  In a topological Riesz space there are two types of bounded subsets: order
bounded subsets and topologically bounded subsets. It is natural to ask (1)
whether an order bounded subset is topologically bounded and (2) whether a
topologically bounded subset is order bounded. A classical result gives a
partial answer to (1) by saying that an order bounded subset of a locally solid
Riesz space is topologically bounded. This paper attempts to further
investigate these two questions. In particular, we show that (i) there exists a
non-locally solid topological Riesz space in which every order bounded subset
is topologically bounded; (ii) if a topological Riesz space is not locally
solid, an order bounded subset need not be topologically bounded; (iii) a
topologically bounded subset need not be order bounded even in a locally
convex-solid Riesz space. Next, we show that (iv) if a locally solid Riesz
space has an order bounded topological neighborhood of zero, then every
topologically bounded subset is order bounded; (v) however, a locally
convex-solid Riesz space may not possess an order bounded topological
neighborhood of zero even if every topologically bounded subset is order
bounded; (vi) a pseudometrizable locally solid Riesz space need not have an
order bounded topological neighborhood of zero. In addition, we give some
results about the relationship between order bounded subsets and positive
homogeneous operators.
