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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, 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!

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!

I have two observations, which stop short of a full solution.

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 $y$.

My second observation is that 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.

I don't know the answer to the fully general question you asked, but I find it very interesting.

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, 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!

I have two observations, which stop short of a full solution.

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 squence 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 $y$.

My second observation is that 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.

I don't know the answer to the fully general question you asked.

I have two observations, which stop short of a full solution.

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 $y$.

My second observation is that 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.

I don't know the answer to the fully general question you asked, but I find it very interesting.