Let \(X\) be the infinite earring (see Example 1 of §80). Show that \(X\) is a compact Hausdorff space with a countable basis whose fundamental group is uncountable. [Hint: Let \(r_n: X \to C_n\) be a retraction. Given a sequence \(a_1, a_2, \ldots\) of zeros and ones, show there exists a loop \(f\) in \(X\) such that, for each \(n\), the element \((r_n)_*[f]\) is trivial if and only if \(a_n = 0\).]
Solution
Let the infinite earring be \[ X=\bigcup_{n\ge1}C_n, \] where the circles \(C_n\) are tangent at the common point \(p\) and have radii tending to \(0\).
The set \(X\) is bounded. It is closed in \(\mathbb R^2\): away from \(p\), only finitely many circles meet a sufficiently small neighborhood, and each is closed; the only possible accumulation point of the family is \(p\in X\). Hence \(X\) is compact. As a subspace of \(\mathbb R^2\), it is Hausdorff and second countable.
For each \(n\), let \[ r_n:X\to C_n \] be the retraction equal to the identity on \(C_n\) and sending every other circle to \(p\). It is continuous: continuity is obvious away from \(p\), and near \(p\) all sufficiently small circles lie in any prescribed neighborhood of \(p\).
Now let \(a=(a_1,a_2,\ldots)\in\{0,1\}^{\mathbb N}\). Divide \([0,1]\) into intervals \[ I_n=[1-2^{-(n-1)},\,1-2^{-n}] \] accumulating at \(1\). On \(I_n\), let \(f_a\) traverse \(C_n\) once if \(a_n=1\), and stay at \(p\) if \(a_n=0\); set \(f_a(1)=p\). This defines a continuous loop because \(\operatorname{diam}(C_n)\to0\).
For each \(n\), \[ (r_n)_*[f_a]= \begin{cases} 0,&a_n=0,\\ 1\in\pi_1(C_n,p)\cong\mathbb Z,&a_n=1. \end{cases} \] Therefore if \(a\ne b\), choose \(n\) with \(a_n\ne b_n\). Then \[ (r_n)_*[f_a]\ne(r_n)_*[f_b], \] so \([f_a]\ne[f_b]\) in \(\pi_1(X,p)\). Thus \[ \{0,1\}^{\mathbb N}\hookrightarrow\pi_1(X,p). \] The left side is uncountable, hence \(\pi_1(X,p)\) is uncountable.