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?


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.]

  • $\begingroup$ I got it, but can I assume that $U$ is a small nbd of $P$? Then the connected components will be 4, but if $U$ is strangely shaped, the components can be more than 4. $\endgroup$ – Gobi Oct 22 '12 at 12:55
  • 2
    $\begingroup$ @Gobi: $P$ has arbitrarily small $+$-shaped nbhds; no point of $\Bbb R^n$ has such a nbhd. See this answer and the comments following it. $\endgroup$ – Brian M. Scott Oct 22 '12 at 12:58

Not the answer you're looking for? Browse other questions tagged or ask your own question.