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Let $A\subseteq\mathbb{R}$ such that $0$ is a limit point of $A$.

Is the set $ZA:=\{ka : k\in\mathbb{Z}, a\in{A}\}$ necessarily dense in $\mathbb{R}$?

P.S. Please on my current stage of learning I wish to and need to learn writing down formal proofs and not just informal hints.

TCHuang
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2 Answers2

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The answer is yes. For example let $A =\{ 1, \frac{1}{2}, \ldots , \frac{1}{n}, \ldots \}$. Consider the orbits of each element separately under $\mathbb Z$. So fox each fixed $n \in \mathbb N$ let $O_n = \{m \frac{1}{n} \ \colon m \ \in \mathbb Z\}$

$O_1 = \{\ldots, -2, -1, 0, 1, 2, \ldots \}$. Every interval of length $1$ contains an element of $O_1$.

$O_2 = \{\displaystyle \ldots, -1, -\frac{1}{2}, 0, \frac{1}{2}, 1, \ldots \}$. Every interval of length $\frac{1}{2}$ contains an element of $O_2$

$O_3 = \{\displaystyle \ldots, \frac{1}{2}, -\frac{1}{4}, 0, \frac{1}{4}, \frac{1}{2}, \ldots \}$. Every interval of length $\frac{1}{4}$ contains an element of $O_3$

$\vdots$

The set you defined is $O_1 \cup O_2 \cup \ldots$ . Can you see why it is dense in this case? How does that proof generalise to a general set $A$ which is dense at zero.

Edit: As demanded a complete formal proof.

Suppose as asked that $A \subset \mathbb R$ has a limit point at $0$. What this means is there is a sequence $\{a_1,a_2, \ldots\} \subset A-\{0\}$ such that $\lim a_n = 0$. We will define a subsequence $b_n$ that is easier to work with. By definition of convergence there exists, for each $n \in \mathbb N$ some sequence element $a_m$ such that $|a_m| < 1/n$. Choose $b_n = a_m$. Then each $|b_n| \le 1/n$. We will show the set $B = \{k b_n \colon k \in \mathbb Z$ and $m \in \mathbb N \}$ is dense in $\mathbb R$. Since $B \subset A$ this will imply that $A$ is dense as well.

To that end let $x \in \mathbb R$ be arbitrary. We will show $x$ is a limit point of $B$.

For each $n \in \mathbb N$ define $O_n = \{\ldots -2b_n, -b_n, 0, b_n, 2b_n, \ldots \}$. Clearly each $O_n \subset B$. These elements are evenly spaced along the real line, and the real number $x$ is an element of some interval $[kb_n, (k+1)b_n]$ which has length $|b_n|$. This implies that $|x-kb_n| \le |b_n| \le 1/n$ by construction. Define $c_n = kb_n$ then $c_n \in O_n \subset B$.

Observe that for each $n \in \mathbb N$ we have $|x-c_n| \le 1/n$. This means $c_n$ is a sequence in $B$ with limit $x$. Since $x$ was arbitrary this completes the proof.

Edit: The role of a math-learner is to read formal proofs and understand not just what the author is doing but, more importantly, why the author is doing it. This skill can not be developed by just reading proofs from a book. You must solve problems yourself and construct your own proofs, formal or otherwise, and either succeed or try to understand what went wrong. Unless you are doing this you are not learning maths. You might just be "learning off" maths though.

Daron
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  • Thanks for your answer but the original question is modified. – TCHuang Oct 17 '15 at 07:50
  • @Tien-ChengHuang so? Daron's point still stands. His hint should be enough to get you going. – guest Oct 17 '15 at 08:05
  • I have modified the notation to be more in line with the question :P – Daron Oct 17 '15 at 14:53
  • @Daron: This is not a homework problem and I don't know how to write down a formal proof dealing different cases. So either I have to modify/simplify the question again or I would rather delete this post. I have got enough hints but they are not so helpful for creating a formal proof to me. Would you like to present a formal proof? Excuse me I am currently a learner on undergraduate level, not an experienced MSE user/mathematician with high reputation. – TCHuang Oct 17 '15 at 16:47
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    Hi Tien-Cheng Huang, it is not clear from the question that what you are asking for is help writing a formal proof for this question. All you have written down is the question itself. Usually just writing the question is interpreted as "I don't know how to solve this. Can someone give me a hint or tell me how?" It might also be helpful if you gave your version of an informal proof so we can better see what specific trouble you are having and how we might help. – Daron Oct 17 '15 at 16:53
  • @Daron: The informal proof is assuming A is a set of non-negative real numbers and only claiming that any non-negative real number can be approximated by elements of ZA as close as desired. That is what I can write down formally for now. – TCHuang Oct 17 '15 at 17:06
  • @Daron Please on my current stage of learning I wish to and need to learn writing down formal proofs and not just informal hints. – TCHuang Oct 17 '15 at 17:58
  • I do not understand what you are asking for. If you like I can modify my answer into a formal proof of the statement? However I hardly see how this will help you to write down your own formal proof. All it will achieve is giving you one more the read, – Daron Oct 17 '15 at 18:45
  • @Daron: how did you learn math of undergraduate level or above (ex. baby Rudin) from the beginning? you learn form baby Rudin's informal hints or from baby Rudin's formal proffs? – TCHuang Oct 17 '15 at 18:51
  • @Tien-ChengHuang: You didn't answer his question. Do you want Daron to change this answer to be a complete proof? –  Oct 17 '15 at 18:54
  • @MikeMiller: yes! i really don't know how to write a formal proof that also deal with negative cases. I will appreciate. – TCHuang Oct 17 '15 at 18:57
  • I will appreciate a formal/complete proof like appreciating Rudin's formal proofs. Also I believe that interpreting/digesting a formal/complete proof is a math learner's job. – TCHuang Oct 17 '15 at 19:04
  • @Tien-ChenhHuang You're not going to learn how to write proofs from just reading proofs, just like you're not going to learn how to paint from just looking at paintings. A better way to learn proof-writing is to try to write a formal proof yourself, and then having it checked by someone else (possibly here on this site by opening a new question and tagging it with 'proof-verification'). – Marc Oct 17 '15 at 19:41
  • @Daron: "What this means is...": It is important that the $a_i$ are non-zero. Otherwise your proof doesn't work. – TonyK Oct 17 '15 at 19:53
  • @MarcPaul: you make sense and I always try to do some exercises after digesting relevant formal proofs. but in fact i have a 'proof-verification' post (http://math.stackexchange.com/q/1473222/275935) got downvotes without any healthy comment or editing. so that discouraged me a little. – TCHuang Oct 17 '15 at 19:58
  • The definition of a limit point means the sequence can be chosen so no $ a_n = 0$ itself. This is now in the answer. – Daron Oct 17 '15 at 20:01
  • @Tien-ChengHuang I don't think there was any reason to downvote that, but if you wanted help with your proof you should have put in in the question, not as an answer. – Kevin Carlson Oct 17 '15 at 21:37
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It suffices to show first that $ZA$ is dense in $[0, \infty)$ $\iff$ $\forall$ p $\in$ $[0, \infty)$ p is a limit point of $ZA$ $\iff$ $\forall$ p $\in$ $[0, \infty)$ $\exists$ a sequence $(k_n*a_n)$ of elements of $ZA$ such that $k_n*a_n \to p$ as $n \to \infty$.

Fix arbitrary p $\in$ $[0, \infty)$. For each large $j \in \mathbb{N}$, since $0$ is a limit point of $ZA$, there are $k_j \in \mathbb{Z}$ and $a_j \in A$ such that $0 < |k_j*a_j| < 1/j$. By Archimedean property of R, $\exists$ $n \in \mathbb{N}$ such that $p-\frac{1}{j} < n|k_j*a_j|$. Put $n_j = min\{n\in\mathbb{N} : p-\frac{1}{j} < n|k_j*a_j|\}$. We claim that $n_j|k_j*a_j| < p+\frac{1}{j}$.

Suppose by way of contradiction that $n_j|k_j*a_j| \ge p+\frac{1}{j}$. Then $(n_j-1)|k_j*a_j| = n_j|k_j*a_j|-|k_j*a_j| \ge p+\frac{1}{j}-|k_j*a_j| > p+\frac{1}{j}-\frac{1}{j}=p > p-\frac{1}{j}$. Hence $n_j$ is not the minimum of the set $\{n\in\mathbb{N} : p-\frac{1}{j} < n|k_j*a_j|\}$ which is a contradiction.

Hence $p-\frac{1}{j} < n_j|k_j*a_j| = n_j|k_j|sgn(a_j)*a_j < p+\frac{1}{j}$ $\forall j \in \mathbb{N}$, or $|n_j|k_j|sgn(a_j)*a_j-p| < \frac{1}{j}$ $\forall j \in \mathbb{N}$. Note that $k_j' := n_j|k_j|sgn(a_j)$ is still in $\mathbb{Z}$. Hence we have $|k_j'*a_j-p| < \frac{1}{j}$ $\forall j \in \mathbb{N}$. Hence $k_j'*a_j \to p$ as $j \to \infty$.

Since p $\in$ $[0, \infty)$ is arbitrary, the above argument showed that $ZA$ is dense in $[0, \infty)$.

Now for each $p < 0$, consider $q = -p > 0$. By the above argument we have a sequence $(k_n*a_n)$ of elements of $ZA$ such that $k_n*a_n \to q$ as $n \to \infty$. Hence $-k_n*a_n \to -q = p$ as $n \to \infty$. This proved that $ZA$ is dense in $\mathbb{R}$.

TCHuang
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