Let $X$ be the subspace of $\mathbb{R}^2$ that is the union of the circles $C_n$ of radius $n$ and center $(n,0)$ for $n \in \mathbb{N}$. Show that $X$ and $\bigvee_\infty S^1$ are homotopy equivalent, but not homeomorphic.

It is unclear to me why the "obvious" maps between the two spaces do not give a homeomorphism. It seems like the difference would have to be a result of the behavior of the map near the origin. Could I get a hint?

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    $\begingroup$ This isn't the Hawaiian earring: the circles get bigger with n. $\endgroup$ – Carl Feb 20 '12 at 23:39
  • $\begingroup$ Whoops. Thanks for catching that! $\endgroup$ – Qiaochu Yuan Feb 20 '12 at 23:44
  • $\begingroup$ I don't see why the second space doesn't have a countable local base in the common point, maybe I'm taking the wrong topology in the space or something, please correct me if I'm wrong, but, for each circle I can take a local basis in the common point thinking the circle as a subspace of the plane. So by making the union of all these basis, I think I can get a countable local basis (I don't make all the possible unions between those sets). Is the topology in the wedge sum, just the topology generated by all the open sets in each circle? $\endgroup$ – Frank Jun 12 '12 at 4:25
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    $\begingroup$ @Frank Yes, the topology is generated by the open sets in each circle. This can be seen using a diagonal argument. Assume there is some countable basis at the common point. Thus, we have a collection of basis elements $\{B_i\}_{i \in \mathbb{N}}$, each one containing the common point. For each $i$, we choose some open set $V_i \in S^1$ such that $V_i$ is not contained in the $i$th circle of $B_i$. Then the inclusion $\bigvee_{\infty} V_i \subset \bigvee_{\infty} S^1$ is not contained in any basis element in our collection. Let me know if you need more explanation. $\endgroup$ – Carl Jun 12 '12 at 7:18
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    $\begingroup$ Correction: You choose $V_i$ such that the $i$th component of $B_i$ is not contained in $V_i$. Basically, you choose a neighborhood of the common point of $\bigvee_\infty S^1$ such that the $i$th part is small enough to prevent the $i$th basis element from being in the neighborhood. $\endgroup$ – Carl Jun 12 '12 at 8:06

HINT: $X$ is first countable. The common point in the wedge does not have a countable local base.


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