Proving Convergence of f_n(a_n) to f(a) Given Uniform Convergence of f_n on I

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SUMMARY

The discussion focuses on proving that if a sequence of functions \( f_n \) converges uniformly to a function \( f \) on an interval \( I \), then for a sequence \( a_n \) in \( I \) that converges to \( a \), the values \( f_n(a_n) \) converge to \( f(a) \). The key point is that uniform convergence guarantees that for any \( \epsilon > 0 \), the difference \( |f_n(x) - f(x)| < \epsilon \) holds for all \( x \) in \( I \). The confusion arises in distinguishing between \( f_n(a_n) \) and \( f(a_n) \), emphasizing the importance of the limit point \( a \).

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  • Understanding of uniform convergence in real analysis
  • Familiarity with sequences and limits in mathematical analysis
  • Knowledge of continuity of functions on intervals
  • Basic proficiency in mathematical notation and epsilon-delta arguments
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  • Study the properties of uniform convergence in detail
  • Explore the implications of continuity in the context of limits
  • Investigate the relationship between pointwise and uniform convergence
  • Review examples of sequences of functions and their convergence behavior
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"Suppose f_n are defined and continuous on an interval I. Assume that f_n converges uniformly to f on I. If a_n in I is a sequence and a_n -> a, prove that f_n(a_n) converges to f(a)."

I don't understand the question. Doesn't uniform convergence imply that for all x in I and e>0, | f_n(x) - f(x) | < e? So in particular, |f_n(a_n) - f(a) | < e.
 
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Wait, nevermind. | f_n(a_n) - f(a_n) | <e, not f(a).
 

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