Clean Proof that a set is bounded in multiple dimension

Question

Say $M:=\{(x,y)\in \mathbb R^{2}:x^2+y^2\leq 1\}$

I am confused on how to prove the a multidimensional set is bounded. This is important in order to be able to prove that $M$ is indeed compact. It one dimension it was rather simple, say $M$ was in one dimension in order to show it was bounded, simply find an $c \in \mathbb R$, such that $\forall a \in M: |a|\leq c$ but this obviously won't work in multiple dimensions:

Therefore my question is: What is a clean way of proving $M$ as a multidimensional set is bounded.

My idea: Let $(x,y) \in M$, it follows that: $\vert \vert (x,y) \vert \vert=\sqrt{x^2+y^2}\leq\sqrt{1}<2$

And therefore $M$ is bounded?

Is there a better way of formulating this?

Answer 1

Just a few thoughts. You have described the set $\mathbf{M}$ which is nothing more than the unit disk on the plane as a subset of $\mathbb{R}^2$. Your very definition of $\mathbf{M}$ implies it is bounded since you wrote the condition that $x^2+y^2\leq 1$. Is that what you meant?.

Answer 2

For a non-empty metric space $(X,d)$ and any $Y\subset X,$ we say that $Y$ is bounded (with respect to $d$) iff there exists $p\in X$ and $r>0$ such that $\forall q\in Y\;(d(p,q)\leq r).$

Equivalently, if $Y$ is not empty, that there exists $q_1\in Y$ and $r_1>0$ such that $\forall q\in Y\;(d(q_1,q)\leq r_1).$

To prove these 2 conditions are equivalent for non-empty $Y$:

(i). If the 2nd condition holds, let $p=q_1$ and let $r=r_1,$ and the 1st condition is seen to hold.

(ii). If the 1st condition holds, choose any $q_1\in Y$ and let $r_1=2r.$ Then $\forall q\in Y\;(d(q_1,q)\leq d(q_1,p)+d(p,q)\leq r+r=r_1)$ so the 2nd condition holds.

For your Q with the metric $d((x,y),(x',y'))=\sqrt {(x-x')^2+(y-y')^2}\;$ let $p=(0,0)$ and $r=1.$

Or, if you prefer, let $p=(\pi,42)$ and $r=10^{100}.$