18.100pset2

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

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Unformatted text preview: 18.100B/C: Fall 2008 Solutions to Homework 2 1. Choose an infinite subset S = { ( x ,y ) | A } of E F . If { x | A } contains infinitely many distinct members then we have by Rudin 2.41 that the set { x | A } has some limit point x E . Choose indices { i | i N } such that | x- x i | < 1 i for each i . If instead { x | A } has only finitely many members, there is some value repeated in the set infinitely many times, so we can actually take the i such that | x- x i | = 0, and we can weaken this to just assume | x- x i | < 1 i for each i and reduce the number of cases considered. Now consider { ( x i ,y i ) | i N } , an infinite subset of our original infinite set. Applying the same idea as in the first paragraph to F , the y i have some limit point y Y , so we can choose { i j | j N } such that | y- y 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 i j ,y i j ) k 2 = q ( x- x i j ) 2 + ( y- y i j ) 2 < q 2 j 2 . Given any r > 0, we can choose j > 2 r and then the preceding sentence implies ( x i j ,y 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 | A } be an open cover of X . Choose a sequence of triples ( x,G ,V ) in the following way: for each x X , choose a set G from our cover such that x G and choose a set V from our base such that x V G . Then, among all the triples with third entry V i , choose a single representative ( x i ,G 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 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 i are an at most countable subcover of the...
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18.100pset2 - 18.100B/C: Fall 2008 Solutions to Homework 2...

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