sol-06 - MATH 453 SOLUTIONS TO ASSIGNMENT 6 OCTOBER 16,...

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M  453 S   A  6 O  16, 2004 Exercise 4 from Section 20, page 127 (a) Since each of their component functions are continuous, all three of f , g and h are continuous when R ω is given the product topology. On the other hand, if R ω is given the box topology, none of them is continuous. To see this, look at the open set U = Q n = 1 ( - 1 / n 2 , 1 / n 2 ) in R ω . The inverse image of U under each of f , g and h is { 0 } and hence is not open. Things are a little more interesting in the uniform topology: f is not continuous, since f - 1 ( V ) = { 0 } if V is the product of countably many copies of ( - 1 , 1). Let k denote either g or h . Then ¯ ρ ( k ( s ) - k ( t )) = ¯ d ( s - t ), so if we have an ε -ball B ¯ ρ ( f ( s ) , ε ) about f ( s ) in the ¯ ρ metric, the ε -ball around s (i.e. the interval ( s - ε, s + ε )) in R will be mapped into B ¯ ρ ( f ( s ) , ε ). Hence both g and h are continuous in the uniform topology. (b) First observe that if ( a n ) denotes any of the given sequences, then π i ( a n ) 0 for each i . It follows from Exercise 18.6 (and a similar proof for the uniform topology) that the only point the sequences can possibly converge to is 0 = (0 , 0 , . . . ). Furthermore,
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sol-06 - MATH 453 SOLUTIONS TO ASSIGNMENT 6 OCTOBER 16,...

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