Hi: Just curious: a continuous function f:X-->Y ; X,Y topological spaces, can fail to be an embedding because it is not 1-1, or, if f is 1-1 , f can fail to be an embedding because, for U open in X f(U) is not open in f(X). Can anyone think of a "reasonable" example of the last case, by reasonable I mean no spaces with finitely many points, the inclusion map from a space into itself with a different topology (i.e.: we have (X,T) and (X,T') , and we use i:X-->X : i(x)=x ) , or maybe or something one could find in "Counterexamples" book. If A is a strict subspace of X , then the subspace topology on A guarantees that i:A-->X , the inclusion, is an embedding. Is this true for other topologies on A.? Thanks.