Let $M$ a manifold and $X \subset M$. Let $N \subset X$ such that $N, X \backslash N$ are submanifolds of $M$. Can I conclude that $X$ is a submanifold of $M$?
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The answer is no.
Consider $M = \mathbb C$ as a $2$-dimensional real manifold. Take for $N \subseteq M$ the real axis. Take for $X\setminus N$ the "open upper unit circle", i.e. $\{z \in \mathbb C\ |\ |z|=1,\ Im\ z > 0\}.$ Yes, it's a bit unconventional to "define" $X\setminus N,$ but bear with me. Both $N$ and $X\setminus N$ are submanifolds of $M$, but $X = N \cup (X\setminus N)$ is not, because of the two points $\pm 1$ where $N$ and $X\setminus N$ come infinitely close to each other. At those points, $X$ doesn't have a well-defined dimension.
jflipp
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But you should mention, that compactness would be a reasonable assumption on the submanifolds which would make the answer yes. – Daniel Valenzuela Nov 28 '14 at 08:25