- #1
pantin
- 20
- 0
suppose f:R^m -> R^n is a map such that for any compact set K in R^n, the preimage set f^(-1) (K)={x in R^m: f(x) in K} is compact, is f necessary continuous? justify.
The answer is no.
given a counterexample,
function f:R->R
f(x):= log/x/ if x is not equal to 0
f(x):= 0 if x=0
note, /x/ is the absolute value of x.
I don't quite get how to draw the image log/x/
and anyone can explain why ?
The answer is no.
given a counterexample,
function f:R->R
f(x):= log/x/ if x is not equal to 0
f(x):= 0 if x=0
note, /x/ is the absolute value of x.
I don't quite get how to draw the image log/x/
and anyone can explain why ?