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?

smooth4-manifolds (well, except for the ones that just spit out a triangulated 4-manifold immediately!), and in dimension 4 smooth and PL are the same. The second question is trivial since all 3-manifolds can be triangulated. This question is more appropriate for math.se, and I have voted to close. $\endgroup$ – Andy Putman Jul 21 '18 at 2:57