Analysis problem and continuity
- From: "Carl R." <solrac140@xxxxxxxxxxx>
- Date: Mon, 8 Sep 2008 05:13:18 -0700 (PDT)
Let f: R - > C be a map from the set of real numbers to the comple
numbers.
Define phi(x,delta):= sup { | f(s) - f(t) | : s, t in (x-delta, x
+delta)}
Let r(x):= inf { phi(x,delta) : delta >0 }
Prove that
a) f is continuous in x iff r(x) = 0.
So here's my work.
Asumme f is continuous in x then for all e>0 we can find delta >0 such
that
if |y-x|< delta then |f(y) - f(x) | < epsilon.
But if |y-x|< delta then y,s are in (x-delta,x+delta) , thus
phi(x,delta): = sup { epsilon } = epsilon.
Thus r(x) := inf { epsilon : epsilon > 0} = 0 because epsilon>0.
Where I'm stuck is the part that if r(x) = 0 then f is continuous in
x. Can you please help?
Also, one more question, how can you show that the set of points of
discontinuity of a map from R to C
is a G_delta set?
.
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