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About LinkBacksLetbe continuous at
and let
. How would you show that there is a neighborhood
of
, then
 > 0)
Forgive my lack of syntax here
if f is continuous at c, then for e> 0 there exists a d>0 st
0< l x - c l < d --> l f(x) - f(c) l < e
but l f(x) - f(c) l < e is the same thing as:
f(c) - e < f(x) < f(c) + e
so as long as 0 < e < f(c) we have f(x) > 0 for all 0 < l x - c l < d
and the conclusion follows
The exact same concept as southprkfan1 but using open ball notation. SinceOriginally Posted by CrazyCat87
Letis an open set we know there exists some
but by continuity there exists some
such that
. The conclusion follows.
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