Answer: Hawaiian Earring vs Infinite Wedge: Compact vs Not

Key Idea / Intuition

The Hawaiian Earring and the infinite wedge look identical combinatorially — both are "infinitely many circles glued at a point." The difference is purely topological: how open sets near the basepoint behave. In the Hawaiian Earring, every neighborhood of the origin must contain entire small circles (since those circles shrink to the origin), forcing compactness. In the infinite wedge, you can choose open sets that contain only a small arc of each circle, so no finite subcover works, and the space is not compact. Compactness is a topological invariant, so they cannot be homeomorphic.

---

Formal Proof / Solution

Step 1: The Hawaiian Earring $H$ Is Compact

Recall $H = \bigcup_{n=1}^{\infty} C_n \subset \mathbb{R}^2$, where $C_n$ is the circle of radius $1/n$ centered at $(1/n, 0)$.

Claim: $H$ is compact.

$H$ is a subspace of $\mathbb{R}^2$. We show it is closed and bounded.

• Bounded: Every point of $H$ lies on some $C_n$, which is contained in the closed disk of radius $2/n \leq 2$ centered at the origin. So $H \subset \overline{B(0,2)}$.

• Closed: Let $(x_k, y_k) \in H$ with $(x_k, y_k) \to (x, y)$. If infinitely many terms lie on a single $C_n$, then $(x,y) \in C_n \subset H$. Otherwise, the terms lie on circles $C_{n_k}$ with $n_k \to \infty$; since $C_{n_k}$ has diameter $2/n_k \to 0$ and all pass through the origin, we get $(x,y) = (0,0) \in H$.

So $H$ is closed and bounded in $\mathbb{R}^2$, hence compact by Heine–Borel.

---

Step 2: The Infinite Wedge $W$ Is Not Compact

Give $W = \bigvee_{n=1}^{\infty} S^1$ the standard quotient/CW topology: a set is open in $W$ if and only if its preimage in $\bigsqcup_{n=1}^\infty S^1$ is open in the disjoint union.

Claim: $W$ is not compact.

Construct an open cover with no finite subcover. For each $n$, let $U_n$ be the open set in $W$ consisting of:

$U_n = \{\text{basepoint}\} \cup \bigcup_{k=1}^{n} S^1_k \cup \bigcup_{k > n} \left(S^1_k \setminus \{q_k\}\right)$

where $q_k$ is some non-basepoint on $S^1_k$. More directly:

For each circle $S^1_n$ in $W$, pick a non-basepoint $q_n \in S^1_n$. Let

$V_n = W \setminus \{q_n\}.$

Each $V_n$ is open in $W$ (since $\{q_n\}$ is closed), and $\bigcup_{n=1}^\infty V_n = W$ because each point $q_n$ is covered by $V_m$ for any $m \neq n$.

However, no finite subcollection covers $W$: any finite set $\{V_{n_1}, \ldots, V_{n_k}\}$ fails to cover the point $q_{n_j}$ for $j$ not in $\{n_1, \ldots, n_k\}$...

Let us give a cleaner argument. Consider the open cover $\{U_n\}_{n \geq 1}$ where

$U_n = \{\text{basepoint}\} \cup \bigcup_{k=1}^{n} S^1_k \cup \bigcup_{k > n} A_k$

with $A_k = S^1_k \setminus \{\text{antipodal point of basepoint on } S^1_k\}$ being a proper open arc. Alternatively, the simplest argument:

Direct argument: The CW topology on $W$ makes each circle $S^1_n$ a closed subspace. For each $n$, pick a point $p_n \in S^1_n$ distinct from the basepoint. The set $\{p_n : n \geq 1\}$ is closed and discrete in $W$ (each $S^1_n$ contributes exactly one point, and they don't accumulate anywhere in $W$ because any compact subset of $W$ meets only finitely many circles non-trivially). An infinite closed discrete set in a compact space is impossible (it would have a limit point, but there is none). Hence $W$ is not compact.

---

Step 3: Conclusion

Since compactness is a topological invariant:

$H \text{ is compact}, \quad W \text{ is not compact} \implies H \not\cong W.$

---

Bonus Intuition: What Goes Wrong at the Basepoint?

In $H$: any open neighborhood of the origin must contain all of $C_n$ for sufficiently large $n$ (since the circles shrink to the origin). This "swallowing" of small circles is exactly what forces compactness.

In $W$: a neighborhood of the basepoint can be chosen to contain only a tiny arc of each $S^1_n$, with no obligation to "swallow" anything. The circles don't shrink, so they remain independent — and this independence is what prevents compactness.

This is a beautiful illustration of how the same combinatorial data (countably many circles at a point) can yield topologically inequivalent spaces depending on how limits behave near the gluing point.