1873_solutions

And that a is strictly subequivalent to s prove that

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Unformatted text preview: 6. Assuming the continuum hypothesis, prove that there is a subset S of the unit square 0, 1 0, 1 such that for every horizontal line L in the square, the set L S is countable and for every vertical line L in the square, the set L S is countable. We assign a well order  to the set 0, 1 in such a way that 0, 1 is order isomorphic to 1 . We observe that every initial segment of 0, 1 relative to the order  must be countable. We now define S  x, y 0, 1 0, 1 xy and observe that S has the desired properties. 7. Suppose that A and B are uncountable sets and that B is strictly subequivalent to A. By a horizontal line in A B we mean a set of the form x, y x A where y is any member of B. Similarly, a vertical line in A B is a set of the form x, y y B where x is any member of A. Prove that if S is any subset of A B whose intersection with every horizontal line is subequivalent to B then there is a vertical line L such that S L . For each y B we define Sy  x A x, y S. Since each set S y is subequivalent to B we deduce from the theorem on unions that  Sy ß B yB and therefore A  Sy . yB If we choose x A  y B S y then the vertical line x, y y B is disjoint from S. 8. Suppose that A is an infinite set. Prove that there is a subset S of A A such that for every horizontal line L in A A the set L S is strictly subequivalent to A and for every vertical line L in A A, the set L S is equivalent to A. We assign a cardinal order to the set A and we define S  x, y A A x y. It is clear that if L is any horizontal line than L S is strictly subequivalent to A and it follows from Exercise 2 that if L is any vertical line then L S ß A. 9. A subset E of 1 is said to be closed if for every x no member of E lies between y and x. 1 E, there is a member y of 1 such that y  x and a. Prove that if A and B are closed uncountable subsets of 1 then A B is a closed uncountable subset of 1. As we know, a subset of 1 is uncountable if and only if it has an upper bound. Suppose that A and B are closed uncountable subsets of 1 . In order to prove that A B is uncountable we shall show that this set has no upper bound. Suppose that y 1 . We shall find a member of A B such that y  x. We begin by defining x 1 be the least member of A that is greater than y. More precisely, we define x 1 to be the least member of the set t A y  t . In order to make this definition we have made use of the fact that the set t A y  t is nonempty; which we know because A has no upper bound. We define x 2 to be the least member of B that is greater than x 1 . We define x 3 to be the least member of A greater than x 2 and we continue in this fashion. For each positive integer n we have x n  x n1 and x n A when n is odd and x n B when n is 72 even. Since the set x n n Z  is countable, it has upper bounds. We define x to be the least upper bound of the set x n n Z  . Given any member u of 1 such that u  x we know that u fails to be an upper bound of the set x n n Z  and so the inequality u  x n...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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