Does Convergence in Open-Compact Topology Ensure Pointwise Convergence?

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Discussion Overview

The discussion revolves around the question of whether convergence in the open-compact topology of continuous functions from one topological space to another implies pointwise convergence at every point in the domain. Participants explore the implications of this convergence and seek clarification on the proof process.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a proof attempt, focusing on the need to show that for every neighborhood of f0(x), there exists an N such that f_n(x) is contained in that neighborhood for sufficiently large n.
  • The same participant questions the existence of a compact set containing x, suggesting that this might be a barrier to completing the proof.
  • Another participant points out that the singleton set {x} is a compact set containing x, implying that it could be used in the proof.
  • A further reply suggests that the original proof attempt may be overly complicated and proposes a simpler approach using open sets in the context of the topology.

Areas of Agreement / Disagreement

Participants express differing views on the complexity of the proof and the necessity of compact sets. There is no consensus on the best approach to proving the statement, indicating that multiple perspectives and methods are being considered.

Contextual Notes

The discussion highlights potential limitations in the proof process, particularly regarding the assumptions about compact sets in the space X and the nature of the open-compact topology.

Who May Find This Useful

This discussion may be useful for those studying topology, particularly in understanding the implications of different convergence concepts in function spaces.

MathematicalPhysicist
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Let C(X,Y) be the continuous functions space between the topological spaces X,Y, with the open-compact topology. prove that if the sequence {f_n} of C(X,Y) converges to f0 in C(X,Y) then for every point x in X the sequence {f_n(x)} in Y converges to f0(x).

here's what I did, let x be in X and for every U_x, neighbourhood, of f0(x) i need to prove that there exists N s.t for every n>=N f_n(x) is in U_x.

but because f_n converges to f_0 in C(X,Y), then for every U open in C(X,Y), s.t f_0 is in U then there exists N s.t for every n>=N f_n is in U, but U is open in C(X,Y) which means that U containes a set W(C,V) which contains f_n, i.e C is compact in X and V is open in Y, s.t f_n(C) is contained in V, but C is compact then so is f_n(C) (cause it's continuous), and thus for every collection of open set in Y which cover f_n(C) also finite union of them covers it.
Now how to relate those matters, if f0(x) in U_x, then let's take a compact set which contains x then from what is given we know that there exists N s.t for every n>=N s.t f_n in W(C,U_x) then f_n(C) is contained in U_x which means that f_n(x) is in U_x.

but here's the catch how do i know that there exists a compact set in X which contains x, unless X is a union of compact sets i don't see how to do this question, any hints?

thanks in advance.
 
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I haven't really read your entire post, but in regards to the last line, surely {x} is a compact set that contains x!
 
yes i know, but can you or someone else check my work or suggest another way to prove this statement?

thanks in advance.
 
I think you're making it much more harder than it is.

If U is an open set containing f0(x) in Y, then W({x},U) is an open set containing f0 in C(X,Y).
 

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