User talk:194.197.235.210

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Happy editing! Mooncrest (talk) 03:44, 3 April 2008 (UTC)[reply]

God meant that to put Him first then people because people will turn away from you.Mooncrest (talk) 00:25, 14 January 2008 (UTC)[reply]

I have to mention that the proof given is false. Althought the statement is true. Namely to consider that continuity of f and g in an simply connencted open domain and f=g on some open neighbourhood of z, does not imply that the set is closed. Actually the set may be closed, open or neither. So how is this informative... Anyway the proof can be compleated as follows:

Let A be the set of points that are limiting points of zeroes of h=f-g and B be the complement of A in D. Now A and B may be made subsets of D. Now, what we actually want to prove is that both sets are open. Namely on definition of connected is that the set can not be covered by two open sets. Anyway A is open, because we can give h a power series representation an prove that it must be zero, because it agrees with 0 on a set of points with an accumulation poin. Now B is as well open, because it contains the points that are not limiting points of zeroes of h. Thus there is a neighbourhood C of each points in B such that it hase no zero of h (and ths each point in this C belongs to B). This is a contradiction.

Please leave you name

Please leave your name on the discussions you have made so people will know who you are. Jus type in four textiles so your name will appear.Mooncrest (talk) 03:44, 3 April 2008 (UTC)[reply]

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