborhood of \(p\). This rules out even connected neighborhoods that are not open, and so is stronger than merely ruling out connected open neighborhoods.

Two Connectedness Facts We Will Use

A set is connected if it cannot be written as the union of two disjoint, nonempty sets that are open in its subspace topology. We need two consequences of this definition: continuous images of connected sets are connected, and a connected subset of the real line contains every number between any two of its points.

Proof. Suppose first that \(f(C)\) is not connected. Then it has a separation \(f(C)=A\cup B\), where \(A\) and \(B\) are disjoint, nonempty, and open in \(f(C)\). By continuity, \(f^{-1}(A)\) and \(f^{-1}(B)\) are open in \(C\). They are disjoint, nonempty, and have union \(C\), contradicting connectedness of \(C\). Therefore \(f(C)\) is connected.

For the second claim, suppose some \(c\) satisfies \(u<c<v\) but \(c\notin D\). Then \(D\cap(-\infty,c)\) and \(D\cap(c,\infty)\) are disjoint, nonempty, relatively open subsets of \(D\), and their union is \(D\). This contradicts connectedness. Hence every \(c\in[u,v]\) belongs to \(D\). \(\square\)

For the sine curve, the lemma has a useful consequence. If a connected subset \(C\subseteq X\) contains a point on the vertical segment and a graph point with horizontal coordinate \(a>0\), then \(\pi(C)\) is connected and contains \(0\) and \(a\). It must therefore contain every \(t\in[0,a]\). Since the only point of \(X\) with first coordinate \(t>0\) is \((t,\sin(1/t))\), the set \(C\) must contain the graph point at every such \(t\). In particular, it must contain graph points where the sine takes both extreme values, provided those coordinates lie between \(0\) and \(a\).

The Failure at Every Point of the Segment

Proof. Fix \(y_0\in[-1,1]\). Choose \(r>0\) sufficiently small that the interval \(J=(y_0-r,y_0+r)\) does not contain at least one of \(-1\) and \(1\). Such a choice is possible because \(J\) can be chosen to have length less than \(2\), whereas the two values are distance \(2\) apart. Choose \(s\in\{-1,1\}\) with \(s\notin J\), and choose any \(\delta>0\). Define \(U=X\cap((-\delta,\delta)\times J)\). This is open in \(X\) and contains \(p\).

We first check that every neighborhood of \(p\) meets the oscillating graph. Choose \(\theta\in[0,2\pi)\) with \(\sin\theta=y_0\). For all sufficiently large \(n\in\mathbb{N}\), the number \(x_n=1/(2\pi n+\theta)\) satisfies \(0<x_n\leq1\). Also \(x_n\to0\) and \(\sin(1/x_n)=y_0\), so \((x_n,\sin(1/x_n))\to(0,y_0)=p\). Therefore every open set containing \(p\) contains a graph point.

Suppose, for a contradiction, that \(U\) contains a connected neighborhood \(N\) of \(p\). Since \(N\) is a neighborhood, it contains an open set \(O\) with \(p\in O\). By the convergence just proved, \(O\), and hence \(N\), contains a graph point \((a,\sin(1/a))\) with \(a>0\). It also contains \(p\), so \(\pi(N)\) is connected and contains \(0\) and \(a\). The lemma implies that \([0,a]\subseteq\pi(N)\).

There are positive numbers \(t_k\to0\) with \(\sin(1/t_k)=s\): for \(s=1\), take \(t_k=1/(\pi/2+2\pi k)\); for \(s=-1\), take \(t_k=1/(3\pi/2+2\pi k)\), with \(k\in\mathbb{N}\). For sufficiently large \(k\), \(0<t_k<a\), so \(t_k\in\pi(N)\). The unique point of \(X\) with first coordinate \(t_k\) is \((t_k,s)\). But \(s\notin J\), so \((t_k,s)\notin U\). This contradicts \(N\subseteq U\). Thus \(U\) contains no connected neighborhood of \(p\), proving the theorem. \(\square\)

Worked Examples on the Segment

Why Small Rectangles Do Not Solve the Problem

A common mistake is to take a small rectangle around a point of the vertical segment and assume that its intersection with \(X\) is connected because it meets both the segment and the graph. Meeting both pieces does not establish connectedness. In fact, even at \(p=(0,0)\), the neighborhood \(U=X\cap((-1/4,1/4)\times(-1/2,1/2))\) from the first example is disconnected.

To verify this, project \(U\) onto the horizontal coordinate. The projection contains \(0\), and for every sufficiently large \(k\) it contains \(x=1/(2\pi k)\), because \(\sin(1/x)=0\) and \(0<x<1/4\). But it omits \(t=1/(\pi/2+2\pi k)\), since the graph height there is \(1\), outside the vertical interval. Moreover, \(0<t<x\): the inequality follows from \(\pi/2+2\pi k>2\pi k\), which makes \(1/(\pi/2+2\pi k)<1/(2\pi k)\). Hence the projection is not an interval and is not connected. If \(U\) were connected, its continuous image under \(\pi\) would be connected, contradicting the lemma.

The distinction is important: the full sine curve \(X\) is connected, but this does not imply that its points have connected neighborhoods of arbitrarily small size. The obstruction here is local and comes from oscillation at every positive horizontal scale. A connected set that contains a point of the segment and reaches the graph must, by its projection, include graph points at every intermediate horizontal coordinate. Those coordinates force it to encounter heights outside any chosen proper vertical band.