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There's a continuous function $f: \mathbb{Q} \rightarrow \mathbb{R}$

Prove that $ \exists t \in \mathbb{R} \setminus \mathbb{Q} \ \ \exists g \in \mathbb{R} :\ \ \lim _{q \rightarrow t, \ q\in \mathbb{Q}}f(q)=g$.

Prove that there are $2^{\aleph _0}$ such numbers $t$.

I know that if a function is continuous on rational points, then it's continuous on whole $ \mathbb{R}$, but that isn't relevant to the problem, is it?

I would appreciate all your help.

Thanks.

Bilbo
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  • What does $f$ have to do with anything else in your question. Typo? – Alex Youcis Mar 09 '13 at 07:46
  • ... and limit of what? – Kaster Mar 09 '13 at 07:49
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    "I know that if a function is continuous on rational points, then it's continuous on whole R, but that isn't relevant to the problem, is it?" What about $\tan$ restricted to $\mathbb{Q}$ ? – Selim Ghazouani Mar 09 '13 at 07:50
  • Each continuous function is exactly determined by its values on $\mathbb{Q}$. If we have $f,g$ real cont. functions and $f(x)=g(x), \ x \in \mathbb{Q}$ and for any $c \in \mathbb{R}$ there exists a Cauchy sequence convergent to a rational number. Due to $f, g$ 's continuity... – Bilbo Mar 09 '13 at 07:58

3 Answers3

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Here is my attempt; I hope that there is a simpler solution. (But if you have already seen the facts I mention below, it is not that complicated.)

Let us define $$\omega(f,x)=\limsup\limits_{\substack{t\to x\\t\in\mathbb Q}} f(t)-\liminf\limits_{\substack{t\to x\\t\in\mathbb Q}} f(t).$$ (This make sense for $x\in\mathbb R$ too; even though the function $f$ is only defined for rationals.)

The continuity on $\mathbb Q$ implies that $\omega(f,x)=0$ if $x\in\mathbb Q$.

The set $\{x\in\mathbb R; \omega(f,x)<\varepsilon\}$ is open in $\mathbb R$ for each $\varepsilon>0$.

If we define, for example, $$g(x)=\liminf\limits_{\substack{t\to x\\t\in\mathbb Q}} f(t),$$ then $$\omega(f,x)=\limsup\limits_{t\to x} g(t)-\liminf\limits_{t\to x} g(t)=\omega(g,x).$$ The fact that the function $\omega(g,x)$ is upper semicontinuous and the above set is open is often used in the proof that the set of continuity points of any function is $G_\delta$.

See, for example:

Hence $M=\{x\in\mathbb R; \omega(f,x)=0\}$ is a $G_\delta$ set containing rationals. If we show that $|M|=2^{\aleph_0}$, then we have $2^{\aleph_0}$ points in $M\setminus\mathbb Q$ with the required properties.

So it suffices to show that:

Every dense $G_\delta$ subset of reals is uncountable.

By a result of Mazurkiewicz (mentioned, e.g., in this answer) such subset can be endowed with a metric which makes it into complete metric space.

If we have complete metric space with no isolated points then it is uncountable by Cantor-Bendixson theorem.

Some arguments showing that such sets must be uncountable have been given also here:

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Martin Sleziak
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Let $\Bbb P=\Bbb R\setminus\Bbb Q$, and for $x\in\Bbb R$ let $B(x,\epsilon)=(x-\epsilon,x+\epsilon)$. Suppose that the first result is false. Then for each $x\in\Bbb P$ there is an $n(x)\in\Bbb N$ such that for each $\epsilon>0$ there are $p,q\in B(x,\epsilon)\cap\Bbb Q$ with $|f(p)-f(q)|\ge 2^{-n}$. For $k\in\Bbb N$ let $A_k=\{x\in\Bbb R\setminus\Bbb Q:n(x)=k\}$; by the Baire category theorem there are an $m\in\Bbb N$ and an open interval $(a,b)$ in $\Bbb R$ such that $A_m$ is dense in $(a,b)\cap\Bbb P$ and hence in $(a,b)$. Fix $p\in(a,b)\cap\Bbb Q$. Since $f$ is continuous, there is an $\epsilon>0$ such that $|f(q)-f(r)|<2^{-m}$ for all $q,r\in B(p,\epsilon)\cap\Bbb Q$, and it follows that $A_m\cap(a,b)\cap B(p,\epsilon)=\varnothing$, which is a contradiction.

In fact this shows that the sets $A_k$ are nowhere dense in $\Bbb P$. They are also closed in $\Bbb P$. To see this, suppose that $x\in\operatorname{cl}_{\Bbb P}A_k$, and let $\epsilon>0$ be arbitrary; then there is a $y\in A_k\cap B(x,\epsilon/2)$, so there are $p,q\in B(y,\epsilon/2)\cap\Bbb Q\subseteq B(x,\epsilon)\cap\Bbb Q$ with $|f(p)-f(q)|\ge 2^{-k}$, and the claim follows. For $k\in\Bbb N$ let $U_k=\Bbb P\setminus A_k$; $U_k$ is a dense open subset of $\Bbb P$, so $G=\bigcap_{k\in\Bbb N}U_k$ is a dense $G_\delta$-set in $\Bbb P$, and clearly $f$ has a continuous extension to each $x\in G$. Thus, to finish the problem we need only show that $|G|=2^\omega$; this is done in this answer to an earlier question, also noted by Martin Sleziak in his answer.

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Brian M. Scott
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HINT:

Question 2, because $R$ has $2^{\aleph_0}$ numbers, and $Q$ has countable numbers, so $R\setminus Q$ still has $2^{\aleph_0}$ numbers...

ADDED: Suppose that $|R\setminus Q|=\kappa<2^{\aleph_0}$, then $|R|=|R\setminus Q \cup Q|=\kappa+\aleph_0=\kappa<2^{\aleph_0}$, which leads a contradiction.

Paul
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  • I was thinking just the same thing ;). By the continuous function $f \ \ \lim _{n \rightarrow \infty} f(r_n) = f(t)$ – Bilbo Mar 09 '13 at 08:02
  • Could you give me some more detaild about the 2nd question? – Bilbo Mar 09 '13 at 08:04
  • A function isn't continuous in $x_0 \ \iff \ \lim _{x \rightarrow x_0^+} f(x) \neq \lim _{x \rightarrow x_0^-} f(x)$. These two determine an interval. There is a rational number in each interval and there are $\aleph _0$ rational numbers. Is that what you mean? – Bilbo Mar 09 '13 at 08:10
  • @Bilbo: have Added. – Paul Mar 09 '13 at 08:15
  • Wait. So in question 2 I just have to probe that there are $2^{\aleph _0}$ irrational numbers? – Bilbo Mar 09 '13 at 08:42
  • Paul: Continuity means $r_n\to t$ $\Rightarrow$ $f(r_n)\to f(t)$ only $t\in\mathbb Q$. (If $t\notin\mathbb Q$ then $f(t)$ is not even defined. If I understand the problem statement correctly, the function domain is rationals and the function is continuous on $\mathbb Q$.) – Martin Sleziak Mar 09 '13 at 09:27
  • Yes, that's right. Do you mean that there is something wrong with the solution of the first part of the problem? – Bilbo Mar 09 '13 at 09:50
  • @Martin: Opps. Sorry. I've edded it. – Paul Mar 09 '13 at 10:04
  • But you cannot be sure that the limit exists. A counterxample given in this comment is restriction of $\arctan$ to $\mathbb Q$. (This function is continuous on $\mathbb Q$.) – Martin Sleziak Mar 09 '13 at 10:07
  • Could you say more? – Paul Mar 09 '13 at 10:18
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    The function $f:\mathbb Q\to\mathbb R$ given by $f(x)=\arctan x$ is continuous. (Each rational number is in some interval $(k\pi-\pi/2,k\pi+\pi/2)$ and $\arctan$ is continuous on this interval.) If you take $t=\pi/2$ and $t_n$ any rational number between $t$ and $t+(-1)^n\frac1n$, then the limit $\lim f(t_n)$ does not exist. This limit is what you suggested to choose as $g$. – Martin Sleziak Mar 09 '13 at 12:58