Question about Cauchy Sequences

[Dark Avorian]

So, I was recently working on a project exploring the Cauchy sequence approach to constructing the real numbers. As I was working through the proofs that multiplication, addition and multiplicative inversion of the reals were well defined within this structure I noticed that the proofs that each operation preserved the Cauchy criterion were basically exact mirrors of the proofs showing that they also preserved equivalence...So here's my question.

Let f be a function defined on the reals. Let F be a function defined on the set of all sequences of reals as follows:

F(\left\{a_n\right\})=\left\{f(a_n)\right\}

And if * is an arbitrary binary operation on the reals... let us define a binary operation on sequences of reals as follows:

\left\{a_n\right\}\cdot\left\{b_n\right\}=\left\{a_n*b_n\right\}

Then are the following two statements equivalent?

F maps all Cauchy sequences to Cauchy sequences

F if two Cauchy sequences are equivalent (their difference converges to zero) then their images are equivalent (note that this merely means the difference approaches zero..not necessarily that they are Cauchy)


And are these statements also equivalent?

* maps any pair of Cauchy sequences to another Cauchy sequence.

If {a} and {b} are equivalent and {c} and {d} are equivalent, then {a*c} is equivalent to {c*d}

[Qaanol]

Let f be a function defined on the reals. Let F be a function defined on the set of all sequences of reals as follows:

F(\left\{a_n\right\})=\left\{f(a_n)\right\}

And if * is an arbitrary binary operation on the reals... let us define a binary operation on sequences of reals as follows:

\left\{a_n\right\}\cdot\left\{b_n\right\}=\left\{a_n*b_n\right\}

Then are the following two statements equivalent?

F maps all Cauchy sequences to Cauchy sequences

F if two Cauchy sequences are equivalent (their difference converges to zero) then their images are equivalent (note that this merely means the difference approaches zero..not necessarily that they are Cauchy)

Yes. Suppose F maps Cauchy sequences to Cauchy sequences. Let A = {a_n} and B = {b_n} be equivalent Cauchy sequences. If F(A) and F(B) were not equivalent, then the Cauchy sequence given by interleaving terms C = a₁b₁a₂b₂… would have an image that is not Cauchy. This proves the forward implication.

Suppose instead that F sends equivalent Cauchy sequences to equivalent sequences. Let A = {a_n} be any Cauchy sequence, and let B = {b_n = b} be the unique constant sequence equivalent to A, which must exist by completeness of the reals. Clearly F(B) is Cauchy, since it is a constant sequence with every term equal to f(b). But F(A) is equivalent to F(B), so it must also be Cauchy. This proves the reverse implication, hence equivalence.

And are these statements also equivalent?

* maps any pair of Cauchy sequences to another Cauchy sequence.

If {a} and {b} are equivalent and {c} and {d} are equivalent, then {a*c} is equivalent to {c*d}

Yes. Look at Cauchy sequences (A, B) in ℝ², whence * is just a function f:ℝ²→ℝ. Now the previous result applies.

[Qaanol]

I was thinking about this some more, wondering if it holds for incomplete spaces. The forward implication clearly holds, as my proof uses only properties of Cauchy sequences. But in proving the reverse implication, “If F sends equivalent Cauchy sequences to equivalent sequences, then F sends Cauchy sequences to Cauchy sequences”, I used completeness.

My intuition says it should be true even for incomplete spaces, because every metric space is a subset of its formal completion by Cauchy sequences, and the statement has to be true in that superspace. Does anyone have a concise proof that stays within the original incomplete space? (Or a counterexample?)

[Nitrodon]

Suppose {x_n} is Cauchy, but {f(x_n)} is not Cauchy. Then we can find some ε > 0 and an increasing sequence {n_k} such that d(f(x_{n_k}), f(x_{n_{k+1}})) > ε for all k. Then {x_{n_k}} and {x_{n_{k+1}}} are equivalent Cauchy sequences since {x_n} is Cauchy, but their images {f(x_{n_k})} and {f(x_{n_{k+1}})} are not equivalent sequences.

This should prove the reverse implication without assuming completeness, assuming I didn't make any dumb mistakes at this late hour.

[Qaanol]

Suppose {x_n} is Cauchy, but {f(x_n)} is not Cauchy. Then we can find some ε > 0 and an increasing sequence {n_k} such that d(f(x_{n_k}), f(x_{n_{k+1}})) > ε for all k. Then {x_{n_k}} and {x_{n_{k+1}}} are equivalent Cauchy sequences since {x_n} is Cauchy, but their images {f(x_{n_k})} and {f(x_{n_{k+1}})} are not equivalent sequences.

This should prove the reverse implication without assuming completeness, assuming I didn't make any dumb mistakes at this late hour.

Nicely done, thanks.

[Token]

Then we can find some ε > 0 and an increasing sequence {n_k} such that d(f(x_{n_k}), f(x_{n_{k+1}})) > ε for all k.

Is this obvious? I'm prone to missing simple things at the best of times and I've not done any proper maths in a while, so bear with me. I know that directly from the definition of not-Cauchy we get two increasing sequences {n_k} and {m_k} such that d(f(n_k), f(m_k)) > ε for all k, but I'm pretty sure you can't stitch these together in an obvious way (in that it's not too hard to come up with two sequences satisfying this condition for a given ε but where all other pairwise distances are < ε). Obviously this is academic since your proof works with the sequences {n_k} and {m_k}, but I'd be interested to know...

[Nitrodon]

You're right that you can easily make the other pairwise distances < ε. However, there will always be enough pairwise distances > ε/2 by the triangle inequality, so it's possible to "stitch together" the sequences by using ε/2 instead of ε.

I somehow didn't think of just using {n_k} and {m_k}, which would have saved me the trouble of having to determine that fact.

[Token]

Ah, of course - if you make sure that you always have n_k < m_k < n_k+1, you can start at n₁ and inductively from n_k either go straight to n_k+1 or go via m_k. I was trying to make the triangle inequality work for me, but didn't think of applying pigeonhole >.<