18.100pset2 - 18.100B/C Fall 2008 Solutions to Homework 2 1...

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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 ( x, y ) - ( x α i j , y α i j ) 2 = ( x - x α i j ) 2 + ( y - y α i j ) 2 < 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 Heine-Borel (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 Heine-Borel again to conclude that E × F is compact. For more detail, see the 18.100C write-up 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 G α , as desired.
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