# Question Re. Simple epsilon proofs

1. Aug 30, 2009

### Newtime

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$$\in$$N has two limits a and b.
By definition:

-For any $$\epsilon$$>0, there exists an N$$\in$$N such that n$$\geq$$N 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$$\geq$$N implies that |xn-a| < $$\epsilon$$, not that n$$\geq$$N 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.

Last edited: Aug 30, 2009
2. Aug 30, 2009

### snipez90

Yes, if the limit exists, then you can replace epsilon with any positive number (so epsilon/2 is of course positive), because that's simply what epsilon typically denotes. Many people want the last step of a proof to be just epsilon, but often times it doesn't matter. In this case, it doesn't matter whether we have epsilon or 2*epsilon at the very end, since epsilon is arbitrary anyways. Note you have a typo in the leftmost inequality at the final step of the proof.

3. Aug 30, 2009

### Newtime

Thanks for the reply, that's what I thought but wanted to hear it from someone who actually knew before I went on doing proofs on an assumption.