Linear transformation from real numbers over rationals to rational Cauchy sequences

Question

Let $\alpha\in\mathbb{R}$. Is there a way to construct a Cauchy sequence of rational numbers $C_\alpha$ that converges to $\alpha$ such that $\forall q\in\mathbb{Q}$, $qC_\alpha=C_{q\alpha}$?

Edit: As John Hughes corrects, I am looking for a map "$f:\mathbb{R}\to S : \alpha\mapsto C_\alpha$, where $S$ is the set of Cauchy sequences of rationals, with the property that $f(q\alpha)=qf(\alpha)$ for all $q\in\mathbb{Q}$."

The answer I'm looking for would be a construction of such a mapping or a proof that no such construction exists.

Answer 1

Construct? I doubt it. But it's easy to show that the map exists (using AC):

Say $S$ is the space of Cauchy sequences of rationals. Define $T:S\to\Bbb R$ by $$Ts=\lim s_j.$$Now $S$ and $\Bbb R$ are both vector spaces over $\Bbb Q$, and AC implies that any surjective linear map has a linear right inverse.

Or so it seems. Skepticism has been expressed in comments. Let's see. Say $T:X\to Y$ is linear and surjective. Say the scalar field is $K$. Let $A$ be the set of pairs $(Y', S)$ such that $Y'$ is a subspace of $Y$, $S:Y'\to X$ is linear, and $TSy=y$ for all $y\in Y'$. Order $A$ in the obvious way (i.e. by extension). Then $A$ has a maximal element $(Y',S)$, and it's enough to show that $Y'=Y$.

Say $y_0\in Y\setminus Y'$. Choose $x_0\in X$ with $Tx_0=y_0$. Let $Y''$ be the span of $Y'$ and $y_0$. There exists a linear map $S'':Y''\to X$ with $S'' y_0=x_0$ and $S'' y=Sy$ for $y\in Y'$. Now $(Y'', S'')$ shows that $(Y',S)$ is not maximal, contradiction.

Answer 2

Define $C_{\alpha}(i) = \alpha (1 - \frac{1}{2^i})$ (for $i \ge 0$). That means, for instance, that $C_1$ is the sequence $$ 0, 1/2, 3/4, 7/8, \ldots $$ which is clearly a Cauchy sequence, for if $i, j > M$, then $|u_i - u_j| < \frac{1}{2^M}$, (where I'm using $u_i$ to denote the $i$th term of the sequence) which can be made as small as necessary.

By the same argument, we can show $C_{\alpha}$ is Cauchy, as the difference of terms as above is no more than $\frac{\alpha}{2^M}$, which can again be made small.

And it evidently has the multiplicative property that you were looking for. Indeed, the map $\alpha \mapsto C_\alpha$ is actually a linear xform, as requested.

I've surely screwed up something here, but this seems as if it's what's needed.

EDIT: I see that I did screw something up --- I've got the kind of sequence I wanted, but the terms aren't all rational. Sigh.

Revised version:

1. pick a basis $\{ r_c | c \in B \}$, for the reals as a vector space over the rationals.
2. Let $r$ be a particular element of that basis. Look at the numbers $r\cdot (1- \frac{1}{2^i})$, $r\cdot (1- \frac{1}{2^{i+1}})$, and pick a rational number $u_i$ in that interval. The distance between $u_i$ and $u_{i+1}$ is no more that $\frac{r}{2^i}$, hence the sequence of $u_i$s is a Cauchy sequence. Define this to be $C_r$.
3. This defines the function $C$ on a basis; extend to all of $\Bbb R$ by linearity.

This follows @DavidCUllrich's idea, but gives the multiplicative structure, which I think his misses, although I could be misreading something there. The map he looks at is linear, but that doesn't (I think) mean that its left-inverse is also linear.