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Show that the union $X$ of the $x$-axis and the $y$-axis in $\mathbb{R}^2$ is not a manifold.

Is the following a valid way of arguing?

Suppose $X$ were a manifold. Then there would be a nbhd $U$ of the origin in $X$ that is homeomorphic to $\mathbb{R}^2$. Then we also have that $U$ with the origin removed is homeomorphic to $\mathbb{R}^2$ with one point removed. But this can't be since $U$ without the origin is not connected, whereas $\mathbb{R}^2$ with one point removed is connected.

iwriteonbananas
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    So you have shown that it is not a $2$ manifold. It could still be a $1$ manifold (or any $n\neq 2$, but I think your argument still work. –  Dec 31 '14 at 15:25
  • @John is right, you seem to ahve only shown it is not a 2-manifold. – JC574 Dec 31 '14 at 15:26
  • For why it is not a $1$-manifold, see the "duplicate" https://math.stackexchange.com/questions/4567600 – aschepler Nov 03 '22 at 19:13

1 Answers1

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You should show that it isn't a manifold of any dimension. Your argument proves that it is not a two-dimensional manifold. If you take any point of $X$ other than the origin, it clearly has a neighbourhood homeomorphic to an interval in $\mathbb{R}$, so if $X$ is a manifold, it is a one-dimensional manifold. You can adapt your argument for this case, but it is slightly less trivial.

Michael Albanese
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    Hmm ok, suppose it were a 1-manifold and $V$ a nbhd of the origin which is homeomorphic to $\mathbb{R}$. Then removing the origin gives us 4 components in $V$ and 2 components in $\mathbb{R}$. Is that enough to prove it's not a 1-manifold? – iwriteonbananas Dec 31 '14 at 15:33
  • Exactly. That's enough because you've shown that any neighbourhood of the origin cannot be homeomorphic to $\mathbb{R}$, so $X$ is not locally Euclidean and therefore not a manifold. – Michael Albanese Dec 31 '14 at 15:35
  • Gotcha, thank u! – iwriteonbananas Dec 31 '14 at 15:49