Prove that $A$ is closed

Question

I am asked to prove the following question:

Let "If $a \in \mathbb{R}$, $a_n \in A$ for all $n \in \mathbb{N}^+$, and $a_n \to a$, then $a \in A$." be a statement $P$. Show that if $P$ is true then $A$ is closed.

I was given a hint that if $a \notin \text{int}(\mathbb{R} \setminus A)$, then for all $n \in \mathbb{N}^+$, there exists $a_n \in (a - 1/n, a+1/n) \cap A$.

But I can't get my head around what is the hint trying to tell me and I don't know how to proceed. Can anyone give a hint? Thanks a lot!

Answer 1

I finally understood the question so here is an attempt at the proof.

Suppose $a \in \textbf{cl}A$, we have that $a \notin \textbf{int}(\mathbb{R} \setminus A)$. Thus, $a$ is not an interior point of $(\mathbb{R}\setminus A)$ and then we have for all $n \in \mathbb{N}^+$, $(a - 1/n, a + 1/n) \nsubseteq \mathbb{R}\setminus A$. This is equivalent to for all $n \in \mathbb{N}^+$, $(a - 1/n, a + 1/n) \cap A \neq \emptyset$. Hence, we let $a_n \in (a - 1/n, a + 1/n) \cap A$. Thus, $a_n \in A$ for all $n \in \mathbb{N}^+$. Moreoever, $a_n \to a$ since $\cap_{n = 1}^\infty(a- 1/n, a+ 1/n) = \{a\}$. Therefore, $a \in A$ by $P$. Hence, $\textbf{cl}A \subset A$ and $\textbf{cl}A = A$ since $A \subset \textbf{cl}A$. Therefore, $A$ is closed since $\textbf{cl}A$ is closed.

Answer 2

First of all, I think it's good to clarify that your definition of closed is as follows :

$\textbf{Definition}$: A point $a$ is said to be adherent to the set $A \subseteq \mathbb{R}$ if for any $\epsilon >0$ we have to : $$ (a-\epsilon,a+\epsilon)\cap A \neq \emptyset $$ Denote as $\overline{A}$ the set of adherent points to the set $A$. We will say that $A$ is closed if $\overline{A}=A$

$\textbf{Theorem}$: A point $a$ is adherent to the set $A \subseteq \mathbb{R}$ if and only if exists $\{x_n\}\subseteq A$ such that $x_n \rightarrow a$. Or in other words : $$ a \in \overline{A} \Leftrightarrow \exists \{x_n \} \subseteq A / x_n \rightarrow a $$

$\textbf{Proof}$: $(\Rightarrow)$ By definition if we consider $\epsilon=1$, exists $x_1 \in (a-1,a+1)\cap A$. For $\epsilon=\dfrac{1}{2}$ exists another $x_2 \in (a- \dfrac{1}{2},a+\dfrac{1}{2})\cap A$. If we reason by induction, then for $\epsilon=\dfrac{1}{n}$ exists $x_n\in (a- \dfrac{1}{n},a+\dfrac{1}{n})\cap A$ and so on successively. So, we get $\{x_n\} \subseteq A$ such that $\vert x_n - a \vert < \dfrac{1}{n}$ in other words $x_n \rightarrow a$ (why?).

$(\Leftarrow)$ Let $\epsilon >0$ by the definition of limit exists $n_o \in \mathbb{N}$ such that for $n>n_o$ we have $\vert x_n - a \vert < \epsilon$. Is clear that $x_n \in (a-\epsilon,a+\epsilon)\cap A,\forall n>n_o$. So $(a-\epsilon,a+\epsilon)\cap A \neq \emptyset$

With all this, the statement $\textbf{P}$ means that given any $a\in \overline{A} \implies a \in A$ so $\overline{A} \subseteq A$. The other inclusion is obvious, because given any $a \in A$ it is enough to take the constant sequence $x_n = a \subseteq A$ that converges to $a$.