By definition, two norms #||*||_1 and ||*||_2# are equivalent if there exists positive constants #a,b in RR^+# such that
#a||*x||_2<=||x||_1<=b||x||_2 . AAx in X#
Now let #T_1 and T_2# be the topologies on X induced by #||*||_1 and ||*||_2#. We require to prove that #T_1=T_2#
Recall that a collection T of open subsets #T_i# of X is a topology for X if the following conditions are satisfied:
- #phi,X in T# (the empty set and the whole set X must be in T)
- #nnn_(i=1)^nT_i in T# (finite intersections of T belong to T)
- #uuu_(i=1)^alphaT_i in T# (arbitrary unions of T belong to T)
Now let #A in T_1# (ie. A is an open subset of #T_1#)
#thereforeAAx in A, EE# an open ball #B(x;epsilon)# such that #B(x;epsilon)sube A#.
#therefore{y in X | ||y-x||_1 < epsilon} sube A.#
But since #||*||_1 and ||*||_2# are equivalent, #=> AA x in X, EE a,b in RR^+# such that #a||y-x||_2<=||y-x||_1<=b||y-x||_2#
#therefore{y in X | ||y-x||_2 < epsilon/a} sube {y in X | ||y-x||_1 < epsilon} sube A #
#therefore # Every point x of A is an interior point in the #T_2# topology #=> A in T_2#
#therefore T_1 sube T_2#
Similarly #T_2 sube T_1# and so #T_1=T_2#
