Does a manifold without boundary necessarily have to be the boundary of some other higher dimensional manifold? Let M be a n-dimensional manifold. It is well known that if M is the boundary of some other (n+1)-dimensional manifold, the manifold M itself has no boundary. 
My question is: Does it work the other way? If some n-manifold M has no boundary, does it imply that there exits some (n+1)-dimensional manifold such that M=boundary of N?
I can not find counterexamples neither proof it (I am studying physics), and have not found any answer anywhere. Thanks beforehand.
 A: Tsemo's comment seems to be irrelevant. A manifold which is the boundary of another manifold is called null-bordant. There are many obstructions to being null-bordant, but every oriented manifold up to dimension 3 does bound. 
However, a Poincare duality argument shows that if $M$ is a boundary, we must have $\chi(M)$ even. So $\Bbb{RP}^2$ and $\Bbb{CP}^2$ do not bound. 
A: Two closed manifolds $M$ and $N$ are cobordant if and only if $M\sqcup N$ is the boundary of a compact manifold $W$. Let $\mathfrak{N}_k$ be the set of $k$ dimensional manifolds up to cobordism. It is easily checked that $\mathfrak{N}_k$ is a group where every element has order at most $2$. Let $\mathfrak N_*=\oplus_k \mathfrak N_k$ be the direct sum of these groups. This turns into a ring where the multiplication is given by the cartesian product of manifolds.
Your question asks if $\mathfrak N_*$ is trivial. It is far from it: I dimension zero a single point does not bound for example (The only compact one dimensional manifolds are unions of circles and intervals, the number of boundary components is even). I give a higher dimensional example below
Thom set out to compute this ring $\mathfrak N_*$. He found that 
$$
\mathfrak{N}_*\cong \mathbb{Z}_2[x_0,x_2,x_4,x_5,\ldots]
$$
where there is one generator $x_i$ in every degree $i$ for which $i\not= 2^k-1$ for some $k$. Explicit generators as manifolds are known (for example the even $x_{2j}$ can chosen to be real projective spaces $\mathbb{RP}^{2j}$.
Moreover one can check if two closed $k$-dimensional manifolds $M$ and $N$ are cobordant. To do this, one needs to compute all the Stiefel-Whitney numbers of the manifolds. These are numbers one can compute for a manifold. If all Stiefel-Whitney numbers agree, there is a compact manifold $W$ with $\partial W=M\sqcup N$. 
The Euler characteristic modulo $2$ is an example of a Stiefel-Whitney number. If it does not vanish, the manifold is not a boundary of a higher dimensional manifold. Examples manifolds with odd Euler characteristic are even dimensional real projective spaces $\mathbb{RP}^{2j}$. 
One can also ask for other types of cobordism: For example what would happen if all manifolds are required to be oriented? There are also answers in this case, but they are more complicated to state. 
