It's a question I've been thinking about but I can't find an easy answer. I think it will be interesting. Can there be a countable collection of real valued functions $f\_1, f\_2 , ... $ such that for any subset $K$ of $\mathbb R$ of cardinality continuum, the set of those $n$ such that $f\_n(K)$ is not the whole of $\mathbb R$ is finite?



Set theoretic question about real valued functions
I have three observations. 


First, I have noticed that there can be no such sequence of
functions $f\_n$, if one insists that every $f\_n$ is a
measurable function. In particular, there can be no sequence of Borel functions with your property. To see this, suppose that
$f\_n:\mathbb{R}\to\mathbb{R}$ is a countable sequence of
measurable functions. If infinitely many $f\_n$ are not
surjective, then clearly the property will fail and we are
done. So we may assume that all but finitely many of the
functions are surjective, and by throwing away finitely
many functions, we reduce to the case where all $f\_n$ are surjective.
For each natural number $n$ and real $y$, let
$A^y\_n=f\_n^{-1}(y)=\{\ x\ \mid\ f\_n(x)=y\ \}$. For any fixed
$n$, the sets $A^y\_n$ for various $y$ partition
$\mathbb{R}$ into continuum many disjoint measurable sets.
Since one cannot have uncountably many disjoint positive
measure sets, it follows that there must be some $y\_n$ (and
in fact many such $y\_n$) with $\lambda(A^{y\_n}\_n)=0$, where
$\lambda$ is the Lebesgue measure. Let
$K=\mathbb{R}-\bigcup\_n A^{y\_n}\_n$, which is the complement
of a measure $0$ set, and so in particular, $K$ has size
continuum. Observe that $f\_n[K]$ omits $y\_n$ by
construction, and so there is no $n$ for which
$f\_n[K]=\mathbb{R}$, showing that the desired property
fails. I guess the argument actually doesn't need that
every $f\_n$ is measurable, but only that $f\_n^{-1}(y)$ is a
measurable set for every $y$.


Second, this idea combines with the idea that Gowers had briefly posted yesterday, namely, the idea of making a large cardinal assumption, and allows us now to prove the following theorem. 


**Theorem.** If the continuum is a real-valued
measurable cardinal (a hypothesis that is equiconsistent
over ZFC with the existence of a measurable cardinal), then
there is no such sequence of functions
$f\_n:\mathbb{R}\to\mathbb{R}$.


Proof. Suppose that the continuum is a real-valued
measurable cardinal. This means that there is a
countably-additive real-valued measure $\mu$, measuring
every subset of $\mathbb{R}$ and giving points measure $0$.
We may assume that $\mu$ extends the Lebesgue measure.
Using this measure, every function $f\_n$ is measurable, of
course, and so the idea of the paragraph above goes
through, using the new measure in place of the Lebesgue measure. QED


In particular, this shows that if large cardinals are consistent with ZFC, then a negative answer to your question is also consistent with ZFC. So one shouldn't expect to be able to build a sequence of functions provably exhibiting the property.


Finally, third, I noticed that if the Continuum Hypothesis holds,
then there is a sort-of-near-positive solution in the sense that there is a sequence of functions
$f\_n:\mathbb{R}\to\mathbb{R}$ such that for every
uncountable set $K\subset\mathbb{R}$, the *combined* range is
onto, meaning $\bigcup\_n f\_n[K]=\mathbb{R}$. (And in fact,
the existence of such $f\_n$ is equivalent to CH.) To see
this, let $\lt$ well-order $\mathbb{R}$ in order type
$\omega\_1$. In particular, every real $x$ has only
countable many predecessors in $\lt$, so enumerate them
as $f\_n(x)$ for $n\in\omega$. If $K\subset\mathbb{R}$ is
any uncountable set, then $K$ is unbounded in the
well-order, and so every real $y$ is a predecessor of some
element of $K$, and so $y=f\_n(x)$ for some $x\in K$. Thus,
$\bigcup\_n f\_n[K]=\mathbb{R}$, as desired.


Your question is very interesting!


