Proving the Uniqueness of inf(A) in a Subset of ℝ | Epsilon Proof Technique

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The discussion centers on proving the uniqueness of the infimum (inf) of a subset A of ℝ, emphasizing the need for clarity in definitions. Participants suggest starting the proof by assuming two infimums, x and y, and demonstrating they must be equal. The epsilon proof technique is mentioned as a potential method, though its application may vary based on the definition of infimum used. A contradiction arises when attempting to assert x < y and y < x, leading to the correct formulation of x ≤ y and y ≤ x. The conversation highlights the importance of precise language and logical consistency in mathematical proofs.
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Homework Statement


Let A be a subset of ℝ, now prove if inf(A) exists, then inf(A) is unique

The Attempt at a Solution


I am using an epsilon proof but i don't think i am going about it the right way, can someone nudge me in the right direction
 
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jaqueh said:

Homework Statement


Let A be a subset of ℝ, now prove if inf(A) exists, then inf(A) is unique

The Attempt at a Solution


I am using an epsilon proof but i don't think i am going about it the right way, can someone nudge me in the right direction

How you do it depends very much on what *definition* you are using for "inf". Different (but equivalent) definitions would need different types of proofs.

RGV
 
Usually, for something like this, it is best to start of saying "Let x and y be inf(S)" and then show that x and y are, in fact, the same. Now, showing that |x - y| < epsilon might be a good strategy, but there are probably more efficient methods. But, as RGV said, it depends on how inf is defined for you. If you have defined it as a greatest lower bound, the method I mentioned works very well.
 
yes i think I am going to revise it as x and y are infimums for A then x<y and y<x so x=y
 
jaqueh said:
yes i think I am going to revise it as x and y are infimums for A then x<y and y<x so x=y

That's a contradiction the way you wrote it. Did you mean x \leq y, \ \text{and} \ y \leq x,\ \text{so} \ x=y?
 
yes that is what i meant and i did write it that way x≥y y≥x
 
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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