Metric space and closed sets Let $\mathbb{R}^n$ have a Euclidian metric $d$ and $\emptyset \neq A \subseteq \mathbb{R}^n$ be a subset such that $(A, d_{|A \times A})$ is a metric space.
(a) Let $x \in A$ and consider the set $D := \{y \in A: d(x, y) \geq r\}$. Is $D$ closed in $A$?
(b) Let $x \in A$ and consider the set $E := \{y \in A: d(x, y) > r\}$. Is $D$ contained in the closure of $E$ in $A$?
Justify your answers.
I am reviewing solutions to various problems for an upcoming examination and this was a problem in a practice set but there are no solutions available. Any assistance with a solution is much appreciated.
 A: $D$ is closed in $A$. If $r = 0$, then $D = A$ which is closed by default. If $r > 0$, then we can prove $D$ is closed as follows. If $y_0 \in A$ is a limit point in $D$, then there is a sequence $(y_n)$ in $D$ which converges to $x_0$. If $\varepsilon > 0$ is small enough that $\varepsilon < r$, then we can choose sufficiently large $N$ so that $n \geq N$ implies that $d(y_n, y_0) < \varepsilon$. The condition on $\varepsilon$ ensures that $d(x,y_n) - d(y_n, y_0) > r - \varepsilon > 0$ so that the absolute value can be removed. Hence, by the reverse triangle inequality $$d(x, y_0) \geq |d(x, y_n) - d(y_n, y_0)| = d(x,y_n) - d(y_n, y_0) > r - \varepsilon$$ which proves that $d(x, y_0) \geq r$ so that $y_0 \in D$ (since $\varepsilon$ can be made arbitrarily small). Thus $D$ contains all its limit points so it is closed.
Since $D$ is closed, it is its own closure. Moreover, if $y \in A$ so that $d(x,y) > r$, then in particular $d(x,y) \geq r$ so that $y \in D$. Thus $A \subseteq D$ which proves (b).
A: Let $x$ be in $A$. We have
$D^c=\{y\in A, d(x,y)<r\}=B(x,r)$
$D^c$ is just the open ball having $x$ as a center and $r$ as a radius. It's an open set so its complementary is closed. Thus $D$ is closed.
Suppose now that $y\in E$, then by definition :
$$y\in E \implies d(x,y)>r$$
Logically speaking, if $d(x,y)>r$ then $d(x,y)\geq r$
So $y\in E \implies y\in D$
And since $D$ is a closed set, we have the following property :
$$ D \text{  is a closed set} \iff \bar{D}=D$$
So D is a closed set, and it's equal to its closure, and since $y\in D$ then
$$y\in \bar{D}$$
