Uniform continuity, cauchy sequences

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SUMMARY

The discussion focuses on proving that if a function f: S -> R^m is uniformly continuous on a set S, then the image of a Cauchy sequence {xk} under f, denoted as {f(xk)}, is also a Cauchy sequence. The proof utilizes the definition of uniform continuity, which states that for every ε > 0, there exists a δ > 0 such that for all x, y in S, if |x - y| < δ, then |f(x) - f(y)| < ε. The participants clarify that it is permissible to replace the variables with sequences, reinforcing the validity of the proof.

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  • Understanding of uniform continuity in mathematical analysis
  • Knowledge of Cauchy sequences and their properties
  • Familiarity with sequences and limits in metric spaces
  • Basic proficiency in mathematical proofs and epsilon-delta arguments
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Homework Statement


If f:S->Rm is uniformly continuous on S, and {xk} is Cauchy in S show that {f(xk)} is also cauchy.


Homework Equations





The Attempt at a Solution


Since f is uniformly continuous,

\forall\epsilon>0, \exists\delta>0: \forallx, y ∈ S, |x-y| < \delta => |f(x)-f(y)| < \epsilon

So I said that let x, y be sequences, {xn} and {xp}

Since {xn} is Cauchy, \forall\epsilon>0, \existsN : \foralln,p \geq N , |xn-xp| < \epsilon

Then using the fact that f is uniformly continuous, |f(xn)-f(xp)| < \epsilon

I don't think this is right.. am I allowed to replace those x's and f(x)'s with the sequences for example?
 
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Yes, basically, that's OK. You only need to be aware of the fact that for exactly this δ > 0 you found N (using the fact that (xn) is Cauchy).
 

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