Copy of s8

Copy of s8 - Solutions to homework # 8. 1. No, this does...

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Solutions to homework # 8. 1. No, this does not imply that f is continuous. Consider the function f ( x ) := ± 1 / | x | if x 6 = 0 , 0 x = 0 . Then f is not continuous at x = 0 even though f ( x ) - f ( - x ) = 0 for all x IR. 2. For every x E , f ( x ) f ( E ) f ( E ), hence x f - 1 ( f ( E )). Thus E f - 1 ( f ( E )). The last set must be closed as the preimage of the closed set f ( E ) by the Corollary to Theorem 4.8, hence it also contains E . So, E f - 1 ( f ( E )) = f ( E ) f ( f - 1 ( f ( E ))) f ( E ) . Here is an example showing that the inclusion f ( E ) f ( E ) may be proper. Take X to be Q and Y to be IR and let f be simply the identity map, identifying rational points as members of IR. Take E to be Q , too. Since E is all of Q , it is closed (in Q ), but its closure in IR is all of IR, hence f ( E ) = IR 6 = Q =
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This document was uploaded on 11/17/2010.

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