A subset of $\mathbb{R}$ is not closed if it does not have any limit points. I just began to study real analysis, and there is this question given by TA confused me.
Let $L\subset\mathbb{R}$, then $L$ is closed if it has no limit points(This is different from it does not contain any limit points according to what he says, and he further adds that in this case $L$ may not contain any convergent sequence as a result).
It is trivial when $L$ is empty since in this case, $L^c$ is $\mathbb{R}$ which is open. But I cannot sort out the case when $L$ is nonempty because the definition of the limit points that
if $x$ is a limit point of $L$ then for every $\epsilon>0$ such that the neighborhood $V_\epsilon(x)$ intercepts some elements of $L$ other than $x$,
forces me to not consider the constant sequence(If we are allowed to consider a constant sequence then it will be handy to construct a counterexample to the claim and so that $L$ can only be an empty set). So I wonder if this claim is justified and so it can be answered with elementary idea.
If possible, no matter this claim is correct or not, I would like to have some examples which address why it is or it isn't. Thanks.
 A: Consider the set $S =$$\{0\}\subset\mathbb{R}$
The complement of $S$, $S^c$$= (-\infty,0)\cup(0,\infty)$
Every point of $S^c$ is an interior point and thus is open. Hence, $S$ is closed. Suppose $x\in\mathbb{R}$ is a limit point of $S$. Choose $r = \frac{1}{2} |x|$. Then a neighborhood of $x$ of radius $r$ does not contain $0$, a contradiction. Therefore, $S$ has no limit point.
A: $L\subset {\mathbb R} $ is closed iff $L$ contains all of it's limit points.
$ L' $ : Set of all limit points of $L$ (Derived Set)
Then, $ L $ is closed in $\mathbb{R} $ iff $L' \subset {L}$
In your question, if $L$ contains no limit points then $L' =\emptyset$
and as we know empty set is a subset of every sets, $L' \subset {L} $
And hence, $L\subset {\mathbb{R}} $ is closed.(It's just a particular case of the definition) .
Now come to the sequential definition of limit point,
a point $x\in \mathbb{R} $ is a limit point of $L$ if $\exists (x_n) \subset L$ and $x_n \neq x,\forall n\in \mathbb{N}$ such that $x_n \to x $.
You can prove this two definitions of limit points are infact equivalent. (Assuming you are inside $\mathbb {R}$)
Then, $L\subset {\mathbb{R}}$
is closed if for every sequence $(x_n) \subset L$ and such that $x_n \to x $ $\implies x\in {L}$.
Now, the question is about the validity of constant sequence.
If we allow constant sequence, then every points of $L$ would be a limit point.
And then the way we define limit point,empty set is the only set having no limit points.
So, it depends on your definition i.e the existence of a ponit in every neighborhood of $ x $ distinct from $x$ or not.
But, the standard way to define a limit point that allows a point other that point.
A: One more thing i wanna added is that if we take a sequence in any set and make range set of it then that sequence will be convergent iff range set contains that limit point. But according to your consideration, you are taking a sequence in $L$ ,   $L$ doesn't contains any limit point then range set which will be subset of $L$ offcourse , so $L$ will also not contains any limit points . So there will no convergent sequence in $L$
