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I want to show that the set $U=\{\left(x, y\right)\in\mathbb R^2 \ | \ \left(x-1\right)^2 + \left(y-1\right)^2 <3\}$ contains the set $S=\{\left(x, y\right)\in\mathbb R^2 \ | \ x \geq1; \ y\geq0; \ x^2+y^2 \leq 4\}$

It is easy to see that $U$ contains $S$ graphically. But I don't know how to show it mathematically. If I can get a little hint, I will appreciate it.

Tarek Badr
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2 Answers2

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Hint

$U$ and $S$ are connected (and convex) subsets of the plane. To prove that $S \subseteq U$, it is enough to prove that:

  1. One point of $S$, i.e. $(1,0)$ belongs to $U$.
  2. And that the boundary of $S$ doesn't intersect the boundary of $U$.

The boundary of $U$ is the circle centered on the origin with radius $\sqrt 3$. And the boundary of $S$ is a portion of a circle plus two line segments.

mathcounterexamples.net
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  • Justifying that description of the boundary of $S$ without resorting to geometric intuition is going to to be some work in itself, though. I think what is really needed is to challenge the OP's premise that a graphical argument is not "mathematical". – Troposphere Oct 09 '21 at 15:11
  • @Troposphere I understand what you mean but only partially agree. Observing the inclusion of a disk into another one can be done graphically. it can also be proven by looking at the position of the centers and the radii. – mathcounterexamples.net Oct 09 '21 at 15:31
  • So by showing that the boundary $\partial S \in U$, then we can conclude that $S \subseteq U$ – Tarek Badr Oct 09 '21 at 15:50
  • Yes because the boundary of $U$ being a circle, the boundary of $S$ should intersect it otherwise. – mathcounterexamples.net Oct 10 '21 at 07:08
  • Please have a look at the other answer. – Arctic Char Dec 31 '21 at 20:09
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$U$ and $S$ are connected (and convex) subsets of the plane. To prove that $S \subseteq U$ or $U \subseteq S$, you need to:

  1. Find one point, i.e. $(1,0)$, that belongs to both of $S$ and $U$.
  2. Prove that the boundary of $S$ doesn't intersect the boundary of $U$.

In addition, to prove specifically that $S \subseteq U$ you need to:

  1. Find one point of $U$ that does not belong to $S$.
branco
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  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Dec 31 '21 at 19:36
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    I am just trying to correct what is wrong with the accepted answer: the poof outlined only proves that $S$ belongs to $U$ or $U$ belongs to $S$, not specifically that $S$ belongs to $U$. And the move to delete the post I find proposterous since I am following the exact framework of the accepted answer which was not proposed for deletion. – branco Dec 31 '21 at 19:50
  • You know, I believe my post is quite useful since it is adding math insight to the discussion that would have otherwise been lost considering I cannot comment on the accepted answer to point what is wrong with my current reputation score. – branco Dec 31 '21 at 20:38