Uncountably Many Loop Classes In The Hawaiian Earring

Loops That Visit Selected Circles

The Hawaiian earring consists of infinitely many circles that meet at one point, with the circles shrinking toward that common point. A loop can traverse some of the circles in sequence, using shorter and shorter time intervals. Shrinking diameters make the resulting path continuous even though infinitely many traversals accumulate before its final time. The key question is whether different choices of circles can give different homotopy classes. We will detect each choice by retracting the whole space onto one circle and measuring a winding number.

Use the Euclidean topology on \(\mathbb{R}^2\) and the subspace topology on the Hawaiian earring \(H\). Write \(p=(0,0)\) for its common point, and, for \(n\in\mathbb{N}\), let \(C_n\) be the circle with center \((1/n,0)\) and radius \(1/n\). Thus \(H=\bigcup_{n\in\mathbb{N}}C_n\). The circles meet pairwise only at \(p\): subtracting the equations for two distinct circles shows that any intersection has first coordinate \(0\), and substitution then gives second coordinate \(0\).

A based loop in \(H\) is a continuous map \(\gamma:[0,1]\to H\) with \(\gamma(0)=\gamma(1)=p\). Two such loops are based-homotopic if there is a continuous map \(F:[0,1]\times[0,1]\to H\) such that \(F(0,t)\) is the first loop, \(F(1,t)\) is the second, and \(F(s,0)=F(s,1)=p\) for all \(s,t\in[0,1]\). We construct a loop for every subset \(S\subseteq\mathbb{N}\).

More explicitly, on an interval where \(n\in S\), use \(c_n((t-t_{n-1})/(t_n-t_{n-1}))\). The interval has length \(t_n-t_{n-1}=2^{-n}\). Each piece begins and ends at \(p\), so adjacent pieces agree at their common endpoint. The only continuity issue not settled by ordinary pasting on finitely many intervals is at \(t=1\).

Why the Infinite Concatenation Is Continuous

Proof. On the interior of each interval \([t_{n-1},t_n]\), the map is either a rescaled continuous parametrization \(c_n\) or the constant map with value \(p\). The formulas meet at \(p\) at each shared endpoint, so \(\gamma_S\) is continuous at every \(t<1\). To check continuity at \(1\), let \(\varepsilon>0\). Choose \(N\) so large that \(2/(N+1)<\varepsilon\). Every point of \(C_n\) is at distance at most \(2/n\) from \(p\), because \(C_n\) has center at distance \(1/n\) from \(p\) and radius \(1/n\). If \(t_N<t<1\), then \(t\) lies in an interval with index \(n>N\), so either \(\gamma_S(t)=p\), or \(\|\gamma_S(t)-p\|\le 2/n\le 2/(N+1)<\varepsilon\). Also \(\gamma_S(1)=p\). This proves continuity at \(1\). The endpoint values are \(p\), so \(\gamma_S\) is a based loop. \(\square\)

Reading One Circle at a Time

For each \(k\), we need a map that retains \(C_k\) and sends every other circle to \(p\). Define \(r_k:H\to C_k\) by \(r_k(x)=x\) for \(x\in C_k\) and \(r_k(x)=p\) for \(x\in C_n\) with \(n\ne k\). This is well-defined because the circles intersect only at \(p\).

Proof. Let \(V\) be open in \(C_k\). If \(p\in V\), then \(C_k\setminus V\) is closed in the compact circle \(C_k\), hence compact and closed in the Hausdorff space \(\mathbb{R}^2\). Since \(r_k^{-1}(V)=H\setminus(C_k\setminus V)\), this preimage is open in \(H\). If \(p\notin V\), then \(V\) is open in \(H\): each point of \(C_k\setminus\{p\}\) has a neighborhood in \(H\) meeting no other circle. Indeed, circles with sufficiently large index lie arbitrarily close to \(p\), away from that point, and the finitely many remaining circles are compact and disjoint from it. Consequently \(r_k^{-1}(V)=V\) is open in \(H\). Both cases show continuity. \(\square\)

Identify \(C_k\) with the unit circle \(S^1=\{z\in\mathbb{C}:|z|=1\}\) by the homeomorphism sending \(c_k(u)\) to \(e^{2\pi i u}\). A loop \(u:[0,1]\to S^1\) based at \(1\) has a continuous angle lift \(\theta:[0,1]\to\mathbb{R}\) such that \(u(t)=e^{2\pi i\theta(t)}\) and \(\theta(0)=0\). Its winding number is the integer \(\theta(1)\). The lift can be obtained by subdividing the interval into pieces on which \(u(t)/u(t_i)\) avoids \(-1\), and using the continuous argument with values in \((-1/2,1/2)\) on each piece. The endpoint is an integer because \(u(1)=1\); the lift is unique, since two lifts differ by a continuous integer-valued function and agree at \(0\).

Proof. First suppose two based loops \(u,v:[0,1]\to S^1\) satisfy \(|u(t)-v(t)|<2\) for every \(t\). Then \(v(t)/u(t)\ne-1\), because the distance between \(1\) and \(-1\) is \(2\). There is a continuous function \(\delta:[0,1]\to(-1/2,1/2)\) with \(v(t)/u(t)=e^{2\pi i\delta(t)}\). Since both loops start and end at \(1\), \(\delta(0)=\delta(1)=0\). If \(\theta\) is the lift of \(u\) starting at \(0\), then \(\theta+\delta\) is the lift of \(v\) starting at \(0\), and its endpoint equals \(\theta(1)\). The winding numbers are equal.

Now let \(F:[0,1]\times[0,1]\to S^1\) be a based homotopy. Uniform continuity on the compact square gives a partition \(0=s_0<s_1<\cdots<s_m=1\) fine enough that \(|F(s_j,t)-F(s_{j+1},t)|<2\) for every \(t\) and every \(j\). The preceding argument shows that the loops \(F(s_j,\cdot)\) and \(F(s_{j+1},\cdot)\) have equal winding numbers. Applying this equality across the finite partition proves that \(F(0,\cdot)\) and \(F(1,\cdot)\) have equal winding numbers. \(\square\)

Composing a based homotopy in \(H\) with \(r_k\), and then with the identification of \(C_k\) with \(S^1\), gives a based homotopy in \(S^1\). The theorem therefore supplies a homotopy invariant for each circle-coordinate of a loop in \(H\).

Uncountably Many Different Classes

Proof. For \(S\subseteq\mathbb{N}\), consider the loop \(\gamma_S\). For each \(k\), applying \(r_k\) gives one traversal of \(C_k\) if \(k\in S\), and the constant loop if \(k\notin S\). Under the identification with \(S^1\), the winding number is therefore \(1\) when \(k\in S\), and \(0\) when \(k\notin S\). If \(S\ne T\), there is some \(k\) belonging to exactly one of them. The corresponding winding numbers differ, so homotopy invariance proves that \(\gamma_S\) and \(\gamma_T\) are not based-homotopic. Thus distinct subsets give distinct loop classes.

For completeness, the set of subsets of \(\mathbb{N}\) is uncountable: if they could be listed as \(S_1,S_2,\ldots\), the subset \(D=\{j\in\mathbb{N}:j\notin S_j\}\) would differ from \(S_j\) at the element \(j\) for every \(j\), so it would not occur on the list. The injection from subsets to loop classes consequently proves that the collection of based loop classes is uncountable. \(\square\)

A Common Pitfall: Shrinking Does Not Mean Homotopically Trivial

It is tempting to reason that the circles become so small that the infinite tail of a loop can be discarded by a homotopy. The diameters tending to zero prove continuity of the concatenation at its limiting time; they do not prove that the tail is homotopically trivial in \(H\). For example, compare \(\gamma_{\mathbb{N}}\) with the loop that traverses only \(C_1,\ldots,C_N\) and then stays at \(p\). At coordinate \(k=N+1\), the first loop has winding number \(1\), while the finite loop has winding number \(0\). The retraction and homotopy-invariance argument proves that these loops are not based-homotopic. The distinction is that smallness controls distances in the plane, whereas winding number records a topological obstruction that remains visible after retracting onto a particular circle.

The coordinate winding numbers are used here to distinguish the constructed family; no claim is being made that they classify every loop in the Hawaiian earring. Their strength is sufficient for the conclusion: each subset of circle indices gives a loop, and any change in the subset changes at least one homotopy invariant.