# My proof of: Given an adherent point, some sequence converges to it.

(I have two specific questions.) Is my proof correct? Are style, wording, and punctuation alright?

$\textbf{Definition 1.}$ Let $X \subseteq \mathbb{R}$. Let $x \in \mathbb{R}$. The point $x$ is an $\textit{adherent point}$ of $X$ iff, for every $\varepsilon \in \mathbb{R}$ with $\varepsilon > 0$, there is an $x' \in X$ such that $|x' - x| < \varepsilon$.

$\textbf{Definition 2.}$ Let $x \in \mathbb{R}$. Let $(x_n)_{ n \in \mathbb{N} }$ be a real sequence. The sequence $(x_n)$ $\textit{converges}$ to $x$ iff, for every $\varepsilon \in \mathbb{R}$ with $\varepsilon > 0$, there is an $N \in \mathbb{N}$ such that, for every $n \in \mathbb{N}$ with $n > N$, we have $|x_n - x| < \varepsilon$.

$\textbf{Theorem.}$ If $x$ is an adherent point of $X$, then there is a sequence $(x_n)_{ n \in \mathbb{N} }$ such that both \begin{equation*} \begin{split} & \text{$(x_n)$ converges to $x$, and,} &\qquad& (\textbf{Property 1}) \\ & \text{for every $n \in \mathbb{N}$, we have $x_n \in X$.} &\qquad& (\textbf{Property 2}) \end{split} \end{equation*}

$\Large\textit{Proof.}$ Let $x$ be an adherent point of $X$. It remains to prove that there is some suitable sequence. We prove this in two steps. First, we define a sequence. Second, we prove that the definiendum features properties 1 and 2.

$\textit{First step.}$ We denote the sequence by $(x_n)_{ n \in \mathbb{N} }$. Below, we define each term $x_n$. First, let $x_0$ be a point in $X$ (necessarily, $X \ne \emptyset$). Below, we choose $x_n$ for every $n \ge 1$. Since $x$ is an adherent point, by definition 1, there is an $x' \in X$ such that $|x' - x| < 1 / n$. We define $x_n := x'$. Saying this, the sequence $(x_n)$ is defined. We note (for future use) that, for all positive natural numbers $n$, \begin{equation*} |x_n - x| < 1 / n. \end{equation*}

$\textit{Second step.}$ Obviously, we have property 2. It remains to prove property 1. To that end, we consider definition 2. Obviously, $x$ is real, and $(x_n)$ is a real sequence. Let $\varepsilon \in \mathbb{R}$ with $\varepsilon > 0$. It remains to prove that there is an $N \in \mathbb{N}$ such that, for every $n \in \mathbb{N}$ with $n > N$, we have $|x_n - x| < \varepsilon$. Let $N$ be some positive natural number such that $1 / N < \varepsilon$. (We could prove that such a number exists.) Let $n \in \mathbb{N}$ with $n > N$. It remains to prove that $|x_n - x| < \varepsilon$. Since $n > N > 0$, we have $1 / n < 1 / N$. Also, as we noted above, $|x_n - x| < 1 / n$. Thus, by transitivity, $|x_n - x| < 1 / N$. Thus, since $1 / N < \varepsilon$, by transitivity, $|x_n - x| < \varepsilon$.

QED

• It’s clear and correct, but very verbose. I can’t say much about punctuation, but it, too, seems correct to me. Your proof also reads easy which is a nice feature. Commented Jan 18, 2015 at 13:29

I find your proof to be correct, but inconvenient. It is indeed clear as k.stm declared, but you would want the proof be concise, quick and to the point ("deadly", if you like). For instance do not declare that you are going to construct a sequence, construct it, or instead of saying "let $\epsilon\in\mathbb{R}$ with $\epsilon>0$" just say "let $\epsilon>0$". Also you might consider adopting some logical quantifiers, like $\forall\epsilon,\exists N,\forall n\geq N:|x_n-x|<\epsilon$ for convergence. I have come to realize that great ideas are more valuable than well-written proofs in mathematics, contrary to my understanding at the end of my freshman year as a mathematics major. Even though your proof is not extremely formal, rigor is not the most important thing in mathematics either, it's the creativity (See https://terrytao.wordpress.com/career-advice/there%E2%80%99s-more-to-mathematics-than-rigour-and-proofs/).
As a final remark, if you are willing to be crystal clear about your line of reasoning, I believe it is more valuable to say "by Archimedean Property of $\mathbb{R}$" instead of saying "we could prove that such a number exists" than your careful choice of words (or even proving it).