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I am studying real analysis through Terence Tao's book named "Analysis I" and I am struggling to understand the intution behind the definition of limit points. Precisely speaking, I present here the sequence of definitions he proposes related to the subject:

Let $(a_{n})_{n=m}^{\infty}$ be a sequence of real numbers, let $x$ be a real number, and let $\varepsilon > 0$ be a real number. We say that $x$ is $\varepsilon$-adherent to $(a_{n})_{n=m}^{\infty}$ iff there exists a natural number $n\geq m$ such that $a_{n}$ is $\varepsilon$-close to $x$. We say that $x$ is continually $\varepsilon$-adherent to $(a_{n})_{n=m}^{\infty}$ iff it is $\varepsilon$-adeherent to $(a_{n})_{n=N}^{\infty}$ for every $N\geq m$. We say that $x$ is a limit point or adherent point of $(a_{n})_{n=m}^{\infty}$ iff it is continually $\varepsilon$-adherent to $(a_{n})_{n=m}^{\infty}$ for every $\varepsilon > 0$.

Could someone please help me to understand it properly?

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    $\begingroup$ Just to clarify: you understand the technical definition and are trying to understand the intuition? or are you also confused about the technical definition? $\endgroup$
    – angryavian
    Apr 18, 2020 at 0:55
  • $\begingroup$ I am more interested in the intuition. As far as I have understood, to be $\varepsilon$-adherent means there exists at least one natural $n\geq m$ for which the distance $|a_{n} - x| \leq \varepsilon$. To be continually $\varepsilon$-adherent means that, whenever we discard the first natural numbers from the set $\{m,m+1,m+2,\ldots\}$, there still exists at least one natural number $n$ such that $|a_{n} - x|\leq \varepsilon$. Finally, to be a limit point, the above relation must hold for any $\varepsilon$. Am I on the right track? $\endgroup$
    – user0102
    Apr 18, 2020 at 1:01

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I will give an intuitive answer.

What is a limit point of a seqeuence?

Like for example, we say that the sequence $(\frac{1}{n})_{n=2}^{\infty}$ $\frac{1}{2}, \frac{1}{3},\cdots, \frac{1}{n},\cdots$ converges to a limit $L$ if for all $m\geq 1 ~ \exists M\geq m$ such that for all $n_1 \geq M$, you can guarantee $|{a_n}_1-L| \leq \epsilon$ occurs for all $\epsilon >0$.

But what about the sequence $\frac{1}{2}, 2-\frac{1}{2}, \frac{1}{3}, 2-\frac{1}{3},\cdots$?

You notice that the terms in the odd places converge to $0$ while the ones in the even place converge to $2$. So this sequence is not convergent. Why? Because it doesnt satisfy the definition for convergence, we cannot fix an $M$, because for all(the important thing to note) $n\geq M$ we will not have that either the sequence is $\epsilon$ close to $2$ or $0$ for all $\epsilon$. But you can note that if we can ignore the "for all" condition and replace it with "there exists" things change, because then, we need to show that for all $M$ we can find an $n_2\geq M$ such that it is $\epsilon$ close to either $2$ or $0$. If you make a diagram, you will find the elements of this sequence "cluster/crowd" around the points $0,2$ (due to existence of subsequences of this sequence converging to these limits). This are precisely limit points, and what makes them different from the notion of convergence is that difference in "for all" and "there exist" conditions, as in convergence you know there is only one such point in the real line around which the elements cluster, but the limit points are thus generalised versioons of limits, as there can be more than two such points on the real line, the there exists condition tells you that you can pick one for all $M$(notice that you can vary $M$, suppose you have chosen one element $n_1\geq M$, then you can shift $M \to M+10^{23}$ and still you can find one $n_3\geq M+10^{23}$, because each element of the sequence occurs infinitely many times, this is what the definition wants to mean rather than that of convergence).

Now you can go back to the definition and analyse.

Geometrically, you can now say that $x_0$ is a limit point of a sequence $(a_n)$ as (informally obviousky for intuition)-

If you fix the point $x_0$ in the real line, you can always fix a small(very very small)$\delta$ such that in the region $(x_0-\delta,x_0+\delta)$ you can always find one element $a_k$ of the sequence such that $a_k \in (x_0-\delta,x_0+\delta)$

You might be thinking this is the same as convergence, i would say it is similar, but the major distinction is that it does not say that all the elements in the sequence after $a_k$ need to appear in this interval(unlike convergence), it only says you can always find one, and the reason behind this is that as there can be two different limit points (as in the example $0,2$) thus you need to also have the same phenomenon going about $0$ and $2$ as well, so obviously it is not possible to have all the elements after $a_k$ lying in a small neighborhood around them both, so it is in fact a kind of "generalisation of limits as that in convergence" (reacall that there cannot be two limits of a convergent sequence, and so there is only one limit point and thus the condition that all elements of the sequence after $a_k$ appear in the same neighborhood is okay as there is only one limit point)

Phew! I hope you got the intuition! :) If you still didnt, do tell me! These all writings above are my own intuition of the limit points, and you too should think of them like this because then it will be impossible to forget about them later when you do more sophisticated things in math.

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  • $\begingroup$ Nice explanation! (+1) for sure. $\endgroup$
    – user0102
    Apr 18, 2020 at 16:21
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  • "$x$ is continually $\epsilon$-adherent: you are right that you can think of this as "even if we remove some finite number of terms from the beginning of the sequence, there is still some element $a_n$ left in the sequence such that $|a_n - x| \le \epsilon$." Another way to think about it is "there are infinitely many terms in the sequence that are $\epsilon$-close to $x$."
  • Thus, for $x$ to be a limit point of the sequence, it must satisfy the following: no matter how small you choose $\epsilon > 0$ to be, the sequence still contains infinitely many points that are $\epsilon$-close to $x$.
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  • $\begingroup$ Thank you for the contribution. It helped understanding the concept. $\endgroup$
    – user0102
    Apr 18, 2020 at 17:02
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I hadn't seen Tao's definition before reading this question; like you, I found it hard to digest. I'm used to thinking in terms of limits of subsequences, but as that concept isn't mentioned in the question (I see it's on pp.171-173 of a PDF (cough) of the first edition of Tao's book) and as it takes some setting up, it seems best to avoid it here. The following terminology is far from standard (I just made it up), but I found it helpful in getting to grips with Tao's definition:

Define a lifted point of the sequence $a = (a_n)_{n=m}^{\infty}$ to be an ordered pair $(n, a_n),$ for some $n \geqslant m.$ (Think of it as a point on the "graph" of $a.$) Define the position of the lifted point $(n, a_n)$ to be $a_n.$ (Think of this as the projection of $(n, a_n)$ onto the vertical axis.) Then an adherent point of $a$ is any real number that for all $\varepsilon > 0$ is $\varepsilon$-close to the positions of infinitely many lifted points of $a.$


Perhaps this is also helpful. (If not, ignore it. It's just a thought that's been nagging away at me ever since I read Tao's definition in the question. I had to scratch that itch, but no-one is obliged to read the resulting waffle!)

It is evident that if $a_n = x$ for infinitely many values of $n,$ then $x$ is an adherent point of $a.$

Suppose, then, that $a_n = x$ for only finitely many values of $n$ (perhaps for no values of $n$ at all). Then $x$ is an adherent point of $a$ if and only if for every $\varepsilon > 0$ there is at least one value of $n$ such that $a_n \ne x$ and $a_n$ is $\varepsilon$-close to $x.$

Given our assumption about $x,$ this condition is clearly necessary in order for $x$ to be an adherent point of $a.$

But it is also strong enough (even without our assumption about $x$) to ensure that $x$ is an adherent point of $a$.

Proof. Define $m_1 = 1.$ By hypothesis, there exists $n_1$ such that $a_{n_1} \ne x$ and $a_{n_1}$ is $(1/m_1)$-close to $x$, i.e. $0 < |a_{n_1} - x| \leqslant 1/m_1.$ Choose an integer $m_2 > m_1$ such that $1/m_2 < |a_{n_1} - x|.$ By our hypothesis again, there exists $n_2$ such that $a_{n_2} \ne x$ and $a_{n_2}$ is $(1/m_2)$-close to $x$, i.e. $0 < |a_{n_2} - x| \leqslant 1/m_2.$ Continuing in this way, we find integers $m_1 < m_2 < m_3 < \cdots$ and distinct integers $n_1, n_2, n_3, \ldots$ such that: $$ 0 < |a_{n_{k+1}} - x| \leqslant \frac1{m_{k+1}} < |a_{n_k} - x| \leqslant \frac1{m_k} \quad (k = 1, 2, \ldots). $$ Given any $\varepsilon > 0,$ there exists $k \geqslant 1$ such that $1/m_k \leqslant \varepsilon,$ whence: $$ 0 < |a_{n_j} - x| \leqslant \frac1{m_j} \leqslant \frac1{m_k} \leqslant \varepsilon \quad (j = k, k+1, \ldots). $$ Therefore $a_n$ is $\varepsilon$-close to $x$ for all $n$ in the infinite set $\{n_k, n_{k+1}, \ldots\},$ therefore $x$ is an adherent point of $a.$ $\square$

To summarise: the set of adherent points of $a$ is the union of two sets (not necessarily disjoint): (A) the set of all points $x$ such that $a_n = x$ for infinitely many values of $n$; (B) the set of all points $x$ such that for every $\varepsilon > 0$ there is at least one value of $n$ such that $a_n \ne x$ and $a_n$ is $\varepsilon$-close to $x.$ Later (pp.244-247 of the first edition, I don't know about other editions) Tao defines limit points (in a related but different sense) of subsets of $\mathbb{R}.$ A limit point of a sequence can be described as a point that either occurs infinitely often in the sequence or is a limit point of the set of points in the sequence. Update: one textbook (the only one I've managed to find so far) that proves this result is Mícheál Ó Searcóid, Metric Spaces (2000), Theorem 6.7.2 (equivalence of conditions (ii) and (v)).

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  • $\begingroup$ Later (pp.244-247 of the first edition, I don't know about other editions) Tao defines limit points (in a related but different sense) of subsets of $\mathbb{R}.$ A limit point of a sequence can be described as a point that either occurs infinitely often in the sequence or is a limit point of the set of points in the sequence. $\endgroup$ Apr 18, 2020 at 20:11

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