[**Edit:** I have added some details, and a more explicit example by Milnor. I also ask a question of my own at the end.]

I will present a couple of examples verifying the conditions required in the question.$\newcommand{\RR}{\mathbb{R}}
\newcommand{\TT}{\mathbb{T}}
\newcommand{\ZZ}{\mathbb{Z}}$


### $H$-cobordant manifolds ###

Most importantly, it is known that if $M$ and $N$ are two $h$-cobordant closed topological manifolds of dimension greater than 3, then $M\times\RR$ is homeomorphic to $N\times\RR$. Here are a couple of quick references for this fact, both making use of the $s$-cobordism theorem:

- In a related mathoverflow question, Oscar Randal-Williams provides a fairly direct argument in a <a href="http://mathoverflow.net/questions/95129/manifolds-with-homeomorphic-interiors/95136#95136">comment to the answer</a>.

- The article "<a href="http://arxiv.org/abs/1212.5618">*Contact manifolds with symplectomorphic symplectizations*</a>" states such a result for smooth manifolds as corollary 2.5. Nevertheless, the proof described there applies also to topological manifolds.

[*Aside*: Incidentally, the converse also holds &mdash; although we will not make use of it. In another mathoverflow question, Igor Belegradek remarks in <a href="http://mathoverflow.net/questions/64029/if-a-manifold-suspends-to-a-sphere/64036#64036">his answer and the ensuing comments</a> that if $M\times\RR$ is homeomorphic to $N\times\RR$, then $M$ and $N$ are $h$-cobordant.]

So it suffices to find closed manifolds $M$ and $N$ which are $h$-cobordant, yet not homeomorphic. As required by the question, we also want $M$ to be null-cobordant; then $N$ will be null-cobordant as well.


### An example by Farrel and Hsiang ###

One such example is given by Farrel and Hsiang in their article "<a href="http://projecteuclid.org/euclid.bams/1183529046">*$H$-cobordant manifolds are not necessarily homeomorphic*</a>". There, the manifold $M$ is a product $M=L\times\TT^n$, where $L$ is any $3$-dimensional lens space such that $\pi_1(L)\simeq\ZZ/p^2$ for some prime $p$, and $\TT^n$ is a torus of dimension $n\geq 3$. Note that $M$ is null-cobordant since the torus $\TT^n$ is null-cobordant. Unfortunately, Farrel and Hsiang do not explicitly describe $N$.


### Milnor's examples ###

A more explicit example is given in the celebrated article "<a href="http://www.maths.ed.ac.uk/~aar/papers/milnhaup.pdf">*Two complexes which are homeomorphic but combinatorially distinct*</a>" by John Milnor. Milnor essentially proves that we can take $M=L(7,1)\times S^{2n}$ and $N=L(7,2)\times S^{2n}$ for any $n>0$, where $L(p,q)$ denotes the lens space of type $(p,q)$. This is *only partially* stated as theorem 4 in that article. By the way, observe that $M$ and $N$ are null-cobordant since spheres are null-cobordant.

I will briefly describe how to conclude from Milnor's article that $M$ and $N$ above actually meet our requirements. In section 2, Milnor sketches a proof that $M$ and $N$ are $h$-cobordant. At the end of section 4, in corollary 2, he uses the fact that $L(7,1)$ and $L(7,2)$ have distinct Reidemeister torsion to conclude that $M$ and $N$ have distinct Reidemeister torsion themselves, and thus are *not diffeomorphic*. But more is true: since the Reidemeister torsions of $M$ and $N$ differ, $M$ and $N$ are *not homeomorphic*. This follows from the topological invariance (i.e. invariance under homeomorphisms) of Reidemeister torsion, which is itself a simple consequence of the topological invariance of Whitehead torsion &mdash; proved by Chapman after the publication of Milnor's article.

In general, the results in sections 2 and 4 of Milnor's article actually prove that we can take $M=K\times S^n$ and $N=L\times S^n$, where:

- $K$, $L$ are parallelizable, smooth, closed $k$-manifolds;

- $K$, $L$ are homotopy equivalent;

- $n$ is even and $n\geq k$;

- the Reidemeister torsions of $K$ and $L$ are defined (for some complex-valued representation of their common fundamental group), and have different absolute values.

For instance, we can take $K$ and $L$ to be $3$-dimensional lens spaces, and not just $L(7,1)$ and $L(7,2)$. The homotopical classification and Reidemeister torsions of lens spaces are well-known: see http://www.map.mpim-bonn.mpg.de/Lens_spaces.


### A question ###

I now pose a question of my own, motivated by the prevalence of lens spaces above. I am not very familiar with lens spaces, so the answer may well be quite simple.

Can we actually take $M$ and $N$ to be some three dimensional lens spaces? In other words, are there non-homeomorphic three dimensional lens spaces which become homeomorphic after taking product with $\RR$? Observe that these lens spaces are necessarily null-cobordant, like all other $3$-manifolds.