18.100pset6 - 18.100B/C Fall 2008 Solutions to Homework 6...

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18.100B/C: Fall 2008 Solutions to Homework 6 1. (Rudin page 98 problem 1) No, f need not be continuous. For a counter-example, consider the function δ : R R defined by δ ( x ) = braceleftBigg 1 , x = 0 , 0 , otherwise. For any x R we have lim h 0 ( δ ( x + h ) - δ ( x - h )) = 0 . In particular, this is true at x = 0 since h = 0 is excluded in the limit for h 0. However, δ is not continuous at 0. Note that the converse of the problem statement is true, i.e., if f is continuous then lim h 0 ( f ( x + h ) - f ( x - h )) = 0. For additional discussion, see the 18.100C writeup of this problem when it becomes available. 2. (Rudin page 98 problem 3) The set { 0 } is closed in R , so by the Corollary to Rudin 4.8 we have that f - 1 (0) = Z ( f ) is closed in X . (Direct proofs of this result probably amount to a special case of the proof of Rudin 4.8.) 3. By Rudin 4.7, it’s enough to show that the function f : X R defined by f ( x ) = d ( x, x 0 ) is continuous. To show this let any ε > 0 and x X be given. Then for any y X we have by the triangle inequality that d ( x, x 0 ) d ( x, y ) + d ( y, x 0 ) and d ( y, x 0 ) d ( y, x ) + d ( x, x 0 ) . Now, if d ( x, y ) < ε then we can rewrite these equations as f ( x ) - f ( y ) < ε and f ( y ) - f ( x ) < ε, so | f ( x ) - f ( y ) | < ε . This proves that f is continuous, as desired. (Quick sanity check: what was the value of δ we used?) One might summarize this result by the statement, “a radially symmetric function is continuous whenever its section is continuous.” 4.(a) In the discrete metric, the only convergent sequences are those that are eventually con- stant. (From the definition of convergence we get that s n s iff N N : n N : s n = s .) Thus for any sequence ( s n ) n N converging to s 0 in S , we have that ( f ( s n )) n N is eventually constant and equal to f ( s 0 ). Then ( f ( s n )) n N surely converges to f ( s 0 ) and so f is continuous at s 0 . Since s 0 was arbitrary, f is continuous. (b) Choose any s 0 S and any ε > 0 and let δ = 1 2 . Then d ( s, s 0 ) < 1 2 implies that s = s 0 and so that d ( f ( s ) , f ( s 0 )) = 0 < ε . Thus f is continuous at s 0 and so f is continuous.
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