325 HW 03 - MAT 325: Topology Professor Zoltan Szabo...

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Unformatted text preview: MAT 325: Topology Professor Zoltan Szabo Problem Set 3 Rik Sengupta rsengupt@princeton.edu February 24, 2010 1. Munkres, p. 118, problem 6 Let x 1 , x 2 ,... be a sequence of the points of the product space Q X . Show that this sequence converges to the point x if and only if the sequence ( x 1 ) , ( x 2 ) ,... converges to ( x ) for each . Is this fact true if one uses the box topology instead of the product topology? Solution. We will prove both directions. (= :) Suppose the sequence x 1 , x 2 ,... converges to some x in the product topology. For a fixed let U be a neighborhood of ( x ). It remains only to check that U contains ( x i ) for i larger than some N . This follows from the fact that U = Q V , where V = X for 6 = and V = U is a neighborhood of x , and hence contains x i for i larger than some N . ( =:) Conversely, suppose that { ( x i ) } converges to ( x ) for all . Let U be a basis element that covers x , so that U = Q U and U = X except when = k , for k = 1 , 2 ,...,K . Now, for each k , we know { k ( x i ) } converges to k ( x ), which means that there is some N k such that k ( x i ) U k whenever i > N k . Set N = max { N k : k = 1 , 2 ,...,K } . Then for all i > N , we have x i U , and therefore, { x i } converges to x . This is not true in the box topology. Take your space as R , and let ( x i ) be an infinite sequence consisting of zeroes everywhere but 1 at the i th position precisely. But then, it immediately follows that each { ( x i ) } converges to zero, but the open neighborhood Q n (- 1 n , 1 n ) around contains none of the x i s. One direction is still true. The only if part is still true, and the above proof works without any change. 2. Munkres, p. 118, problem 7 Let R be the subset of R consisting of all sequences that are eventually zero, that is, all sequences ( x 1 ,x 2 ,... ) such that x i 6 = 0 for only finitely many values of i . What is the closure of R in R in the box and product topologies? Justify your answer. Solution. We consider the two cases separately. Product Topology Choose any x R . There is a basis element B containing x . Then, we may write B as B = Y i =1 U i , where U i = R for all but finitely many indices i . So now, we may define z R as: 1 z i = 0 if U i = R x i if U i 6 = R Therefore, obviously, z B . It follows that the closure R is the whole space R ....
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This note was uploaded on 05/07/2010 for the course MAT 325 taught by Professor Zoltánszabó during the Fall '09 term at Princeton.

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325 HW 03 - MAT 325: Topology Professor Zoltan Szabo...

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