$(X,d)$ is a metric space. We fix a point, $a \in X$, and we let $A = \bigcap_{n\in\mathbb{N}} \left\{x: d(x,a) < r + \frac{1}{n} \right\} \in X$.
Is $A$ open or closed? If it is closed. What is the proof that it is closed?
Thank you for your time.
Kind regards,
Marius
$A$ is closed if $A^c$ is open. Remember that in a metric space, a set is open if for each point $y$ there is an $\epsilon$ such that the $\epsilon$-ball around $y$ lies completely in the set.
So let $y$ be any point in $A^c$. There is an $n_0 \in \mathbb{N}$ such that $d(y,a) \geq r + \frac{1}{n_0}$. For any point $x \in A$ it holds that $d(x,a) < r+ \frac{1}{2n_0}$. With $\epsilon := \frac{1}{2 n_0}$, you see from the triangle inequality that $d(y,a) \leq d(x,a) + d(y,x)$, from which $$d(y,x) \geq d(y,a) - d(x,a) > r + \frac{1}{n_0} - r - \frac{1}{2 n_0} = \epsilon$$ follows. Therefore, the $\epsilon$-ball $K_\epsilon := \{y' \in X: d(y',y) < \epsilon\}$ lies completely in $A^c$, which means that $A^c$ is open and $A$ is closed.
$\forall_n d(x,a) < r + \frac{1}{n}$ means the same thing as $d(x,a) \le r$.
If the right hand side holds, certainly the left side does, and if the right hand side does not hold, then $d(x,a) > r$ and so $d(x,a) - r > 0$, and for some $k$, $d(x,a) - r > \frac{1}{k}$. But then $d(x,a) > r + \frac{1}{k}$ and so the left hand side does not hold either.
So $A = \{x: d(x,a) \le r \}$. Show that a set of this form is closed in any metric space. Then find a (natural) example where a set of this form is not open.
