Uniform continuity, cauchy sequences

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Uniform continuity of a function f implies 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. Given that the sequence {xk} is Cauchy, for any ε > 0, there exists an N such that for all n, p ≥ N, |xn - xp| < δ. By applying the uniform continuity condition, it follows that |f(xn) - f(xp)| < ε, demonstrating that {f(xk)} is also Cauchy. The key point is that the uniform continuity allows the transition from the Cauchy condition of {xk} to that of {f(xk)}. Thus, the conclusion is valid and the reasoning is sound.
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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.


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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).
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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