# Math 396. Product topology - Department of Mathematics - Stanford

6 Pages · 2005 · 120 KB · English

Math 396. Product topology The aim of this handout is to address two points: metrizability of nite products of metric spaces, and the abstract characterization of the

## Math 396. Product topology - Department of Mathematics - Stanford free download

Math 396 Product topology The aim of this handout is to address two points: metrizability of nite products of metric spaces, and the abstract characterization of the product topology in terms of universal mapping properties among topological spaces This latter issue is related to explaining why the de nition of the product topology is not merely ad hocbut in a sense the ight" de nition In particular, when you study topology more systematically and encounter the problem of topologizing in nite products of topological spaces, if you think in terms of the universal property to be discussed below then you will inexorably be led to the right de nition of the product topology for a product of in nitely many topological spaces (it is not what one would naively expect it to be, based on experience with the case of nite products) 1Metrics on finite products Let X 1; : : : ; X dbe metrizable topological spaces The product set X=X 1     X d admits a natural product topology, as discussed in class It is natural to ask if, upon choosing metrics  j inducing the given topology on each X j, we can de ne a metric on Xin terms of the  j's such that induces the product topology on X The basic idea is to nd a metric which describes the idea of \coordinatewise closeness", but several natural candidates leap out, none of which are evidently better than any others: max ((x 1; : : : ; x d) ; (x 0 1 ; : : : ; x 0 d )) = max 1 j d j( x j; x 0 j )  Euc ((x 1; : : : ; x d) ; (x 0 1 ; : : : ; x 0 d )) = v u u t d X j =1  j( x j; x 0 j ) 2  1(( x 1; : : : ; x d) ; (x 0 1 ; : : : ; x 0 d )) = d X j =1  j( x j; x 0 j )  p(( x 1; : : : ; x d) ; (x 0 1 ; : : : ; x 0 d )) = 0 @ d X j =1  j( x j; x 0 j )p 1 A 1 =p ; p 1 When X j= R for all j, with  j the usual absolute value metric, these recover the various concrete norms we've seen on X=Rd Our rst aim will be to show that all of these rather di erentlooking metrics are at least bounded above and below by a positive multiple of each other (which is the best we can expect, since they sure aren't literally the same), and so in particular they all de ne the same topology In fact, we will see that the common topology they de ne is the product topology We rst axiomatize the preceding examples Let N:R d ! Rbe any norm which satis es the property that on the orthant [0 ;1 )d with nonnegative coordinates it is a monotonically increasing function in each individual coordinate when all others are held xed Examples of such N's include our old friends k  kmax; k  k Euc; k  k 1; k  k p(for p 1) where we recall that k(a 1; : : : ; a n) k p = 0 @ d X j =1 j a jjp 1 A 1 =p : 1 2 Here is the general theorem which shows that many metrics (including all those mentioned above) on a product space are bounded above and below by a positive multiple of each other and hence determine the same theory of open sets, closed sets, and convergence of sequences Theorem 11 LetN:R d ! Rbe any norm as considered above Then for metric spaces (X j;  j) for 1 j d, with product space X=X 1     X d, the function  N : X X ! Rde ned by  N (( x 1; : : : ; x d) ; (x 0 1 ; : : : ; x 0 d )) = N( 1( x 1; x 0 1 ) ; : : : ;  d( x d; x 0 d )) is a metric on X, and al l such  N 's are bounded above and below by a positive multiple of each other Proof Let's rst check that each  N really is a metric Since Nis

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