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Let $A = \{(x,y): 5x+4y \leq 20, x \geq 2, y \geq 1\}$. With the usual Euclidean metric $d$, prove that $A$ is bounded and closed on $(X,d)$. Is $A$ compact?

My attempt: $A = \{(x,y): 2 \leq x \leq \frac{16}{5} , 1 \leq y \leq \frac{5}{2} \}$. So let $r_0 = \sqrt{(\frac{16}{5})^2 + (\frac{5}{2})^2 }$, choose $r> r_0$, then the open ball S with center $(0,0)$ and radius $r$ always contains $A$. Thus $A$ is bounded.

$\mathbb{R}^2 \backslash A = \{(x,y): x <2 \ \text{or} \ x> \frac{16}{5} , y <1 \ \text{or} \ y >\frac{5}{2} \}$. Since this set is open, $A$ is closed.

$A$ is compact because $\mathbb{R}^2$ is a complete space, then any bounded and closed set is compact.

Is the proof correct and rigorous enough?

dunguyen
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  • Wrong! The last part of your proof is wrong. In a complete metric space a closed and bounded set need not be compact. Instead of bringing in completeness you have to quote the well-known result that closed and bounded subsets of $\mathbb R^{n}$ are compact. – Kavi Rama Murthy May 03 '18 at 05:38
  • @астонвіллаолофмэллбэрг A metric space is compact iff it is complete and totally bounded...so I don't believe such a proof exists. – Math1000 May 03 '18 at 06:09
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    @Math1000 You are right. I phrased it incorrectly. I will remove the previous comment. – Sarvesh Ravichandran Iyer May 03 '18 at 06:16

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