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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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    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.2012-06-12

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