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Here's an example that helps illustrate my question:
Prove: A sequence in R can have at most one limit.
Proof:
Assume a sequence {xn}n\inN has two limits a and b.
By definition:
-For any \epsilon>0, there exists an N\inN such that n\geqN implies that |xn-a| < \epsilon/2.
-A similar argument can be made for the limit b.
Thus:
|a-b| \leq |xn-a| - |xn-b| < \epsilon/2 + \epsilon/2 = \epsilon
Thus a=b.
Now here's my question...in the step immediately following "By definition," the actual definition of a limit of a sequence shows that n\geqN implies that |xn-a| < \epsilon, not that n\geqN implies that |xn-a| < \epsilon/2. So is it acceptable to put \epsilon over any number when convenient in a proof? As in the above proof, it is convenient to put \epsilon/2 instead of just \epsilon so that in the final step the two add up to \epsilon and show that a=b.
Thanks.
edit: sorry for the awkward formatting, if anything is unclear, let me know I'll explain it in words.
Prove: A sequence in R can have at most one limit.
Proof:
Assume a sequence {xn}n\inN has two limits a and b.
By definition:
-For any \epsilon>0, there exists an N\inN such that n\geqN implies that |xn-a| < \epsilon/2.
-A similar argument can be made for the limit b.
Thus:
|a-b| \leq |xn-a| - |xn-b| < \epsilon/2 + \epsilon/2 = \epsilon
Thus a=b.
Now here's my question...in the step immediately following "By definition," the actual definition of a limit of a sequence shows that n\geqN implies that |xn-a| < \epsilon, not that n\geqN implies that |xn-a| < \epsilon/2. So is it acceptable to put \epsilon over any number when convenient in a proof? As in the above proof, it is convenient to put \epsilon/2 instead of just \epsilon so that in the final step the two add up to \epsilon and show that a=b.
Thanks.
edit: sorry for the awkward formatting, if anything is unclear, let me know I'll explain it in words.
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