Contrapositive of statement regarding open sets

Question

I am taking a Real Analysis class and i am having problems with the following:

"A set $A \subset \mathbb{R}$ is open $\Leftrightarrow$ the following holds: if a sequence $\left(x_{n}\right)$ converges to a point $a \in A$, then $x_{n} \in A$ for $n \in \mathbb{N}$ sufficiently large"

I am trying to prove $(\Leftarrow)$. I think i can prove the contrapositive statement. However, i am not sure how to properly write the statement in the contrapositive form.

Attempt 1 to write the contrapositive:
If $A \in \mathbb{R}$ is not open, then there exists a sequence $\left(x_{n}\right)$ such that $\lim x_{n}=a$ and $\mathrm{x}_{\mathrm{n}} \notin \mathrm{A}$.

Attempt 2 to write the contrapositive:
If $A \in \mathbb{R}$ is not open, then there exists a sequence $\left(x_{n}\right)$ such that $\lim x_{n}$ is not equal to $a$ and $\mathrm{x}_{\mathrm{n}} \notin \mathrm{A}$.

Why should (or should not) the statement $\lim x_{n}=a$ be negated?

Thanks in advance!

Answer 1

Neither of them is quite right, even apart from the use of $\in$ instead of $\subseteq$ at the beginning of each of them. A correct statement of the contrapositive is:

If $A\subseteq\Bbb R$ is not open, then there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $\Bbb R$ that converges to a point $a\in A$, but for each $m\in\Bbb N$ there is an $n\ge m$ such that $x_n\notin A$.

Equivalently, one could say:

If $A\subseteq\Bbb R$ is not open, then there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $\Bbb R$ that converges to a point $a\in A$ but has infinitely many terms that are not in $A$.