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Unformatted text preview: Basic Topology Written by MenGen Tsai email: b89902089@ntu.edu.tw 1. Prove that the empty set is a subset of every set. Proof: For any element x of the empty set, x is also an element of every set since x does not exist. Hence, the empty set is a subset of every set. 2. A complex number z is said to be algebraic if there are integers a ,...,a n , not all zero, such that a z n + a 1 z n 1 + ... + a n 1 z + a n = 0 . Prove that the set of all algebraic numbers is countable. Hint: For every positive integer N there are only finitely many equations with n +  a  +  a 1  + ... +  a n  = N. Proof: For every positive integer N there are only finitely many equa tions with n +  a  +  a 1  + ... +  a n  = N. (since 1 n N and 0  a  N ). We collect those equations as C N . Hence S C N is countable. For each algebraic number, we can form an equation and this equation lies in C M for some M and thus the set of all algebraic numbers is countable. 3. Prove that there exist real numbers which are not algebraic. Proof: If not, R 1 = { all algebraic numbers } is countable, a contra diction. 1 4. Is the set of all irrational real numbers countable? Solution: If R Q is countable, then R 1 = ( R Q ) S Q is countable, a contradiction. Thus R Q is uncountable. 5. Construct a bounded set of real numbers with exactly three limit points. Solution: Put A = { 1 /n : n N } [ { 1 + 1 /n : n N } [ { 2 + 1 /n : n N } . A is bounded by 3, and A contains three limit points  0, 1, 2. 6. Let E be the set of all limit points of a set E . Prove that S is closed. Prove that E and E have the same limit points. (Recall that E = E S E .) Do E and E always have the same limit points? Proof: For any point p of X E , that is, p is not a limit point E , there exists a neighborhood of p such that q is not in E with q 6 = p for every q in that neighborhood. Hence, p is an interior point of X E , that is, X E is open, that is, E is closed. Next, if p is a limit point of E , then p is also a limit point of E since E = E S E . If p is a limit point of E , then every neighborhood N r ( p ) of p contains a point q 6 = p such that q E . If q E , we completed the proof. So we suppose that q E E = E E . Then q is a limit point of E . Hence, N r ( q ) where r = 1 2 min( r d ( p,q ) ,d ( p,q )) is a neighborhood of q and contains a point x 6 = q such that x E . Note that N r ( q ) contains in N r ( p ){ p } . That is, x 6 = p and x is in N r ( p ). Hence, q also a limit point of E . Hence, E and E have the same limit points. 2 Last, the answer of the final subproblem is no. Put E = { 1 /n : n N } , and E = { } and ( E ) = . 7. Let A 1 , A 2 , A 3 , ... be subsets of a metric space. (a) If B n = S n i =1 A i , prove that B n = S n i =1 A i , for n = 1 , 2 , 3 ,... (b) If B = S i =1 , prove that B S i =1 A i . Show, by an example, that this inclusion can be proper....
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This note was uploaded on 12/16/2011 for the course STAT 5446 taught by Professor Frade during the Fall '09 term at FSU.
 Fall '09
 Frade

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