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I am working on homework for my intro to analysis class and I was assigned a problem to find all accumulation points of the set $S=\{x\mid x\in[0,1]$ and $x$ is rational$\}$. I hodge podged a solution but I feel like it may be incorrect. If anyone could let me know, mostly in that can I assume that the following approximation even exists? I feel like using the fact that $\mathbb{R}$ is ordered should work but I'm not sure.

Let k be some number in the interval [0,1] . Then we can do the following, write an approxmation of k as $0.d_{1}$ where $k-0.1<0.d_{1}<k$ and $0\le d_{1}\le9$ Which must be possible as $\mathbb{R}$ is an ordered field. We know $0.d_{1}=\frac{d_{1}}{10}$ which is rational. Then we can find a better approxmiation by writing $k$ as $0.d_1 d_2$ where $k-0.01<0.d_1 d_2<k$ and $0\le d_1,d_2\le9$ . Once again $0.d_1 d_2=\frac{d_1 d_2}{10^{2}}$ which is rational. Continue by writing $k-10^{-n}<0.d_{1}d_{2}d_{3}\ldots d_{n}<k$ where $\frac{d_1 d_2 d_3\ldots d_n}{10^n}$ is obviously rational.

Rewrite $ 0.d_1 d_2 d_3\cdots d_n$ as $\sum_{i=1}^n 10^{-i}\ldots d_i$. Then, if we take the limits of the RHS and LHS of the inequality we get,

$$\lim_{n\to\infty}k-10^{-n}=k\text{ and }\lim_{n\to\infty}k=k$$ which by the squeeze theorem says that $$\lim_{n\to\infty}\sum_{i=1}^{n}10^{-i}\cdot d_{i}=k$$ Repeating this process with $k+0.1>0.d_{1}>k$ gives the limits:

$$\lim_{n\to\infty}k+10^{-n}=k\text{ and }\lim_{n\to\infty}k=k$$ which by the squeeze theorem says that $$\lim_{n\to\infty}\sum_{i=1}^{n}10^{-i}\cdot d_{i}=k.$$

Therefore this shows that every number in the range $[0,1]$ is bounded above and below by an infinite sum of rational numbers. Which then implies that any number $k\in[0,1]$ has infinitely many $x\in Q$ within the range $\left\{ k-\epsilon,\:k+\epsilon\right\} $ . Therefore the accumulation points of $S=\{x \mid x\in[0,1]$ and x is rational} is all of $[0,1]$

Note: I don't actually know the answer to this question if my solution is wrong.

Michael Hardy
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EvHi
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1 Answers1

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I think you did more than was necessary. An accumulation point is simply a point that is the limit of a sequence. Trivially we know that any $x \in S$ is an accumulation point, as it is the limit of the constant sequence $\{x\}_{i=1}^\infty$. This means we are interested in the elements of $[0,1]\setminus S$. Given any irrational $q \in [0,1]$, there exists a sequence of rational numbers $\{r_i\}_{i=1}^\infty$ such that $r_i \to q$. So every $q \in [0,1] \setminus S$ is an accumulation point of $S$. You may also want to verify that there isn't an accumulation point outside of $[0,1]$.

When you define the process for "writing an approximation for a $k \in [0,1]$" I think you are getting at the idea that a rational sequence can converge to an irrational number. Your sequence $\{d_1,\space d_1d_2, \space d_1d_2d_3,\ldots\}$ is such a sequence. But I don't think you need to be so explicit. There is also no need to invoke an infinite sum.

graydad
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  • Thanks for the response. It was really helpful. That's what I was trying to get at but I didn't actually know beforehand that every irrational had a rational sequence whose limit was the irrational. Kind of figured it out as I went. I mean it's obvious while thinking of it as a decimal expansion but guess I just never thought about it. – EvHi Jun 16 '15 at 16:48
  • @EvHi Happy to help, I could tell you were on the right track! It isn't necessarily obvious that every irrational number can be the limit of a strictly rational sequence. But it is a very useful thing to remember as you go through analysis. – graydad Jun 16 '15 at 16:50