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Unformatted text preview: 18.100B/C: Fall 2008 Solutions to Homework 2 1. Choose an infinite subset S = { ( x α ,y α )  α ∈ A } of E × F . If { x α  α ∈ A } contains infinitely many distinct members then we have by Rudin 2.41 that the set { x α  α ∈ A } has some limit point x ∈ E . Choose indices { α i  i ∈ N } such that  x x α i  < 1 i for each i . If instead { x α  α ∈ A } has only finitely many members, there is some value repeated in the set infinitely many times, so we can actually take the α i such that  x x α i  = 0, and we can weaken this to just assume  x x α i  < 1 i for each i and reduce the number of cases considered. Now consider { ( x α i ,y α i )  i ∈ N } , an infinite subset of our original infinite set. Applying the same idea as in the first paragraph to F , the y α i have some limit point y ∈ Y , so we can choose { i j  j ∈ N } such that  y y α i j  < 1 j for all j . Moreover, we can choose i j ≥ j for each j by Rudin 2.20. Thus k ( x,y ) ( x α i j ,y α i j ) k 2 = q ( x x α i j ) 2 + ( y y α i j ) 2 < q 2 j 2 . Given any r > 0, we can choose j > 2 r and then the preceding sentence implies ( x α i j ,y α i j ) ∈ B (( x,y ) ,r ), so ( x,y ) is a limit point of S . Since x ∈ E and y ∈ F we have ( x,y ) ∈ E × F . Since S was an arbitrary infinite subset of E × F , E × F is compact by Rudin 2.41. Alternatively, we have by HeineBorel (also Rudin 2.41) that E and F are closed and bounded, so one can seek to show that E × F is both closed and bounded and then apply HeineBorel again to conclude that E × F is compact. For more detail, see the 18.100C writeup of this problem when it becomes available. 2. Let our countable base for X be V 1 ,V 2 ,... and let { G α  α ∈ A } be an open cover of X . Choose a sequence of triples ( x,G α ,V ) in the following way: for each x ∈ X , choose a set G α from our cover such that x ∈ G α and choose a set V from our base such that x ∈ V ⊆ G α . Then, among all the triples with third entry V i , choose a single representative ( x i ,G α i ,V i ). Note that this collection is at most countable, since to each V i we have at most one triple associated. Also, the union of the associated G α i contains the union of all the V i , and this must contain all of X (since X is open in X and so is a union of some of the V i ). Thus, these G α i are an at most countable subcover of the...
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This note was uploaded on 12/07/2011 for the course MATH 18.100B taught by Professor Prof.katrinwehrheim during the Fall '10 term at MIT.
 Fall '10
 Prof.KatrinWehrheim

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