So, the way that I understand your argument, L ranges over the values L,L+ϵ,L+2ϵ,…,L+Nϵ. (Of course there is a circular dependency here but as far as I can tell this was what you were thinking.) Of course, all of these possible values of L are clearly distinct. I’m not really sure what your point is about what happens if you assume that a sequence is eventually close to two different numbers. I think part of the issue is that your proof is a proof by contradiction, but in a proof by contradiction you are only allowed to introduce the contradiction hypothesis once, not an arbitrary number of times.
The fundamental question that any proof of this theorem should answer is: given a bounded monotonic sequence, how do we find the L that it converges to? Your proof doesn’t appear to address this question at all, which makes me think you need a completely different idea.
Note: to address your “strong intuition” perhaps I should say that of course if you have two real numbers L1 and L2 such that for all ϵ1,ϵ2>0, there is a number that is ϵ1-close to L1 and ϵ2-close to L2, then L1=L2. But I don’t think this fact has the significance you want it to have in the broader argument.
So, the way that I understand your argument, L ranges over the values L,L+ϵ,L+2ϵ,…,L+Nϵ. (Of course there is a circular dependency here but as far as I can tell this was what you were thinking.) Of course, all of these possible values of L are clearly distinct. I’m not really sure what your point is about what happens if you assume that a sequence is eventually close to two different numbers. I think part of the issue is that your proof is a proof by contradiction, but in a proof by contradiction you are only allowed to introduce the contradiction hypothesis once, not an arbitrary number of times.
The fundamental question that any proof of this theorem should answer is: given a bounded monotonic sequence, how do we find the L that it converges to? Your proof doesn’t appear to address this question at all, which makes me think you need a completely different idea.
Note: to address your “strong intuition” perhaps I should say that of course if you have two real numbers L1 and L2 such that for all ϵ1,ϵ2>0, there is a number that is ϵ1-close to L1 and ϵ2-close to L2, then L1=L2. But I don’t think this fact has the significance you want it to have in the broader argument.