Building a Sequence of Tangent Circles
The Hawaiian earring is a planar space made by arranging infinitely many circles so that they shrink toward a single common point. The construction is simple, but it combines two features that matter in topology: infinitely many pieces approach the same point, and the pieces remain joined there. We will define the space precisely and prove that it is compact and path-connected.
We use \(\mathbb{R}^2\) with its Euclidean topology and Euclidean norm \(\|(x,y)\|=\sqrt{x^2+y^2}\). Each circle below has center on the positive horizontal axis and passes through the origin. The space itself has the subspace topology inherited from \(\mathbb{R}^2\).
Every \(C_n\) contains the origin, because substituting \((0,0)\) into its defining equation gives \((0-1/n)^2+0^2=1/n^2\). Expanding the equation gives a convenient equivalent test for membership: $$ x^2+y^2=\frac{2x}{n}. $$ The centers and radii both tend to zero as \(n\) increases. We will use a stronger geometric bound: every point of \(C_n\) is at distance at most \(2/n\) from the origin.
Worked Example: Checking Points on the First Two Circles
The expanded equation for \(C_1\) is \(x^2+y^2=2x\). For \(p=(2,0)\), the left side is \(2^2+0^2=4\), and the right side is \(2\cdot2=4\). Thus \(p\in C_1\).
For \(C_2\), the equation is \(x^2+y^2=x\). Let \(q=(1/2,1/2)\). The left side is \((1/2)^2+(1/2)^2=1/4+1/4=1/2\), and the right side is \(1/2\), so \(q\in C_2\). The same point is not on \(C_1\): its left side is \(1/2\), whereas the right side \(2x\) is \(2(1/2)=1\).
Finally, the origin satisfies \(x^2+y^2=2x/n\) for every \(n\), since both sides equal zero. These checks show that the circles share at least the origin; in fact, they share no other points, as the next argument verifies.
How the Circles Fit Together
The circles are tangent at the origin: the radius from \((1/n,0)\) to the origin is horizontal, so the tangent line there is vertical, \(x=0\), for every \(n\). Their pairwise intersection can also be verified directly from the equations.
Proof. Suppose \((x,y)\in C_m\cap C_n\). The expanded equations give \(x^2+y^2=2x/m\) and \(x^2+y^2=2x/n\). Subtracting yields \(2x(1/m-1/n)=0\). Since \(m\ne n\), the factor \(1/m-1/n\) is nonzero, so \(x=0\). Substituting \(x=0\) into either circle equation gives \(y^2=0\), hence \(y=0\). Conversely, the origin belongs to every circle, so it belongs to their intersection. \(\square\)
Worked Example: The Circles Shrink Toward the Origin
For \(z=(x,y)\in C_n\), write \(c_n=(1/n,0)\). The distance from \(z\) to \(c_n\) is exactly the radius \(1/n\), so the triangle inequality gives \(\|z\|\leq\|z-c_n\|+\|c_n\|=1/n+1/n=2/n\).
This bound is attained at the rightmost point \(r_n=(2/n,0)\). Indeed, \((2/n-1/n)^2+0^2=1/n^2\), so \(r_n\in C_n\), and \(\|r_n\|=2/n\). For example, when \(n=4\), the circle has center \((1/4,0)\) and radius \(1/4\); its rightmost point is \((1/2,0)\), whose distance from the origin is \(1/2=2/4\).
As \(n\) increases, the upper bound \(2/n\) tends to zero. In particular, for any \(\varepsilon>0\), choosing \(N\in\mathbb{N}\) with \(N>2/\varepsilon\) ensures that every point of \(C_n\) has norm at most \(2/n\leq2/N<\varepsilon\) whenever \(n\geq N\).
Compactness and Path-Connectedness
A space is compact if every cover of it by open sets has a finite subcover. A space is path-connected if for every pair of its points \(p,q\), there is a continuous map \(\gamma:[0,1]\to E\) with \(\gamma(0)=p\) and \(\gamma(1)=q\). The following result establishes both properties for this construction.
Proof. We first prove compactness. The bound proved above gives \(\|z\|\leq2/n\leq2\) for every \(z\in C_n\), since \(n\geq1\). Thus \(E\) is bounded in \(\mathbb{R}^2\).
To prove that \(E\) is closed, take any point \(p\in\mathbb{R}^2\setminus E\), and put \(d=\|p\|\). Since the origin belongs to \(E\), we have \(d>0\). Choose \(N\in\mathbb{N}\) such that \(2/N<d/2\). If \(n\geq N\) and \(z\in C_n\), then \(\|z\|\leq2/n\leq2/N<d/2\). Therefore \(\|p-z\|\geq\|p\|-\|z\|>d/2\), so the open ball \(B(p,d/2)\) misses every \(C_n\) with \(n\geq N\).
The remaining set \(F=\bigcup_{n=1}^{N-1}C_n\) is a finite union of closed sets. Each \(C_n\) is closed because it is the preimage of the closed set \(\{1/n^2\}\subseteq\mathbb{R}\) under the continuous function \((x,y)\mapsto(x-1/n)^2+y^2\). Thus \(F\) is closed. As \(p\notin F\), the open set \(\mathbb{R}^2\setminus F\) contains \(p\). The intersection \(B(p,d/2)\cap(\mathbb{R}^2\setminus F)\) is an open neighborhood of \(p\) disjoint from all of \(E\). This proves that \(\mathbb{R}^2\setminus E\) is open, so \(E\) is closed. By the Heine–Borel theorem for Euclidean space, a closed and bounded subset of \(\mathbb{R}^2\) is compact; hence \(E\) is compact.
For path-connectedness, take \(p\in E\). It lies on at least one circle \(C_n\). Subtracting the center and multiplying by \(n\) turns its coordinates into a point \((u,v)\) on the unit circle, so there is an angle \(\theta\in\mathbb{R}\) with \(u=\cos\theta\) and \(v=\sin\theta\). Define \(\alpha(t)=(1/n+(1/n)\cos((1-t)\theta+t\pi),(1/n)\sin((1-t)\theta+t\pi))\) for \(0\leq t\leq1\). This map is continuous, lies in \(C_n\) at every \(t\), starts at \(p\), and ends at the origin because \(\cos\pi=-1\) and \(\sin\pi=0\). Thus every point of \(E\) can be joined to the origin by a path in \(E\).
Now let \(p,q\in E\), and let \(\alpha\) be a path from \(p\) to the origin and \(\beta\) a path from \(q\) to the origin. Define \(\gamma(t)=\alpha(2t)\) for \(0\leq t\leq1/2\), and \(\gamma(t)=\beta(2-2t)\) for \(1/2\leq t\leq1\). Both formulas give the origin at \(t=1/2\); the first gives \(p\) at \(t=0\), and the second gives \(q\) at \(t=1\). Each piece is continuous, and the pieces agree at their common endpoint. The finite pasting lemma therefore gives a continuous path \(\gamma\) in \(E\) from \(p\) to \(q\). The lemma applies because the two pieces have closed domains \([0,1/2]\) and \([1/2,1]\); preimages of closed sets are closed on each domain, and their finite union is closed in \([0,1]\). Hence \(E\) is path-connected. \(\square\)
Worked Example: An Explicit Path Between Two Circles
Take \(p=(2,0)\in C_1\) and \(q=(1/2,1/2)\in C_2\). A path from \(p\) to the origin is \(\alpha(t)=(1+\cos(\pi t),\sin(\pi t))\). At \(t=0\), this is \((1+\cos0,\sin0)=(2,0)=p\). At \(t=1\), it is \((1+\cos\pi,\sin\pi)=(0,0)\). At every \(t\), \((\alpha_1(t)-1)^2+\alpha_2(t)^2=\cos^2(\pi t)+\sin^2(\pi t)=1\), so \(\alpha(t)\in C_1\).
A path from \(q\) to the origin on \(C_2\) is \(\beta(t)=(1/2+(1/2)\cos(\pi/2+t\pi/2),(1/2)\sin(\pi/2+t\pi/2))\). At \(t=0\), the angle is \(\pi/2\), so \(\beta(0)=(1/2+0,1/2)=(1/2,1/2)=q\). At \(t=1\), the angle is \(\pi\), so \(\beta(1)=(1/2-1/2,0)=(0,0)\). Its displacement from the center \((1/2,0)\) has coordinates \((\cos(\pi/2+t\pi/2)/2,\sin(\pi/2+t\pi/2)/2)\); the sum of their squares is \(1/4\), the squared radius of \(C_2\).
Combining the paths gives \(\gamma(t)=\alpha(2t)\) on \(0\leq t\leq1/2\) and \(\gamma(t)=\beta(2-2t)\) on \(1/2\leq t\leq1\). At the joining time, both values are the origin, and the endpoints are \(p\) and \(q\), respectively. Thus the computation checks both the path’s endpoints and its continuity at the join.
A Common Pitfall
When checking circle membership, use the full equation \(x^2+y^2=2x/n\). For instance, a tempting but incorrect check for \((1,0)\in C_1\) is to write \((1-1)^2+0^2=0=1-1\). That calculation does not test the defining equation: the right side of the expanded equation is \(2x=2\), while the left side is \(1^2+0^2=1\). Therefore \((1,0)\notin C_1\). The point \((2,0)\) is a valid point of \(C_1\), since its two sides are both \(4\).
There is also a general compactness trap: an arbitrary union of compact sets need not be compact. Each singleton \(\{n\}\) is compact in \(\mathbb{R}\), but their union \(\mathbb{N}\) is not compact in its subspace topology. The sets \((-1,k)\cap\mathbb{N}\), for \(k\in\mathbb{N}\), form an open cover of \(\mathbb{N}\); any finite subcollection has a largest index \(K\) and misses the integer \(K\). The Hawaiian earring is compact for a different reason: its infinitely many circles shrink toward the origin, and the closedness proof controls all sufficiently small circles together rather than relying on compactness of each circle separately.
The construction is useful because it gives a compact, path-connected space in which infinitely many distinct circles accumulate at one point. The shrinking estimate and the shared point serve different purposes: shrinking makes it possible to prove closedness, while the shared point supplies paths between points on different circles.
Check Your Understanding
Use the defining equations and the compactness and path arguments above to answer the following.
- Write the expanded equation for \(C_3\), and test whether \((2/3,0)\) belongs to \(C_3\).
- Prove directly from the circle equations that \(C_2\) and \(C_5\) meet only at the origin.
- What upper bound does the shrinking estimate give for the distance to the origin of any point on \(C_6\)?
- Explain why proving that each \(C_n\) is closed does not, by itself, prove that \(E\) is closed.
- Construct a path from an arbitrary point of \(C_n\) to the origin using an angle parameter.