Halp

How do I proved this?

Assume h : R -> R is continuous on R. Let K = {x : h(x)=0}. Show that K is a closed set.

>>1
Let L be the compliment of K, so L = {x : h(x) != 0}. Let x0 be a point in L. Then |h(x0)| > e for some positive real number e. By the definition of continuity, there exists a positive real number d such that for all x satisfying |x0 - x| < d, |h(x0) - h(x)| < e. From the above fact that |h(x0)| > e, we have |h(x)| > 0. Hence the open ball of radius d around x0 is in L, and L is open. K is the compliment of an open set, and thus is a closed set.

>>2
Well done, doing his homework for him.

>>2
Thanks though I came up with a more convoluted proof on my own.

I'm confused. Why does L have to exist at all?

Say h(x)=0 for all x. Then h is continuous, but K=R which is open. What am I missing here?

>>5
R is also closed.

>>6
R is closed? Fuck. No wonder I got a C in Real Analysis.

Riddle me this, guys.

There exists 0 < c < 1 such that abs( f(x) - f(y) ) < c * abs(x-y).
Let y_n be the sequence (y_1, f(y_1), f(f(y_1)), ...).
Show that y_n is a Cauchy sequence first. Let y = lim y_n. Then show y = f(y) and that y is unique. And then show x_n = (x, f(x), f(f(x)), ...) converges to y using definition y = lim y_n.

R iz clozed and iz open also dude

>>8 abs( f(x) - f(y) ) <= c * abs(x-y).

Help me with my homework guyz

>>8
This is the Banach Fixed Point Theorem. You can find the proof on google or wikipedia probably, it's a bit lengthy to type out from memory.

>>11
Thanks. I wonder if my teacher expects us to cheat through google or what.

test

Ah crap, we use n>N, N in reals, instead of n>=N, N is an integer,  for the limit of a sequence proof in our class. So I cheated and still got it wrong.