6

Let

  • $C=C([0,1])$ be the space of continuous functions from $[0,1]$ to $\mathbb{R}$ (or $\mathbb{C}$)
  • $B=B([0,1])$ be the space of bounded funciton from $[0,1]$ to $\mathbb{R}$ (or $\mathbb{C}$)
  • $l_\infty$ the space of bounded sequences of real (or complex numbers)
  • $\{q_n\}_{n\in\mathbb{N}}$ an enumeration of the rational numbers $\mathbb{Q}\cap[0,1]$

The function $$\begin{align} \varphi:B & \longrightarrow l_\infty\\ f & \longmapsto \varphi(f)=(f(q_1),f(q_2),\dots,f(q_n),\dots) \end{align}$$ is onto but clearly not one-to-one. However, $\varphi$ restricted to $C$ is one-to-one because $\{q_n\}_{n\in\mathbb{N}}$ is dense in $[0,1]$ and the functions are continuous. But when restricted to $C$, $\varphi$ is not onto because something like $(0,0,0,1,1,1,1,1,1,1,1,\dots)$ is not the image of any continuous function.

So is there any intermediate space $C\le X\le B$ such that the restriction $\varphi\vert_{X}$ is bijective?

I have thought vaguely of some things that may work, like making the function of $(x_n)_{n\in\mathbb{N}}$ with some limit process but now i don´t have time to check a lot of those ideas. There should be some difficulties with the fact that $\{q_n\}_{n\in\mathbb{N}}$ has zero measure in $[0,1]$ so maybe instead of $B$ we should be using $L^\infty([0,1])$. Does someone have an idea or a complete solution? Also $\varphi$ is (i strongly think but cant check now) continuous and even norm preserving on $C$ but obviously not on $B$. Maybe we shouldn't be thinking of a subspace of $B$ but rather of $l_\infty$?

Also, how does the enumeration change $\varphi$?

Thanks!

edit: maybe we are looking for bounded functions continuous only on the irrational numbers?

augustoperez
  • 3,216
  • 1
    Restricting to bounded functions continuous only on the irrational numbers is not going to help. There are sequences, the pre-image of which are nowhere continuous. As an example, you can split the enumeration into 2 disjoint dense sets in $[0,1]$, and let a function take $0$ for rationals in the first set, and equal to $1$ for rationals of the second set. – Paul Jul 02 '19 at 12:47

1 Answers1

5

We do this with ultrafilters. Thus it is very "non-constructive".

For each point $x \in [0,1]$ choose an ultrafilter $\mathcal U_x$ on $\mathbb N$ such that $$ \lim_{n,\mathcal U_x} q_n = x , $$ that is, the limit of $q_n$ according to $\mathcal U_x$ is $x$. Another restriction: if $x$ is rational, say $x=q_m$, then $\mathcal U_x$ must be chosen as the fixed ultrafilter at $m$.

Write $\mathfrak U$ for this system $(\mathcal U_x)_{x \in [0,1]}$.

Definition. A function $f : [0,1] \to \mathbb R$ is said to be $\mathfrak U$-continuous iff: for all $x \in [0,1]$, $$ f(x) = \lim_{n,\mathcal U_x} f(q_n) . $$

Our space $X$ is the set of all bounded $\mathfrak U$-continuous functions. $C \subset X \subset B$.

What about extension? Existence: Let $f : \mathbb Q\cap[0,1] \to \mathbb R$ be bounded. Define $g : [0,1] \to \mathbb R$ by: $$ \text{for all $x \in [0,1]$, let}\quad g(x) = \lim_{n,\mathcal U_x} f(q_n) . \tag{1}$$ For $x$ rational, $\mathcal U_x$ is fixed, so we get $g(x)=f(x)$. That is, $g$ is an extension of $f$. Next, $g$ defined in this way is $\mathfrak U$-continuous. Indeed, let $x \in [0,1]$. For all $n \in \mathbb N$, the ultrafilter $\mathcal U_{q_n}$ is fixed at $q_n$. Then $$ \lim_{n,\mathcal U_x} g(r_n) =\lim_{n,\mathcal U_x} \;\lim_{m,\mathcal U_{r_n}} f(r_m) =\lim_{n,\mathcal U_x} f(r_n) = g(x) . $$ So $g$ is $\mathfrak U$-continuous.

Injective: Let $f_1, f_2$ be two bounded functions on $\mathbb Q\cap[0,1]$. Suppose $f_1 \ne f_2$. Then there is $m$ such that $f_1(q_m) \ne f_2(q_m)$. Let the extensions as defined be $g_1, g_2$. We have $g_1(r_m) = f_1(r_m) \ne f_2(r_m) = g_2(r_m)$, so $g_1 \ne g_2$.

Surjective: Let $h \in X$ be a bounded $\mathfrak U$-continuous function. Let $f$ be the restriction to $\mathbb Q \cap [0,1]$. We claim the extension $g$ defined by $(1)$ is $h$. Indeed, for any $x \in [0,1]$, $$ h(x) = \lim_{n,\mathcal U_x} h(r_n)= \lim_{n,\mathcal U_x} f(r_n) = g(x) . $$

GEdgar
  • 111,679
  • Thanks a lot! I have not yet accepted your answer because I am not at all familiar with the ultrafilter thing but the more I learn the more it convinces me. Do you know if this $\mathcal{U}$-continuous functions are riemann-integrable? Also, is $\phi:X\to l_\infty$ continuous? isometric? thanks again! – augustoperez Jul 02 '19 at 15:11
  • Riemann integrable? No. Start with a function on $\mathbb Q \cap [0,1]$ that is $0$ on a dense set and $1$ on a dense set. Its extension to $X$ is not Riemann integrable. Isometric? Yes, I think so. – GEdgar Jul 02 '19 at 15:53
  • how do you know $\mathcal{U}_x$ exists for a given $x$? – augustoperez Jul 02 '19 at 17:04
  • Suppose $x$ is irrational. Choose a sequence of rationals that converges to $x$. This gives you a subset of $\mathbb N$. Let $\mathcal U_x$ be any ultrafilter that contains that set. – GEdgar Jul 02 '19 at 17:40
  • also, how do we know that the extensions to U-continuous functions are bounded? – augustoperez Jul 02 '19 at 17:40
  • If $|f(q)| \le M$ for all rationals $q$, then $|g(x)| \le M$ for all reals $x$. The set ${t \in \mathbb R : |t| \le M}$ is a closed set, so the limit of things in it is still in it. – GEdgar Jul 02 '19 at 17:45