# Interior points are limit points in $\mathbb{R}$?

I have read another question, and know that interior points are not limit points in general topology space.

But when we talk about any subset $\mathbb{A}$ of $\mathbb{R}$, can I say that $\operatorname{int}\mathbb{A} \subset \mathbb{A}'$?

If the subsets are continuous, it seems obvious that $\operatorname{int}\mathbb{A} \subset \mathbb{A}'$. If the subsets are discrete, $\operatorname{int}\mathbb{A} = \emptyset$ and $\mathbb{A}' = \emptyset$, thus $\operatorname{int}\mathbb{A} \subset \mathbb{A}'$

Please correct me if I am wrong.

Thanks

• To confirm, does $\Bbb{A}'$ denote the set of limit points? Commented Sep 3, 2015 at 7:22
• What is a continuous subset? Commented Sep 3, 2015 at 7:22
• I guess continuous stands for connected here. Commented Sep 3, 2015 at 7:26
• $\mathbb{A}'$ denotes limit points of the set $\mathbb{A}$ I use continuous subset to represents sets like (0,1), sorry for causing confusion. Commented Sep 3, 2015 at 7:35

You are right. If $\DeclareMathOperator{\int}{int}x ∈ \int(A) \setminus A'$ then there is an open set $U$ such that $U ∩ A = \{x\}$. Hence, $U ∩ \int(A) = \{x\}$ and $x$ is an isolated point of whole space $X$. But that is not possible if $X$ has no isolated points – which is the case with $\mathbb{R}$.
• how do we know there is an open set $U$ such that $U \cap A = \{x\}$? Commented Sep 3, 2015 at 7:39
• @user1292919: By the definition of the limit point: $x ∈ A'$ iff for every open $U$ such that $x ∈ U$ we have $U ∩ A \setminus \{x\} ≠ ∅$. Commented Sep 3, 2015 at 7:48