Can an $n$-dimensional manifold live inside $\mathbb{R}^m$ for $m < n$? This may be a silly question but I want to make sure my general intuition is correct. I don't think it is possible for an $n$-dimensional manifold to live inside $\mathbb{R}^m$ for $m < n$ because in the "best case scenario" we can move in all directions at a given point of a manifold but this movement cannot be mirrored in its coordinate chart image where a small enough ball centered at the point's image has more dimensions for movement. Am I being completely silly? 
 A: Let $n>m$.
If you could embed an $n$-manifold in $\Bbb R^m$, then you could
embed $\Bbb R^n$ in $\Bbb R^m$ (take a coordinate neighbourhood).
If you could embed $\Bbb R^n$ in $\Bbb R^m$, then you could
embed $\Bbb R^{m+1}$ in $\Bbb R^m$ (as $n\ge m+1$).
If you could embed $\Bbb R^{m+1}$ in $\Bbb R^m$, then you could
embed the $m$-sphere $S^m$ in $\Bbb R^m$.
But that contradicts the Borsuk-Ulam theorem.
Alternatively, consider $\Bbb R^m$ as a hyperplane in $\Bbb R^{m+1}$.
If you can embed $S^m$ in $\Bbb R^m$ then you can embed it in a hyperplane
in $\Bbb R^{m+1}$. But the image of such an embedding will not
separate $\Bbb R^{m+1}$, thus contradicting the Jordan-Brouwer separation theorem.
A: Your intuition is right. There are rigorous ways to show that. I've seen it done often the following way:
By definition $n$-dimensional manifold is covered by open subset of $\mathbb{R}^{n}$, e.g. open balls. Suppose there is an inclusion $i: B \to \mathbb{R}^{m}$, where $B$ is a unit ball in $\mathbb{R}^{n}$. 
Then $i(B) \setminus i(0)$ is homeomorphic to $B\setminus 0$, we have $H_{n}(B\setminus 0, \mathbb{Z}) = \mathbb{Z}$, but $H_n(i(B) \setminus i(0), \mathbb{Z}) = 0$ for dimensional reasons.
A: Your intuition is right, but only if you insist that "live inside" means
that 
$$
f : M^n \to \Bbb R^m.
$$
the function that puts the manifold into $m$-space, is continuous. If you allow arbitrary functions, then there's certainly a map from, say, a sphere like the surface of the earth into the real line, because they have the same cardinality. (The other answers are implicitly assuming that the inclusion map is required to be continuous, which is almost certainly what you intended, but did not say.) 
