If (X, \tau) and (X, \nu) are complete metrizable TVSs (i.e. F-spaces) and if \nu is coarser than \tau then \tau = \nu; this is no longer guaranteed to be true if any one of these metrizable TVSs is not complete. Said differently, if (X, \tau) and (X, \nu) are both F-spaces but with different topologies, then neither one of \tau and \nu contains the other as a subset. One particular consequence of this is, for example, that if (X, p) is a Banach space and (X, q) is some other normed space whose norm-induced topology is finer than (or alternatively, is coarser than) that of (X, p) (i.e. if p \leq C q or if q \leq C p for some constant C > 0), then the only way that (X, q) can be a Banach space (i.e. also be complete) is if these two norms p and q are equivalent; if they are not equivalent, then (X, q) can not be a Banach space.
