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Si $f$ es una función continua y si $f(m/2^n)=0$ para todo entero $m$ y todo natural $n$, ¿cómo demuestro que $f(x)=0$ para todo numero real $x$?


[translation by mixedmath]

If $f$ is a continuous function and if $f\left(\dfrac{m}{2^n}\right) = 0$ for all integers $m$ and all natural $n$, how do I show that $f(x) = 0$ for all real $x$?

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    Lei sin cuidado la pregunta. Como esta escrita, el unico numero que es limite de una sucesion de la forma que usted pone es el $0$. Apuesto a que no escribio lo que queria preguntar, y sugiero que trate con cuidado de escribir correctamente lo que de verdad quiere preguntar. Yo ya me voy a dormir... (Translation: I read the question carelessly. As written, the only number that is the limit of such a sequeence is $0$. I'll bet you did not write what you meant to ask, and I suggest you try carefully to write down what you actually want to ask. Me, I'm going to bed...)2011-09-17
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    The quantifier is surely not in the intended place. The number $m$ is not intended to be fixed.2011-09-17
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    What definition of real number do you have?2011-09-17
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    Is this Andres from the "Topology Exercise" question? You asked the same question as he did in the comment. Andres also wrote his question incorrectly multiple times. Anyway, do you mean that $x = m \sum_{n \in \mathbb{Z}} a_n \frac{1}{2^m}$ where $a_n$ is $0$ or $1$? If this is the case, then as Magidin noted, this can only hold if $x$ and $m$ have the same sign (or both zero). If $m \neq 0$, then just use the binary for $\frac{x}{m}$ to get the necessary $a_n$'s.2011-09-17
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    New version: If $f$ is a continuous function and $f(m/2^n)=0$ for all integers $m$ and all natural numbers $n$, how do I prove that $f(x)=0$ for all real numbers $x$?2011-09-17
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    Did Jerson change this question? Although Jerson wrote his question incorrectly, I don't think this question is in same spirit as the original. I think he wanted something like an expansion of a real number.2011-09-17
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    @William: Yes, he changed it completely. I think that he may have been trying to answer the present question, realized that getting each real as the limit of a sequence of dyadic rationals would do the trick, asked the subsidiary question badly, and changed it to the main question when his wording was challenged.2011-09-17
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    @Arturo Magidin: What is mixedmath? I hoped it was some clever translator but could not find any sense when I googled it.2011-09-17
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    @AD.: http://math.stackexchange.com/users/9754/mixedmath2011-09-17
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    @William: Yes, he changed it; if the original question was meant to be what most people think it was (with $m$ not fixed), or something similar, then answering the original question implies this one (since you would be proving the dyadic rationals are dense).2011-09-17

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Let $x\in\mathbb{R}$, and let $\epsilon\gt 0$. Then there exists $\delta\gt 0$ such that for all $z$, if $|x-z|\lt\delta$ then $|f(x)-f(z)|\lt\epsilon$. Let $n$ be a positive integer such that $\frac{1}{2^{n}}\lt \delta$. Then there is an integer multiple of $\frac{1}{2^{n+1}}$ that lies in $(x-\frac{1}{2^n},x+\frac{1}{2^n})$, since the interval has length $\frac{1}{2^{n-1}}$. Let $z=m/2^{n+1}$ be that integer multiple. Then $\epsilon\gt |f(x)-f(z)| = |f(x)-0| = |f(x)|$.

Thus, $|f(x)|\lt \epsilon$ for all $\epsilon\gt 0$, hence $f(x)=0$.


Sea $x\in\mathbb{R}$, y sea $\epsilon\gt 0$. Existe $\delta\gt 0$ tal que para todo $z$, si $|x-z|\lt\delta$ entonces $|f(x)-f(z)|\lt\epsilon$. Sea $n$ un entero positive talk que $\frac{1}{2^n}\lt\delta$. Entonces hay un multiplo entero de $\frac{1}{2^{n+1}}$ en el intervalo $(x-\frac{1}{2^n},x+\frac{1}{2^n}$, pues el intervalo tiene longitud $\frac{1}{2^{n-1}}$. Sea $z=m/2^{n+1}$ ese multiplo entero. Entonces $\epsilon\gt |f(x)-f(z)| = |f(x)-0| = |f(x)|$.

Por lo tanto, $|f(x)|\lt\epsilon$ para toda $\epsilon\gt 0$, de manera que $f(x)=0$.

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    y de como seria la prueba usando la expansión binaria?2011-09-17
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    Is this the first time we've ever had a bilingual answer? :)2011-09-17
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    Moral: Classical analysis is beautiful in ANY language. : )2011-09-17
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Added: The question changed while the answer below was being typed. A short note is added at the end to answer the altered question.

The wording of the original question does not accurately convey the intention. I believe that the OP wants a proof that every real number $x$ is the limit of a sequence $(a_k)$, where each $a_k$ is of the form $m/2^n$, with $m$ an integer and $n$ a non-negative integer. (Numbers of the form $m/2^n$ are called dyadic rationals.)

It is enough to show that for any positive integer $k$, we can find a number $a_k$ of the required form such that $|x-a_k|<1/k$.

We prove the result for $x>0$. A small modification takes care of the case $x<0$.

Let $n$ be the smallest integer such that $2^n>k$. Consider the numbers $0/2^n$, $1/2^n$, $2/2^n$, $3/2^n$, and so on.

Let $m$ be the smallest non-negative integer such that $(m+1)/2^n>x$. Then $m/2^n \le x$. Let $a_k=m/2^n$. Since $$\frac{m+1}{2^n} -\frac{m}{2^n}=\frac{1}{2^n} <\frac{1}{k},$$ it follows that $$0\le x-\frac{m}{2^n}=x-a_k <\frac{1}{k}.$$

Added Note: The question was changed and now asks one to show that if $f$ is a continuous function which is $0$ on the dyadic rationals, then $f$ is $0$ everywhere.

Let $x$ be a real number. We show that $f(x)=0$. Let $(a_k)$ be a sequence of dyadic rationals with limit $x$. Such a sequence exists by the calculation above. Then $$f(x)=\lim_{k\to \infty} f(a_k) =0.$$

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    Did you notice the OP changed the question to something somewhat different? (Of course, the answer here implies the answer to the question as currently phrased).2011-09-17
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    @Arturo Magidin: I noticed after posting. Added a couple of lines. Already the question I first answered was different from the original question. My Spanish is still good enough to notice the unintended "fixing" of $m$. Unfortunately, it is no longer good enough for literate translation of the answer above.2011-09-17
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    Would you like a translation into Spanish of your answer? I can do basic analysis in Spanish (I'd have a hard time with Universal Algebra, since I learned *that* in English and have never thought about it in Spanish )2011-09-17
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    @Arturo Magidin: Thank you, but I expect that is not necessary. The OP is undoubtedly quite comfortable *reading* English, just as I am reading Spanish. Understanding colloquial conversations is another matter. That depends very much on *where* the speaker is from.2011-09-17
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    Well, yes. I am a native Spanish speaker (grew up in Mexico), and I remember watching "La Historia Oficial" many years ago (a film from Argentina); there were several scenes where I could simply not follow the dialogue until I saw the movie with English subtitles! Sometimes I miss entire sentences in Almodovar films, too. (-:2011-09-17
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Let $D = \left\{\frac{m}{2^n}:m\in\mathbb{Z} \land n\in\mathbb{N}\right\}$. (Members of $D$ are called dyadic rationals.) Since $f$ is continuous, you know that $\lim\limits_{n\to\infty}f(x_n)=f(x)$ whenever $\lim\limits_{n\to\infty}x_n=x$. Let $x$ be any real number. If we can find a sequence $\langle x_n:n\in\mathbb{N} \rangle$ of dyadic rationals such that $\lim\limits_{n\to\infty}x_n=x$, we’ll be done, because then $f(x) = \lim\limits_{n\to\infty}f(x_n)=0$.

One way to do this is to use the binary expansion of $x$. Suppose that $\displaystyle x = \sum\limits_{k=m}^{\infty} b_k2^{-k}$ for some integer $m\le 1$, where each $b_k$ is either $0$ or $1$. For each $n\in \mathbb{N}$ let $\displaystyle x_n = \sum\limits_{k=m}^n b_k2^{-k}$. Then $$x_n = \frac{\sum_{k=m}^n b_k 2^{n-k}}{2^n} \in D,$$ and it's clear that $\lim\limits_{n\to\infty}x_n=x$.

(I’m sorry that I can’t translate this into Spanish.)

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    Which neatly answers the question now being asked, as well as the likely question that was *meant* to be asked originally...2011-09-17
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    @Arturo: Jerson had edited it into the present form just before I saw it.2011-09-17