I have a question regarding the definition of a bounded set in $\mathbb{R}^n$ here:
Specifically, the definition in the link says that "A set in $\mathbb{R}^n$ is bounded iff it is contained inside some ball $x_1^2+...+x_n^2 \le R^2$ of finite radius $R$".
Is my following understanding (paraphrasing) of this definition correct: A set in $\mathbb{R}^n$ is bounded iff there exists a $R>0$ such that for all $\mathbf{x} =(x_1, \cdots, x_n) \in A$, it follows that $d(\mathbf{x}, 0) < R$ where $0 = (0, \cdots, 0)$ and $d$ is the standard Euclidean metric on $\mathbb{R}^n$.
If my understanding is correct, then a further question I have is why are we forcing the center of the "ball" to be at $0$? In the general definition of a bounded set in a metric space, we need to pick a point in the metric space to have the ball centered upon (not necessarily $0$), so why in $\mathbb{R}^n$ we can just pick $0$?
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$\begingroup$Your paraphrasing is correct. And it turns out that it doesn't matter where we pick our center, because of the triangle inequality. If $R_0$ is a radius which works for $0$ being the center, and you have some point $\mathbf y$ that you like, then $R_{\mathbf y} = R_0 + d(\mathbf y, 0)$ is a radius which works for $\mathbf y$ being the center.
Going the other way, if $R_{\mathbf y}$ is a radius which works for $\mathbf y$ being the center, then $R_0 = R_{\mathbf y} + d(\mathbf y, 0)$ is a radius which works for $0$ being the center.
In summary, whichever definition you're actually using, you can use this to prove that the other definition agrees on whether some working radius exists. Using $0$ as the center has fewer freedoms to handle, so it's usually less messy. However, some times your set is defined in such a way that it has a natural center other than $0$, and in that case you could use that instead.
$\endgroup$ $\begingroup$In a metric space there is no $"0"$. Let $A$ be a set in $ \mathbb R^n$. Then the following statements are eqivalent:
$(1)$ there is $R>0$ such that $d(x,0) \le R$ for all $x \in A$;
$(2)$ there is $r>0$ and $x_0 \in \mathbb R^n$ such that $d(x,x_0) \le r$ for all $x \in A$.
Try a proof !
Statement $(1)$ is easier to handle !
$\endgroup$ $\begingroup$Your understanding is correct. Also, it is totally okay to choose an arbitrary centre for the ball, say $y$, because if a set is within $R$ of $y$, then it must be within $R+d(0,y)$ of $0$ by the triangle inequality.
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