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``inverse image of _any_ U in Y?'' by four on 2008-04-15 13:18:23
I think there is some sort of subtlety I'm missing, because it seems like the following example would be non-continuous under the current definition:

X = [0,1]
Y = Real numbers
f(x) = 0 (constant for all x in [0,1]).

Then if U = (-1,1), U is open and in Y, but yet F^-1(U) = {0} which is not open.

A constant function is not continuous!? What is wrong?
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