Let X and X' denote a single set in the two topologies T and T', respectively. Let i: X' -> X be the identity function. a) Show that i is continuous <=> T' is finer than T. Ok im able to show that for any set in T|X this set is in T'. This is done as follows: Assume i is continuous. For any open set X^U (in X), i^-1(X^U) = X^U is open in X'. Since X' is open in the space with T', X^U is open in that space (i.e. X^U is an element of T'). However, i dont see how this necessarily applies to every open set defined by T. Maybe i interpreted the question wrong but i dont think so, so please help if you can..