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Suppose I have a symplectic manifold $(\mathcal{M}, \omega)$. Does it hold that any open subset of $(\mathcal{M}, \omega)$ is a symplectic submanifold?

The statement trivially holds for smooth manifold but in the symplectic setting I am not sure.

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up vote 5 down vote accepted

It trivially holds for symplectic manifolds as well. Let $(M^{2n}, \omega)$ be a symplectic $2n$-manifold and $X \subset M$ be an open subset. Then we know that $X$ is a $2n$-dimensional smooth submanifold of $M$. This implies that for all $x \in X$, $T_x M = T_x X$ (a $2n$-dimensional subspace of a $2n$-dimensional vector space is the whole vector space). Therefore since $(T_x M, \omega_x)$ is a symplectic vector space by the fact that $(M,\omega)$ is a symplectic manifold, $(T_x X, \omega_x)$ is a symplectic vector space as well. Therefore $X$ is a symplectic submanifold of $(M,\omega)$.

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