# sol-06 - MATH 453 SOLUTIONS TO ASSIGNMENT 6 Exercise 4 from...

This preview shows pages 1–2. Sign up to view the full content.

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,

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

sol-06 - MATH 453 SOLUTIONS TO ASSIGNMENT 6 Exercise 4 from...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online