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Possible Duplicate:
A wedge sum of circles without the gluing point is not path connected

I know that figure eight is not a manifold because its center has no neighborhood homeomorphic to $\mathbb{R}^n$. But how to prove this strictly?

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Suppose that there was a neighborhood $U$ of the center point $P$ that was homeomorphic to $\mathbb{R}^n$. Consider $U \setminus \{P\}$. How many connected components does it have? How many connected components are there in $\mathbb{R}^n \setminus \{\text{point}\}$? [Be careful to note that the answer is different for $n=1$ than for $n > 1$, but that doesn't ultimately cause any trouble.]

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    @Gobi: $P$ has arbitrarily small $+$-shaped nbhds; no point of $\Bbb R^n$ has such a nbhd. See [this answer](http://math.stackexchange.com/a/213538/12042) and the comments following it.2012-10-22