Open set of real valued functions

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SUMMARY

The discussion focuses on proving that the set A = {f ∈ C[0,1] | f(0) > 1} is open in the metric space (X, d), where d(f,g) = sup|f(x) - g(x)| for x ∈ [0,1]. It establishes that a point f in A is an interior point if there exists a neighborhood N of f such that N is entirely contained within A. The participants seek clarification on the nature of neighborhoods in this context and the characterization of interior points within the set A.

PREREQUISITES
  • Understanding of metric spaces and the concept of open sets.
  • Familiarity with the supremum metric and its properties.
  • Knowledge of continuous functions on the interval [0,1].
  • Basic concepts of topology, particularly interior points and neighborhoods.
NEXT STEPS
  • Study the properties of metric spaces, focusing on open and closed sets.
  • Learn about the supremum metric and its implications for function spaces.
  • Explore the concept of neighborhoods in topology and their role in defining open sets.
  • Investigate examples of continuous functions and their behavior in relation to open sets in C[0,1].
USEFUL FOR

Mathematicians, students of analysis, and anyone studying topology or functional analysis, particularly those interested in the properties of function spaces and open sets.

rjw5002

Homework Statement


Consider the set X = {f:[0,1] \rightarrow R | f \in C[0,1]} w/ metric d(f,g) = sup|f(x) - g(x)| (x \in [0,1])
Prove that the set A = {f \in X | f(0) > 1} is open in (X,d).



Homework Equations


E is open if every point of E is an interior point
p is an interior point of E if there is a neighborhood N of p s.t. N\subsetE.
C[0,1] is the set of all real valued functions on [0,1]
supremum.

The Attempt at a Solution



To be honest, I have a great deal of difficulty understanding this question. Any hints or advice to help me in the right direction would be greatly appreciated.
 
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