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$\mathbf{Question:}$ Let $ E \subseteq [a,b] $ be any subset which has only finite number of limit points.
can we say $E$ is a null set ?

$\mathbf{My \hspace{1mm} attempt:}$

If $A \subset \mathbb{R}$ is said to be null set, then for every $\varepsilon > 0$ there exists a countable collection of open intervals $(A_{n})_{n=1}^{\infty}$ such that $A \subseteq \bigcup\limits_{n=1}^{\infty} A_{n}$ and $\sum\limits_{n=1}^{\infty} \ell(A_{n}) < \varepsilon$.

Any countably infinite set $S$ = {${x_{1},x_{2},x_{3},...}$} is a null set. To see this,
let $\varepsilon > 0$ be given. Let $I_{j}$ = $ ( \frac{x_{j} - \varepsilon}{2^{j+1}} , \frac{x_{j} + \varepsilon}{2^{j+1}} ), j=1,2,3,.... $ Then
$S \subseteq \bigcup\limits_{n=1}^{\infty} I_{j}$ $\hspace{1mm}$ and $\hspace{1mm}$ $\sum\limits_{n=1}^{\infty} \ell(I_{j}) < \varepsilon.$

So by the above one can easily say $\mathbb{Q}$ is a null subset of $\mathbb{R}.$

Coming to our question, Since the given set has $\mathbf{finite}$ number of limit points. Thus if we apply sequential approach (since sequence is a function from $\mathbb{N}$ to $\mathbb{R}$) to find those limit points of the set it has got countable number of elements. So then our set $E$ becomes Null set.
Is my approach correct or is there any counterexample to the statement.
Thank you in advance.

V_G
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  • Your approach is correct, and it is true that the set $E$ is countable. But what you write is rather too vague for a complete and detailed proof: ... if we apply sequential approach ... it has got countable number of elements. – Lee Mosher Aug 20 '24 at 20:05

1 Answers1

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An uncountable set of reals will have an uncountable set of limit points: 1 2 3. If your set has only a finite set of limit points, it is countable and therefore null. The converse is not true: see Cantor set.

Lucenaposition
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