It says to prove that the closed ball in $\mathbb{R}^2$ with the Euclidean metric is this, but isn't that the definition of a closed ball?

Question

Prove that in $\mathbb{R}^2$

$$\overline{B_1((0,0))}=\{x\in \mathbb{R}^2:d_2(x,0) \leq 1\}$$

where

$$d_2 (x,y) = \sqrt{\sum_{i=1}^n(x_i-y_i)^2}$$

Answer 1

I will denote $\{x \in \mathbb{R}^2: d_2(x,0) \leq 1\}$ by $A$ in this answer.

By definition,

$$B_1((0,0))=\{x\in\mathbb{R}^2: d_2(x,0) < 1\}$$

To find its closure, let $a \in \overline{B_1((0,0))}$. Therefore, there exists a sequence $\{a_n\}_{n=1}^{\infty}$ in $B_1((0,0))$ such that $\lim_{n\to\infty}a_n=a$. For each $a_n$, we have that

$$d_2(a_n,0) < 1 $$

Taking limit from both sides and noting that $d_2(x,y)$ is continuous, we can write

$$\lim_{n\to\infty}d_2(a_n,0) =d_2(\lim_{n\to\infty}a_n,0)\leq 1 $$ $$d_2(a,0) \leq 1$$

This shows that $\overline{B_1((0,0))} \subseteq \{x \in \mathbb{R}^2: d_2(x,0) \leq 1\}=A$. Conversely, to prove the other direction, let $x \in A$, then either $d(x,0) < 1$ or $d(x,0) = 1$. The former case implies that $x \in B_1((0,0)) \subseteq \overline{B_1((0,0))}$.

In the later case, take $\{a_n\}_{n=1}^{\infty}$ to be the sequence $a_n=(1-\frac{1}{n})x$. You can easily see that $$\forall n\in\mathbb{N}: d_2(a_n,0)=(1-\frac{1}{n})d_2(x,0) \leq 1 - \frac{1}{n} < 1$$ Hence, $a_n \in B_1((0,0))$ and $\lim_{n\to\infty}a_n=x$. Proving that $A \subseteq \overline{B_1((0,0))}$. Q.E.D.

Answer 2

I believe the question is asking you to show that the closure of the open ball is the closed ball, which is not a trivial exercise.

This might not be the case all the time, for instance if you take $X$ to be any set with the discrete metric, then open ball around $x$ is a singleton $\{x \}$, which is a closed set, so its closure is the same.

The closed ball, however, is the whole of $X$.

However, in the euclidean metric space $\mathbb{R}^n$, it is true.