Isometry composition, (f^n (x)) - increasing sequence

Let be an isometry in a metric space . Suppose

Could you explain to me why the sequence is monotone and why does it converge with respect to the usual topology of and also with respect to and, in particular, it satisfies the condition of Cauchy?

I know that since isometry is a continuous and bijective function, it must be monotone. We put , so it must be increasing. But I can't figure out why

Could you help me with that? I would really appreciate a thorough explanation, because I can't find anything about isometry composition anywhere.

Thank you.

Re: Isometry composition, (f^n (x)) - increasing sequence

Quote:

Originally Posted by

**wilhelm** I know that since isometry is a continuous and bijective function, it must be monotone. We put

, so it must be increasing. But I can't figure out why

If you know that f is monotone, then it's trivial: apply n times the function f to both sides of . And converges by the monotone convergence theorem.

Re: Isometry composition, (f^n (x)) - increasing sequence

Thank you. Could you tell me why ?

Re: Isometry composition, (f^n (x)) - increasing sequence

Quote:

Originally Posted by

**wilhelm** Could you tell me why

?

Good question. I was a little hasty. But does not it follow that f(x) = x? Indeed, since f is an isometry and f(0) = 0, for every 0 <= x <= 1 we have |f(x)| = |f(x) - f(0)| = |x - 0| = |x| = x. But since f(x) ∈ [0, 1], f(x) >= 0, so |f(x)| = f(x) = x.

Re: Isometry composition, (f^n (x)) - increasing sequence

Am I missing something or does it follow straight from that this isometry is identity?

Re: Isometry composition, (f^n (x)) - increasing sequence

I'm sorry there should be commas instead of "-" signs.

Re: Isometry composition, (f^n (x)) - increasing sequence

Quote:

Originally Posted by

**wilhelm** Am I missing something or does it follow straight from

that this isometry is identity?

In addition to this fact, one needs f(0) = 0 and f(x) ∈ [0, 1]. Otherwise, and satisfy the quoted property but are not identities. With these additional properties, a proof that f is the identity is in post #4.

Re: Isometry composition, (f^n (x)) - increasing sequence

I hope I'm not bothering you. I have one more question. Do you know how to prove that the sequence is increasing for an arbitrary metric consistent with natural topology of ?

Re: Isometry composition, (f^n (x)) - increasing sequence

Do I understand right that now we have some metric on [0, 1] that generates the natural topology and that f is still an isometry w.r.t. this metric?

Re: Isometry composition, (f^n (x)) - increasing sequence