TITLE: Any 3-manifold can be realized as the boundary of a 4-manifold
QUESTION [7 upvotes]: We know 

"Any closed, oriented $3$-manifold $M$ is the boundary of some oriented $4$-manifold $B$." See this post: Elegant proof that any closed, oriented 3-manifold is the boundary of some oriented 4-manifold?

I heard this statement is true:


(1) Any closed 3-manifold is a boundary of some compact 4-manifold.


See also this paper p.2's 3rd paragraph uses the fact:


(2) Any 3-manifold $M$ can be realized as the boundary of a
  4-manifold $B$. 


In particular, we know that all 3-manifolds can be triangulable. However for 4-manifolds, there are simply connected non-triangulable manifolds (such as the E$_8$ manifold). (Note: a closed 4-manifold is triangulable if and only if it's smoothable.) See this MO post: Not all manifolds can be triangulated


(3) For any 3-manifold $M_3$ that can be realized as the boundary of a
  4-manifold $B_4$, the $M_3$ must be triangulable. So must the $M_3$ be the boundary of a triangulable 4-manifold $B_4$?
(4) Are there any non-triangulable 4-manifold $B_4'$ with a 3-dimensional boundary (i.e. $B_4'$ is not closed)? Then would the 3-manifold boundary $M_3'$ be triangulable (if $M_3'$ is non-triangulable, isn't that leads to a contradiction)?


Can one show these (1), (2) and explain them as intuitively as possible?

REPLY [3 votes]: In this question it is shown how to see that two smooth manifolds are topologically cobordant if and only if they are smoothly cobordant, which answers some subset of the questions (the second manifold can be $S^3,$ though I am usually abusively think of it as being $\emptyset$), and in arbitrary dimension [the fact that any three-manifold bounds is more elementary and does not require the algebraic machinery].