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.(adsbygoogle = window.adsbygoogle || []).push({});

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 dont quite get how to draw the image log/x/

and anyone can explain why ?

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# The relationship between con't function and a compact set

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