Directions: Please simplify the proofs to make them, short, simple and to the point. I need to memorize these proofs for an upcoming exam and right now these 4 are too long and complicated for me to memorize. So please simplify them in as little text as possible but include all the necessary info. Don’t forget to use screenshots too. Thanks. Problem 6: Firstly, we use mathematical induction to prove that for non-empty finite set S of positive integers has a minimum element. The statement is true for a set S of cardinality 1. Induction hypothesis: Assume that every non-empty finite set S of cardinality n contains a minimum element. Consider a finite set of cardinality n + 1. Let s  S , then by induction hypothesis, the set S \ s contains a minimum element s . If s  s then s is the minimum element of S . Otherwise, if s  s then s is the minimum element of S . Hence, it is already proved by mathematical induction that any non-empty finite set S of positive integers must have a minimum element. Now, consider an infinite non-empty set T of positive integers. Let t  T . Consider the set T ( 0, t  . ) This set is finite, so it has a minimum element t  = min T   0, t  . We will show that t  is also the minimum element of T . Let any x  T . If t  x then t   t  x since t  0, t  . Otherwise, if t  x ( ) then x  T  0, t  , and we again have t   x because of definition t  = min T   0, t  . Hence, we have proved that in any case, a non-empty set of positive integers must contain a smallest element. Problem 7: Firstly, we have to show that the set = 0,1, 2,  is not bounded from above. This can be proved by contradiction, by assuming that is bounded from above. Because  and has the Least Upper Bound Property, therefore has a least upper bound   . So,  is the smallest real number satisfying the property: n   for all n . Hence,  − 1 is not the upper bound of because if  − 1 were the upper bound, then  would not be the least upper bound since  −1   . Therefore, there exists some m such that  − 1  m . But then this implies   k + 1 and k + 1 . This contradicts that  is the upper bound of . We have just proved that = 0,1, 2,  is not bounded from above. y for all n x y would be the upper x y bound of , contradiction). Therefore, there must exist some n such that n  . But this is the x same as there exists some n such that nx  y , which is the Archimedean Property. For given x, y  with x  0 , we cannot have n  Hence, we have just proved the Archimedean Property of (because . Problem 15: The Cantor Intersection Property states that the intersection all of nested sequences of non-empty closed bounded intervals in the real line is non-empty . Proof: For any sequences of non-empty closed bounded intervals n  a , b  n n with the "nested" property  an+1 , bn+1    an , bn  for all n Consider A = an : n   , since we have a1  a2  set A is bounded above. Any bn , n  an  an+1  n , where an , bn  for all .  bn+1  bn   b2  b1 , the , is an upper bound for A . Therefore, according to the Least Upper Bound Axiom, there is a least upper bound for A , denote it by c . Since c is the least upper bound for the set A , we have c  bn for all n definition, an  c for all n . So, we have an  c  bn for all n In addition, if lim ( bn − an ) = 0 then n → , and since c is upper bound for A , then by , or equivalently, c  an , bn  , n  n , that is, c  n  an , bn  . i =1 n  an , bn  contains a single point because if d   an , bn  then i =1 c − d  bn − an for all n . Taking n →  gives c − d = 0, i.e, c = d . i =1 Problem 16: Since Qk 's are closed and bounded for all k , by Heine-Borel Theorem, Qk 's are compact sets for all k .  Proof by contradiction, suppose that  Uk = X k =1 . Let U k = X \ Qk where X =   U k = ( X \ Qk ) = X \  Qk  k =1 k =1  k =1   since k =1 Qk =    k =1 On the other hand, let N be the smallest integer such that Qk  ,  Qk   for all N  , taking N →  gives k =1 N Uk = X k =1 N  U k = ( X \ Qk ) = X \  Qk  k =1 k =1  k =1  N since N Qk =  ( N must exist k =1 k =1 because if otherwise, k =1 N N and k =1 Qk   ). Again, we have Qk =  .  Since Qk is compact set for all k , U k = X \ Qk is open set for all k . And since n Uk = X k =1 U1,U2 , X= n , it implies that U k : k   is an open cover of . Therefore, Qk =  . and N −1 N n Uk = X k =1 and X and it has a finite subcover ,U N  . Therefore, X is compact. By Heine-Borel theorem , X is closed and bounded. But , which cannot be a bounded set, a contradiction. Hence, the generalized version of Cantor Intersection Property is proved.
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