About symplectic embedding I never read about the symplectic embeddings.   While reading a general math note, I have following question:
Does every symplectic manifold $(M,\omega)$ can be symplectically embedded to some $(\mathbb R^n, \omega_{std})$.
Can I have an counter example of a symplectic manifold which can not be embedded(symplectically) 
If questions are stupid and trivial.. I request to provide reference so that I can learn this type of questions.
 A: Your question has a trivial answer and a deeper answer. 
First of all, there is an obvious obstruction. The standard symplectic form on $\mathbb{R}^{2n}$ is exact, so if there is an embedding $M \hookrightarrow \mathbb{R}^{2n}$, the cohomology class of $\omega$ has to vanish. In particular, then, $M$ cannot be a closed symplectic manifold. So for instance, $\mathbb{C}P^1$ cannot be embedded symplectically in $\mathbb{R}^{2n}$ for any $n$.
There is a great theorem of Gromov, however, that such topological obstructions are the only obstructions if $M$ is an open symplectic manifold. Such obstructions are also the only obstructions for a codimension 4 embedding. This has to do with Gromov's h-principle. This is a beautiful area of mathematics, and I high recommend the book by Eliashberg and Mishachev (though be careful! it has a fair number of misprints)

I forgot to mention: if you want to embed in $\mathbb{C}P^n$ for $n$ sufficiently large, you can always do it. See for instance this Mathoverflow question and its references:
https://mathoverflow.net/questions/59733/projective-embedding-of-symplectic-manifolds
