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I have tried to solve the problem : Prove that the union of the two coordinate axes in $\mathbb{R^2}$ is not a manifold.

Let $X = \{(x,y) \in \mathbb{R^2} : x=0~or~ y=0\}$ be the union of the two coordinate axes in $\mathbb{R^2}$.

What happens to a neighborhood of $0$ when $0$ is removed?

I would like to finalize this question - issue by myself. I just need an explanation. Is anyone is able to help me?

3 Answers3

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How many connected components must any neighborhood of $0$ in $X$ have? How many connected components does an open interval in $\mathbb{R}$ with a point removed have? So can there be a homeomorphism between a neighborhood of $0$ in $X$ and an open interval in $\mathbb{R}$?

Michael Joyce
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I can't lead with Micheal Joyce response. Someone could explain his comment in detail or suggest me another easier to use. The number of connected components seems interesting, but I can't use that information. How do I '' count '' the latter.

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    In general, it is better to respond with a comment to an answer rather ask a question about an answer in a different answer. But I'm glad from your other answer that you've figured the problem out. – Michael Joyce Apr 28 '15 at 21:06
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I found the problematic.

Removal of the point $0$ breaks the set into four connected components, something that cannot happen to any manifold. An alternative argument is that if $M \subset \mathbb{R}^2$ is any manifold and $x \in M$ then by the inverse function theorem, in some neighbourhood of $x \in M$, either the projection to the $x_1$-axis, or the projection to the $x_2$-axis is a local diffeomorphism, and in particular is one-to-one. Neither is one-to-one in this case.

Michael Joyce
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