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Unformatted text preview: Mathematics 105 — Spring 2004 — M. Christ 1 Solutions to selecta from problem set #1 No number . Let T : R n → R m be a linear transformation. Show that there exists some finite constant M such that  T ( v )  ≤ M  v  for every v ∈ R n . (This was not assigned, but I used it in class and promised to provide a proof.) Solution. Let { e 1 , ··· , e n } denote the standard basis for R n . Define the finite number B = max(  T ( e 1 )  , ··· ,  T ( e n )  ). Consider any v ∈ R n and write v = ( v 1 , ··· , v n ) = ∑ n j =1 v j e j . Then  T ( v )  =  n X j =1 v j T ( e j )  ≤ n X j =1  v j  ·  T ( e j )  ≤ B n X j =1  v j  . Now  v j  ≤  v  for all j , so we find that  T ( v )  ≤ Bn  v  . Thus it suffices to define M = Bn . (Actually B √ n would work, but there’s no bonus for efficiency here.) I4 . Prove that for any x, y ∈ R n ,  x    y  ≤  x y  . Solution. First of all, if you tried to write this out using square roots and so on, you probably encountered a bit of a mess. A better way to do this is to remember that absolute values are often best reasoned about by cases, and to note that on the lefthand side of the inequality, the outermost symbols denote the absolute value of the real number  x    y  . So split the proof into two cases. If  x  ≥  y  then we’re asked to show that  x    y  ≤  x y  . This is equivalent to  x  ≤  y  +  x y  . Since  x  =  y + ( x y )  , this is a direct consequence of the triangle inequality. The problem is symmetric, in the sense that if the roles of x, y are interchanged, then neither side of the inequality changes at all. Thus the case where  y  ≥  x  is exactly the same as the case where  x  ≥  y  . I14 . Prove that the union of any family of open sets is open. Prove that the inter section of any finite family of open sets is open, and give a counterexample showing that this need not hold for infinite families....
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 Spring '04
 MichaelChrist
 Math, Topology, Empty set, Metric space, Topological space, Closed set

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