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Unformatted text preview: MAT 325: Topology Professor Zoltan Szabo Problem Set 3 Rik Sengupta [email protected] 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 α . Show that this sequence converges to the point x if and only if the sequence π α ( x 1 ) ,π α ( x 2 ) ,... converges to π α ( x ) for each α . 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 α let U α be a neighborhood of π α ( x ). It remains only to check that U α contains π α ( x i ) for i larger than some N . This follows from the fact that U = Q β V β , where V β = X β for β 6 = α and V β = U α is a neighborhood of x , and hence contains x i for i larger than some N . ( ⇐ =:) Conversely, suppose that { π α ( x i ) } converges to π α ( x ) for all α . Let U be a basis element that covers x , so that U = Q α U α and U α = X α except when α = α k , for k = 1 , 2 ,...,K . Now, for each α k , we know { π α k ( x i ) } converges to π α k ( x ), which means that there is some N k such that π α k ( x i ) ∈ U α 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 ω , 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 { π α ( 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 ω 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 ω in the box and product topologies? Justify your answer. Solution. We consider the two cases separately. Product Topology Choose any x ∈ R ω . 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 ω ....
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 Fall '09
 ZoltánSzabó
 Topology, Metric space, Open set, Topological space

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