The Hawaiian Earring Is Not Semilocally Simply Connected

The Local Question at the Common Point

The Hawaiian earring consists of circles of decreasing size that all meet at one point. Their diameters tend to zero, but each circle still carries a loop that can be detected by a retraction onto that circle. The key issue is therefore not whether the circles are small, but whether a neighborhood of the common point can avoid containing an entire circle.

Work in \(\mathbb{R}^2\) with its usual topology. For each \(n\in\mathbb{N}\), let \(C_n\) be the circle with centre \((1/n,0)\) and radius \(1/n\), and let \(p=(0,0)\). The Hawaiian earring is the subspace \(H=\bigcup_{n\in\mathbb{N}}C_n\) of \(\mathbb{R}^2\). Each \(C_n\) contains \(p\), and its farthest point from \(p\) is \((2/n,0)\), at distance \(2/n\).

To disprove this condition at \(p\), it suffices to show that every open neighborhood \(U\) of \(p\) contains a based loop that is not nullhomotopic in \(H\). The proof has two parts: small circles fit inside every such \(U\), and a loop going once around any one of those circles is nontrivial in \(H\).

Every Neighborhood Contains a Whole Circle

Because \(U\) is open in the subspace \(H\), there is an \(\varepsilon>0\) such that \(H\cap B_{\mathbb{R}^2}(p,\varepsilon)\subseteq U\). Here \(B_{\mathbb{R}^2}(p,\varepsilon)\) is the open Euclidean ball. If \(2/n<\varepsilon\), then every point of \(C_n\) is at distance at most \(2/n\) from \(p\), so \(C_n\subseteq U\). The strict inequality ensures the whole circle lies in the open ball.

Detecting a Circle Loop by Retraction

Fix \(n\). Define \(r_n:H\to C_n\) to be the identity on \(C_n\) and to send every point outside \(C_n\) to \(p\). This is a retraction: it maps into \(C_n\) and fixes every point of \(C_n\). We verify continuity, since the behavior at the accumulation point \(p\) matters.

At \(p\), let \(V\) be an open neighborhood of \(p\) in \(C_n\). By the subspace topology there is a \(\delta>0\) such that \(C_n\cap B_{\mathbb{R}^2}(p,\delta)\subseteq V\). Every \(y\in H\cap B_{\mathbb{R}^2}(p,\delta)\) either lies in \(C_n\), in which case \(r_n(y)=y\in V\), or lies outside \(C_n\), in which case \(r_n(y)=p\in V\). This proves continuity at \(p\).

For any \(x\in H\setminus\{p\}\), the circles \(C_m\) with large \(m\) lie close to \(p\): every point of \(C_m\) is at distance at most \(2/m\) from \(p\). Thus a sufficiently small neighborhood of \(x\) misses all sufficiently large circles. Among the remaining finitely many circles, distinct circles meet only at \(p\); hence a neighborhood of \(x\) can be chosen to miss \(C_n\), unless \(x\in C_n\), when it can be chosen to miss every other circle. On the first kind of neighborhood \(r_n\) is constant, and on the second it is the identity. This proves continuity at every \(x\ne p\), and hence continuity on \(H\).

Why Winding Number Rules Out a Nullhomotopy

For a loop \(\alpha:[0,1]\to S^1\) based at \(1\), choose a continuous function \(\theta:[0,1]\to\mathbb{R}\) such that \(\theta(0)=0\) and \(e^{2\pi i\theta(t)}=\alpha(t)\). Such a function is called a lift of the loop’s angle. It exists by subdividing \([0,1]\) into intervals on each of which the values of \(\alpha\) lie in an arc admitting a continuous choice of angle, and choosing each successive angle to agree at the shared endpoint. Any two such lifts starting at \(0\) agree: their difference is continuous, integer-valued, and zero initially. Since \(\alpha(1)=1\), \(\theta(1)\) is an integer. The winding number is this integer.

Proof. First suppose two based loops \(\alpha,\beta\) satisfy \(|\alpha(t)-\beta(t)|<2\) for every \(t\). Then \(\beta(t)/\alpha(t)\ne-1\), so this ratio has a continuous argument \(\delta(t)\in(-1/2,1/2)\), with \(e^{2\pi i\delta(t)}=\beta(t)/\alpha(t)\). Both loops are based at \(1\), so the ratio is \(1\) at \(t=0\) and \(t=1\); consequently \(\delta(0)=\delta(1)=0\). If \(\theta\) lifts \(\alpha\), then \(\theta+\delta\) lifts \(\beta\) and has the same endpoint values as \(\theta\). Therefore the two winding numbers agree.

Now let \(F:[0,1]\times[0,1]\to S^1\) be a homotopy through based loops. Its domain is compact, so \(F\) is uniformly continuous. Subdivide the homotopy parameter into finitely many intervals fine enough that the loops at the two ends of each interval differ in distance by less than \(2\), uniformly over \(t\). The preceding argument gives equal winding numbers at the two ends of each interval. Chaining these equalities proves that the initial and final loops have the same winding number. A constant loop has the constant lift \(0\), hence winding number \(0\). Therefore a loop with nonzero winding number cannot be homotopic through based loops to a constant loop. \(\square\)

We can now prove the central claim. If \(\gamma_n\) were nullhomotopic in \(H\), composing its based nullhomotopy with the continuous map \(g_n\circ r_n:H\to S^1\) would give a based nullhomotopy of \(e^{2\pi it}\) in \(S^1\). But this loop has winding number \(1\), contradicting the theorem.

Proof. Let \(U\) be any open neighborhood of \(p\) in \(H\). Choose \(\varepsilon>0\) such that \(H\cap B_{\mathbb{R}^2}(p,\varepsilon)\subseteq U\), and choose \(n\) with \(2/n<\varepsilon\). Then \(C_n\subseteq U\), so \(\gamma_n\) is a loop in \(U\) based at \(p\). The retraction and winding-number argument shows that \(\gamma_n\) is not nullhomotopic in \(H\). Since this holds for every \(U\), no neighborhood satisfies the definition of semilocal simple connectivity at \(p\). \(\square\)

A Common Mistake

Pitfall. Since the diameters \(2/n\) tend to zero, one might conclude that the loops \(\gamma_n\) become nullhomotopic in \(H\). Small diameter alone does not give a contraction. For example, the tempting radial formula \(K(s,t)=(1-s)\gamma_1(t)\) need not even stay in \(H\). At \(t=1/8\) and \(s=1/3\), it gives \[ \left(\frac{2}{3}\left(1-\frac{\sqrt{2}}{2}\right),-\frac{\sqrt{2}}{3}\right). \] For a nonzero point \((x,y)\) to lie on some \(C_m\), its circle equation requires \(x^2+y^2=2x/m\). At the displayed point, \((x^2+y^2)/(2x)=2/3\), which would require \(1/m=2/3\), or \(m=3/2\). No \(m\in\mathbb{N}\) has that value, so the point is not in \(H\). The radial formula is not a homotopy in \(H\). More decisively, the retraction shows that no alternative based nullhomotopy of \(\gamma_n\) in \(H\) can exist.