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Let $A$ be the subset of points $(x,y) \in \mathbb{R} \times \mathbb{R}$ such that $x^3+y^3=1$. Is $A$ compact?

I think it should be, as $x$, $y$ can't have values greater than $1$. So $A$ is bounded. Also it is closed. So, compact. But answer key says it is not compact.

Najib Idrissi
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aarbee
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2 Answers2

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I think it should be, as $x$, $y$ can't have values greater than $1$. So $A$ is bounded.

Odd powers can be negative, so this doesn't hold. For example: take $x=2$, then $y = \sqrt[3]{1-2^3}$.

Note that your reasoning would work for e.g. $x^2+y^2=1$.

StackTD
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It's not bounded. $x$ or $y$ can be anything greater than $1$. For example, if $x=10$ then $y=\sqrt[3]{1-x^3} = (-999)^{1/3}$.

Try plotting $x^3+y^3=1$ in a graphing calculator or desmos.com, etc., for visual confirmation.

  • Would it not give us complex values? – aarbee Nov 24 '16 at 14:05
  • @Ramit, no. (Removed my previous comment for simplicity.) You can always take the cube root of any real number and get a real number as a result. For example, the real cube root of $-8$ is $-2$. –  Nov 24 '16 at 14:10