Show a unit disk with max metric is closed?

Question

I must show that if $\mathbb R$² is equipped with max metric,
d = (($x$₁, $x$₂),($y$₁, $y$₂)) = max{|$x$₁ - $y$₁| , |$x$₂ - $y$₂|}
then the disk
D = {($x$₁, $x$₂) ∈ $\mathbb R$² : $x$₁² + $x$₂² ≤ $1$ }
is a closed set.

My attempt (I'm just about two weeks of experience with any of this):

To show D is closed, I must show that it contains all of its limit points. So let L = ($x,y$) be a limit point of D. As such, it has some sequence S_n = ( $x$_n , $y$_n ) such that this sequence converges to L and all of its terms remain in D. Since all these terms remain in D, we can say $x$_n² + $y$_n² ≤ $1$ for all $n$. Then, by triangle inequality we have
($x$_n + $y$_n)² ≤ $1$

$x$_n² + $y$_n² ≤$ 1 $
Also, since as the limit of n goes to infinity, S_n $\to$ L , I think we can say
$(x+ y)$² ≤ $1$

$x$_n² + $y$_n² ≤ $1$

And that's where I'm completely stuck....assuming any of this is right at all.

Answer 1

An elaboration on Surb's comment

For clarity, I'll use $d_\infty((x_1,x_2),(y_1,y_2)) = \max\{|x_1 - y_1| , |x_2 - y_2|\}$, and $d_2((x_1,x_2),(y_1,y_2)) = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2}$. Observe that for any $(x_1,x_2),(y_1,y_2) \in \Bbb R^2$, $$d_\infty((x_1,x_2),(y_1,y_2)) \le d_2((x_1,x_2),(y_1,y_2)) \le \sqrt 2 d_\infty((x_1,x_2),(y_1,y_2)).$$ Thus we say that $d_\infty$ and $d_2$ are equivalent.

Problem

I type it once in $\rm \LaTeX$.

Show that $D := \{(x_1,x_2) \in \Bbb R^2 : x_1^2 + x_2^2 \le 1\}$ is closed in $(\Bbb R^2, d_\infty)$.

Since the OP's way is to construct a convergent sequence $(S_n):=(x_n,y_n)$ in $D$, and suppose that $S_n \to L:=(x,y)$ under $d_\infty$, I will show that $L \in D$ in $(\Bbb R^2,d_\infty)$.

Since $(S_n)$ is in $D$, $\forall n \in \Bbb N, x_n^2 + y_n^2 = d_2((0,0),(x_n,y_n)) \le 1$.

By $S_n \to L$ under $d_\infty$, for any $\epsilon > 0, \exists N \in \Bbb N$ such that $\forall n \ge N, d_\infty((x_n,y_n),(x,y)) < \dfrac{\epsilon}{\sqrt2}$. $$\forall n \ge N, d_2((x_n,y_n),(x,y)) \le \sqrt2 d_\infty((x_n,y_n),(x,y)) < \epsilon$$

Using the Triangle Inequality,

$$d_2((0,0),(x,y)) \le d_2((0,0),(x_n,y_n)) + d_2((x_n,y_n),(x,y)) \le 1 + \epsilon$$

Since the choice of $\epsilon$ is arbitrary, we conclude that $d_2((0,0),(x,y)) \le 1$. In other words, $x^2 + y^2 \le 1 \iff L \in D$. Hence $D$ is closed.

$\rm \LaTeX$ Source code