Prove that the Hawaiian Earring $-\{(0, 0\}$ equipped with the subspace topology is not equivalent to the Hawaiian Earring equipped with disjoint union topology

Recall that the Hawaiian Earring (minus the common point) is the set $H_{-0} = \bigcup_{n = 1}^{\infty}X_n = \{(x-1/n)^2+y^2=1/n^2\} - \{(0,0)\}$

Let $\mathcal{S}$ denote the subspace topology on the Hawaiian Earring (minus the common point), and let $\mathcal{D}$ denote the disjoint union topology (as defined here: https://en.wikipedia.org/wiki/Disjoint_union_(topology)) on $\bigsqcup_{n = 1}^{\infty} X_n$

I need to show that $\mathcal{S} \neq \mathcal{D}$. I was given a hint to consider a neighbourhood of the deleted origin in both topologies.

Let $N \subseteq \mathbb{R}^2$ be such a neighbourhood, where $N$ is a ball centered at $(0, 0)$ of some radius $0 < \epsilon \leq 1$. $N$ completely contains infinitely many circles $X_n$ and intersects the rest of them, and since $H_{-0}$ has the subspace topology inherited from $\mathbb{R}^2$, $N$ is open in $H_{-0}$.

Now in each $X_n$ viewed individuallly as a topological space (where $X_n$ has the subspace topology inherited from $\mathbb{R}^2$. Pick $n \geq 1$, for this $n$ we have $N \cap X_n \subseteq \bigsqcup_{n = 1}^{\infty} X_n$ open in the disjoint union space $\bigsqcup_{n = 1}^{\infty} X_n$ (Why? Because $X_n \cap X_m = \emptyset$, so $N\cap X_n \cap X_m = \emptyset $ which is open in $X_m$ for any $m \in \mathbb{Z}$)

Now fix $i > 1$, such that $N \cap X_n$ intersects $X_i$ in two arcs of $X_i$. $N \cap X_n$ is not open in in $H_{-0}$, since it is not an open set of $\mathbb{R}^2$. Hence $\mathcal{D} \not\subseteq \mathcal{S}$

I'm sure my argument is wrong (and perhaps I've made some elementary mistake here that I'm not picking up), since for finitely many circles the disjoint union space and subspaces should agree, in my argument that isn't the case.

  • $\begingroup$ What is the origin in $\mathcal{D}$? $\endgroup$ – user60589 Sep 19 '17 at 11:11
  • $\begingroup$ I'm not sure I understand. What you mean by "let $\mathcal{D}$ denote the disjoint union topology on the Hawaiian ring". This doesn't make much sense since the Hawaiian ring is not a disjoint union of $X_n$ - they all have a common point. So what I think is that you want the wedge sum topology, i.e. $U$ is open in $X$ iff $U\cap X_n$ is open in each $X_n$. $\endgroup$ – freakish Sep 19 '17 at 11:41
  • $\begingroup$ But then they cannot be homeomorphic because the wedge sum is connected while $H-\{(0,0)\}$ is obviously not. The Hawaiian ring $H$ is homeomorphic to a one-point compactification of a disjoint union of open intervals. So $H-\{(0,0)\}$ is a disjoint union of open intervals. $\endgroup$ – freakish Sep 19 '17 at 11:41
  • $\begingroup$ If, on the other hand, you say that $X$ is actually the disjoint union of open intervals, then they are homeomorphic. Finally if $X$ is the disjoint union of circles then it is easy to see that any homeomorphism $X\to H-\{(0,0)\}$ induces a homeomorphism $S^1\to(0, 1)$ where here $(0,1)$ stands for the open interval. This is not possible since $(0,1)$ is not compact. $\endgroup$ – freakish Sep 19 '17 at 11:47

In your notation you need to distinguish between circles and open intervals.

For instance, $H=\bigcup_{n}X_n$ where $X_n=\{(x,y)|(x-1/n)^2+y^2=1/n^2\}$ is the n-th circle.

Each $I_n=X_n\backslash \{(0,0)\}$ is a circle with a point removed and so is homeomorphic to an open interval.

Now $H\backslash \{(0,0)\}\cong \coprod_{n}I_n$ where the right side is a disjoint union. After all, each $I_n$ is open in $H$.


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