Convergent Filter Base and Continuous Function

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SUMMARY

The discussion centers on the equivalence of two statements regarding continuous functions between topological spaces. Specifically, it asserts that for a function f from a topological space (X, T) to (Y, U), the continuity of f (statement 1) is equivalent to the condition that for every convergent filter base F converging to x, the induced filter base f[[F]] converges to f(x) (statement 2). The user has successfully proven the implication from statement 1 to statement 2 but seeks guidance on proving the reverse implication, 2 to 1.

PREREQUISITES
  • Understanding of topological spaces and their properties
  • Knowledge of continuous functions in topology
  • Familiarity with filter bases and convergence in topology
  • Basic proficiency in mathematical proofs and logic
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  • Study the definition and properties of filter bases in topology
  • Learn about the concept of continuity in topological spaces
  • Explore the relationship between convergence of filter bases and continuity
  • Review proof techniques in topology, particularly those involving equivalences
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Mathematicians, students of topology, and anyone interested in the foundational concepts of continuity and convergence in topological spaces.

jetplan
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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 !
 
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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.
 

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