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I'm working with the definition:

A smooth embedding of M into N is an injective immersion $F:M\rightarrow N$ that is also a topological embedding.

I'd like to explain why the klein bottle cant embed in $\mathbb{R}^3$. The immersion is not injective, thus results in self intersections, correct?

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    If you have self-intersection in a point $P$, how do you cope with the non-unicity of tangent space in $P$, for example ? – 2017-02-02
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    This reasoning raises the question why the immersion is not (cannot be) injective. – 2017-02-02
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    Are you asking why the "standard picture" of the Klein bottle is not an embedding? Or, instead, how to prove that there does not exist any embedding? – 2017-02-02
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    It suffices to make it kleiner und kleiner, thus at the end, we will not see the self intersection... – 2017-02-02
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    Argue that $F$ cannot be an immersion since that Klein bottle contains a copy of the Mobius strip. – 2017-02-05

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