i want to prove that $$A=\{(x,y)\in\mathbb{R}^2, x^2+y^2\leq1\}$$ is bounded

my idea is to prove that $diam(A)<\infty$

$$ diam(A)=\sup_{X,Y\in A} d_2(X,Y)=\sup_{X,Y\in A} \sqrt{(x_0-y_0)^2+(x_1-y_1)^2} $$

how to do with $-2(x_0y_0+x_1y_1)$

Thank you

  • 1
    $\begingroup$ What is your definition of bounded set? $\endgroup$ Sep 9 '18 at 20:51
  • $\begingroup$ the diametr is finite $\endgroup$
    – user523857
    Sep 9 '18 at 20:52
  • $\begingroup$ I think it's pretty much obvious that this set is bounded. The Euclidean norm of a vector in $A$ can't be bigger than $1$. So unless you use a different definition for a bounded set I don't see a reason to work with $diam(A)$. $\endgroup$
    – Mark
    Sep 9 '18 at 20:52
  • $\begingroup$ i need an other prove using diameter @Mark $\endgroup$
    – user523857
    Sep 9 '18 at 20:55
  • $\begingroup$ You can equivalently define A to be $ A = \{x \in \mathbb{R}^2 : d_2(x, 0) \leq 1\} $. This definition yields that $ diam(A) \leq 2 $ by the triangle inequality. $\endgroup$ Sep 9 '18 at 21:08

If $(x,y),(x',y')\in A$ then: $$d((x,y),(x',y'))\leq d((x,y),(0,0))+d((0,0),(x',y'))\leq 1+1=2$$

So: $$\text{diam}(A)=\sup\{d((x,y),(x',y'))\mid (x,y),(x',y')\in A\}\leq2$$

If possible then put the definition based on "diameter" aside.

A set $B\subseteq\mathbb R^2$ is by definition bounded if $B\subseteq B((0,0),r)$ for some $r<\infty$.

That works much more handy and is equivalent with the definition based on diameter.

Here $B((0,0),r)$ is an open ball with $(0,0)$ as center and $r$ as radius.


We have $(x,y)\in A\iff \|(x,y)\|\le 1$, where $\|(x,y)\|=\sqrt{x^2+y^2}$ is the euclidean norm. Hence $A$ is trivially bounded. Recall that a set $D$ is bounded in a normed space iff there exists $M>0$ s.t. $\|x\|\le M$ for any $x\in D$.

  • $\begingroup$ i need an other proof with the diameter $\endgroup$
    – user523857
    Sep 9 '18 at 21:01
  • $\begingroup$ please do you have an idea with the diameter? $\endgroup$
    – user523857
    Sep 9 '18 at 21:07
  • $\begingroup$ In the metric space the diameter of the ball is not greater than twice the radius. This is enough for your proof. Nevertheless, in the euclidean metric the diameter is precisely twice the radius. $\endgroup$
    – szw1710
    Sep 9 '18 at 21:15

Let $(x,y), (x',y') \in A$.

By triangle inequality $d((x,y),(x',y') \le d((x,y),(0,0)) + d((x',y'),(0,0))=$

$\sqrt{x^2 + y^2} + \sqrt{x'^2 + y'^2} \le$

$\sqrt 1 + \sqrt 1 = 2$.

So $diam(A) \le 2$.


Or to continue what you started:

$diam(A)=\sup_{X,Y\in A} d_2(X,Y)=\sup_{X,Y\in A} \sqrt{(x_0-y_0)^2+(x_1-y_1)^2}$

$= \sqrt {x_0^2 - 2x_0y_0 + y_0^2 + x_1^2 - 2x_1y_1 + y_1^2} \le$

$\sqrt {x_0^2 + 2|x_0y_0| + y_0^2 + x_1^2 + 2|x_1y_1| + y_1^2} \le$

$\sqrt {(|x_0| + |y_0|)^2 + (|x_1| + |y_1|)^2}$.

Note: If $(a,b) \in A$ then $|a| = \sqrt {a^2} \le \sqrt{a^2 + b^2} \le \sqrt 1 =1$ so

$\sqrt {(|x_0| + |y_0|)^2 + (|x_1| + |y_1|)^2}\le \sqrt { (1+ 1)^2+(1+1)^2} =\sqrt 3 = 2\sqrt 2$

Not as efficient but it'll get you there.


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