By manifold i mean an embedded submanifold of $\mathbb{R}^n$.
The question is exactly the one in the title.
A manifold M is closed if $\partial M = \emptyset$.
I tried to prove it in the case of $k = n$, using the fact that i can put a (at least) continuous unit normal vector field on this "surface", and considering the lines spanned by the opposite of this normal vectors, then taking segments on those lines (wo vertices) defined by the intersections of such lines with the surface. Doing this I intuitively identified "the space inside" this surface, and then i wanted to prove it to be an open with topological boundary consisting of my $(n-1)$-manifold, but here I'm stuck.
I'm not even shure this theorem being true, so if a weaker version is, please let me know. Sorry for my english, I hope the question is clear enough. Thanks in advance.
EDIT:
Since it has been pointed out that the theorem is not true, I ask only for the specific case I need:
It would be sufficient to show that every closed oriented $C^p$ 2-manifold of $\mathbb{R}^3$ is the boundary of some $C^p$ 3-manifold-with-boundary.
Any $C^p$-manifold for $p\geq 1$ has a unique smooth structure, so we can reduce to the $p = \infty$ case. The strong Whitney theorem also guarantees that any smooth manifold $M^n$ can be embedded in $M^{2n}$, so that condition is easy to relax too, unless you want to put more restrictions on $k$ versus $n$.
Even in just the topological category, though, it isn't true that any closed manifold $M$ is a boundary; that is $M = \partial N$ for some compact manifold with boundary $N$. (We need $N$ to be compact to avoid trivial cases like $N = M \times \mathbb{R}^{\geq 0}$). For even $n$, for example, $M = \mathbb{RP}^{n}$ has $S^n$ as a double cover, so $\chi(\mathbb{RP}^n) = 1$. But if some compact $N^{2n+1}$ has $\partial N = \mathbb{RP^n}$, then the connected sum $P = N \#_{\partial N} N$ of two copies of $N$ along their boundary has $\chi(P) = 2\chi(N) - \chi(\mathbb{RP^n})$ odd. But $\chi(P)$ has odd dimension $2n + 1$, so $\chi(P)$ is even by Poincare duality over $\mathbb{Z}_2$.
There are other obstructions as well; take the Stiefel-Whitney classes, for example. Going back to the smooth case, there's a result from Thom that the Stiefel-Whitney numbers of a smooth closed manifold $M$ (the pairing of products of those classes with the fundamental class of $M$ over $\mathbb{Z}_2$) vanish iff $M$ is the boundary of a smooth, compact manifold. With a bit of algebraic topology, this result implies, for example, that any orientable closed $3$-manifold is the boundary of a compact $4$-manifold. (There are also geometric proofs of that.) In the opposite direction, it gives a variety of smooth manifolds that aren't boundaries.
In general, what you're probably looking for is the idea of cobordism. Sticking to the category of smooth, compact, unoriented manifolds for now, manifolds $M, N$ are cobordant if there exists a (smooth, compact) manifold with boundary $W$ such that $\partial W = M \coprod N$. Thus, for example, any $M$ is cobordant to itself via $W = M \times [0, 1]$. Under disjoint union and product, the set of (smooth, compact) manifolds modulo cobordism is a ring graded by dimension. It's also quite small: it turns out to be a polynomial algebra $\mathbb{Z}_2[x_i]$ with $\deg x_i = i$ and $i$ running over all integers $i > 1$ with $i\not = 2^k - 1$. (For even $i$, we can take $i = \mathbb{RP}^i$; for odd $i$, there's also an explicit but more complicated representative.) There are similar results of varying degrees of complexity and explicitness for different kinds of cobordism: oriented (where $W$ has to preserve orientations), complex, etc. The upshot, though, is that there are well-known obstructions to a manifold's being a boundary, and that condition does not generally hold.