Uniformly Bounded Functions: Proving Sequence Convergence

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A sequence of uniformly convergent bounded functions is uniformly bounded, as each function in the sequence is bounded by a specific constant. The proof begins by establishing that each function's norm is less than or equal to a corresponding bound, and for finitely many functions, a maximum bound can be determined. For indices beyond a certain point, a new maximum can be defined to ensure uniform boundedness. The second part of the discussion involves demonstrating that a function defined by the distance metric is bounded, which can be shown using the triangle inequality. Overall, the discussion emphasizes the importance of establishing bounds in the context of uniformly convergent sequences and metric spaces.
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Prove that a sequence of uniformly convergent bounded functions is uniformly bounded.

Attempt at proof:

So first we observe the following: ||fn||\leqMn. Each function is bounded. Also, |fn-f|\leq\epsilon for all n \geq N. First off, we observe that for finitely many fn's, we have them bounded by M1...MN. Then for n\geq N MN+1 acts as a bound.

So now we make a new M = max{M1...MN+1}.

I'm kind of stuck on how to demonstrate uniform boundedness with an inequality.

My second question is as follows:

Let X be a metric space with a metric d. Let a be a fixed point in X. Let p be a point in X. Define fp=d(x,p)-d(x,a).

I want to show this is bounded.

For this one I have no idea.
 
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The second one follows directly from the "triangle inequality": d(x,p)\le d(x,a)+ d(a,p)
 
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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