# Convergent Filter Base and Continuous Function

1. Aug 12, 2009

### jetplan

Hi All,

I can't see how the following is proved.

Given two topological space (X, T), (Y, U) and a function f from X to Y and the following two statements.

1. f is continuous, i.e. for every open set U in U, the inverse image f-1(U) is in T

2. For every convergent filter base F -> x, the induced filter base f [[F]] -> f(x)

it is claimed that statement 1 and statement 2 are equivalent, i.e. 1 if and only if 2

I can prove 1 -> 2
but how do i prove 2 -> 1

thanks a lot !

2. Aug 13, 2009

### g_edgar

Suppose the filter-base thingy holds. Let A be an open set in Y. Let B = f^{-1}(A).
We must show B is open in X. Suppose not. Then there is c in B such that every neighborhood of c meets X minus B. Your filter-base will be all these sets: (neighborhood of c) minus B. This filter-base converges to c. So the image filter-base in Y converges to f(c). But the sets of that image filter-base are outside A and converge to a point of A, contradicting the assumption that A is open.