Re: continuity and open sets,

vsgdp wrote:
Suppose f:R-->R. Let A = {(x,y):y<f(x)} and B={(x,y):y>f(x)}. Show f
continuous IFF A and B are open.

First, I see how if f is continuous then A and B are open. But say
f(x) = 0 on [a,b] and 1 on (b,c]. Is it the vertical jump the part
that causes A and B not to be open (an open ball on that edge could
contain points > f(x) and < f(x) ?

He point (b,1/2) is in B. Is there an open ball containing this point
that is contained in B?

Second. What is the best strategy for showing f continuous implies
A,B open? I tried by letting p in A and looking for an open ball. It
doesn't seem like continuity helps with that approach since
continuity given info "near the curve" of f, and not at some
arbitrary point p in A.

Let p = (x0,y0). So, y0 < f(x0). Let epsilon = |f(x0) - y|/2. So, by
continuity, we can find a delta > 0 such that

|x - x0| < delta implies |f(x) - f(x0)| < epsilon.

--
David Marcus
.

Relevant Pages

  • Re: continuity and open sets,
    ... vsgdp wrote: ... I tried by letting p in A and looking for an open ball. ... seem like continuity helps with that approach since continuity given info ... you have to prove that f: R -> R is continuous iff ...
    (sci.math)
  • Re: continuity and open sets,
    ... I tried by letting p in A and looking for an open ball. ... doesn't seem like continuity helps with that approach since continuity ...
    (sci.math)