Terence Tao defines limit points the following way –
Let $(a_n)_{n = m}^\infty$ be a sequence of real number, let $x$ be a rea l number, and let $\epsilon > 0$ be real number. We say that $x$ is $\epsilon-adherent$ to $\left( a_n \right) _{n=m}^\infty$ iff there exists an $n \ge m$ such that $a_n$ is $\epsilon-close$ to $x$. We say that $x$ is continually $\epsilon-adherent$ to $\left( a_n \right) _{n = m}^\ infty$ iff it is $\epsilon-adherent$ to $\left( a_n \right) _{n=m}^\infty $ for every $N \ge m.$ We say that $x$ is a limit point or adherent point of $\left( a_n \right) _{n=m}^\infty$ iff it is continually $\epsilon-adherent$ to $\left( a_n \right) _{n=m}^\infty$ for every $\epsilon > 0$.
When I read this, there was something unsettling about it. It’s related to the property of $\epsilon$-adherence for all $\epsilon > 0$. If such a property indeed holds, isn’t it continually adherent automatically? It was unsettling to know that for all $\epsilon > 0$, there is eventually some $a_k$ which is $\epsilon$-close to $x$. But is it still not the case that, we can find such an $a_k$ for a given N, such that $k \ge N$? In my notes I wrote,
But $\mathbb{R}$ is not countable. So isn't it the case that $\epsilon$-adherence for all $\epsilon > 0$ is equivalent to continual $\epsilon$-adherence?
Honestly, I am not able to go back to that unsettling intuition, perhaps because it has already settled? I am not able to justify the intuition I had to work towards proving the above fact. But nevertheless, I did and here’s a potential proof.
Before we begin, let me at least describe the context which motivated this train of thought. I was proving the fact – if all the elements of a sequence are non-zero and converge to a non-zero real, then the sequence is bounded away from zero. There’s a simple way to prove this,
Let $(b_n)_{n=m}^\infty$ be a sequence of real numbers such that $b_n \ne 0$, and let $y$ be the non-zero real number it converges to. Then there exists an $N \ge m$ such that for all $n \ge N$, $|a_n - y| \le |y|/2$. Thus, for all $n \ge N$, we have $|y| \le |a_n - y| + |a_n|$ and thus $|a_n| \ge |y|/2 > 0$. Hence, the sequence is bounded away from zero.
But this is not how I approached at the beginning. I tried to assume on the contrary that the sequence is not bounded away from zero, i.e.
for every $\epsilon > 0$ (which fails to bound the sequence from below), there is an $n \ge 1$ such that $|b_n| < \epsilon$. We know that there is an $N \ge 1$ s.t. $|b_n - y| < \epsilon$ for all $n \ge N$. There is already a problem recipe in sight. There is always a $b_n$ arbitrarily close to $0$! But then, $b_n$ is eventually, arbitrarily close to $y$. So is it unreasonable to expect that this means $y=0$? Or probing directly the roots, is it unreasonable to ask whether it is even possible to have such a $|b_n| < \epsilon$ for arbitrary $\epsilon$ such that, it's always the case that $n < N$? Doesn't this unsettlingly question the infiniteness of reals (since $\epsilon \in \mathbb{R}$)? (And perhaps the countability?) If we indeed find a $k \ge N$ for all $N \ge 1$ s.t. $|b_k| < \epsilon$, then we immediately see that $|y| \le 2\epsilon$. We could have started with $\epsilon=|y|/2$ and we see the contradiction.
We now give the proof for the anticipated claim.
$\epsilon$-adherence for all $\epsilon > 0 \iff$ continually $\epsilon$-adherence for all $\epsilon > 0$. Adherence for all $\epsilon$ is the crucial part of this result. Speaking for each $\epsilon$, continual adherence is of course stronger.
The proof precisely speaks rigorously about the intuition I had, and it is as follows.
There is an $\epsilon > 0$ such that for some $N \ge 1$, $|a_n - x| > \epsilon$ for all $n \ge N$. That is we assume on the contrary, $x$ is not continually $\epsilon$-adherent to $(a_n)_1^\infty$. Let $\epsilon' = min(|a_1 - x|, \ldots, |a_{N-1} - x|)$. Since $x$ is $\epsilon$ adherent, we have $|a_j - x| \le \epsilon$ for some $j \le N -1$. That is, $\epsilon' \le |a_j - x| \le \epsilon$. So, $\epsilon' \le \epsilon$. Then $\frac{\epsilon'}{M} < \epsilon$ for all positive integers $M > 1$. So, we have, $|a_i - x| \ge \epsilon' > \frac{\epsilon'}{M} \forall i = 1, \ldots, N-1 $ and $ |a_i - x| > \epsilon > \frac{\epsilon'}{M} \forall i \ge N.$ That means, $x$ is not $\frac{\epsilon'}{M}$-adherent to $(a_n)_1^\infty$, and we arrive at a contradiction.
What we saw is, $x$ must be continually adherent in order to be arbitrarily adherent. Otherwise, we will find a huge collection of reals for which it is not adherent. Basically such a finite restriction ($N$) cannot accommodate to a very dense and arbitrary adherence! That’s the intuition we had, and we made it rigorous. The finiteness played a role in obtaining the contradiction by giving us the $\epsilon’$ (if it was infinite, then we wouldn’t know if it’s bounded) which we used to bound the sequence from below and prevent the arbitrary-adherence!