325 HW 02 - MAT 325: Topology Professor Zoltan Szabo...

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MAT 325: Topology Professor Zoltan Szabo Problem Set 2 Rik Sengupta rsengupt@princeton.edu February 7, 2010 1. Munkres, p. 101, problem 13 Show that X is Hausdorff if and only if the diagonal Δ = { x × x : x X } is closed in X × X . Solution. We will show both directions. (= :) Suppose X is Hausdorff. We want to show that the diagonal is closed, i.e. in other words, the complement of the diagonal, Δ = X × X - Δ, is open. Take any point ( x,y ) in Δ. Then, by definition, x 6 = y , and hence there are open sets U and V in X , containing X and y respectively (because X is Hausdorff by assumption). So, by the definition of the product topology, U × V is an open subset of X × X . Furthermore, obviously, U × V Δ. For, otherwise, there is some point in U × V that is in Δ. This point must be of the form ( x,x ), but that means x U and x V , and then U and V are forced to be not disjoint, which is a contradiction. Hence it follows that given any point ( x,y ) Δ, we can always find some open neighborhood U × V of that point which is completely contained in Δ, i.e. Δ is open. Hence, Δ is closed. ( =:) Suppose Δ is closed. This simply means Δ is open. Let x and y be two distinct elements of X . Then, ( x,y ) Δ by definition. Furthermore, Δ is open, and so there is a basis open set U × V Δ which contains ( x,y ). Now, note that U and V are open by definition of the product topology. Furthermore, they are disjoint subsets of X , containing x and y respectively. Therefore, for any unequal x,y X , we can find disjoint open neighborhoods contained in X , containing x and y respectively. Therefore, X is Hausdorff. Therefore, X is Hausdorff if and only if the diagonal Δ = { x × x : x X } is closed in X × X . ± 2. Munkres, p. 101, problem 16 Consider the five topologies on R given in Exercise 7 of § 13. (a) Determine the closure of the set K = { 1 /n : n Z + } under each of these topologies. (b) Which of these topologies satisfy the Hausdorff axiom? The T 1 axiom? Solution. We will divide this solution into two parts. Preliminary Comments. First of all, notice that in the finite complement topology, the sequence ( x n ) = 1 /n converges to all points in R . This is because ( x n ) is a sequence with infinitely many distinct points. Therefore, any open set in the finite complement topology will contain all but finitely many points of ( x n ). It must follow, then, that ( x n ) converges to every point in R . Formally, take any r R , and any open set U r (in the finite complement topology) containing r . Since U r is not empty, it follows that R - U r is a finite set, by definition. Then, define 1
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M = 1 min { ( { x n } ∩ ( R - U r )) ∪ { 1 }} . Then, it is easy to see that for all
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325 HW 02 - MAT 325: Topology Professor Zoltan Szabo...

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