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I am considering the set $S'$, which extends the original problem to allow non-negative integer exponents: $$ S'=\{(x,y,z)\in \mathbb{R}^3:\exists n_1,n_2,n_3\in\mathbb{N}_{\ge 0},\ x^{n_1}+y^{n_2}=z^{n_3}\} $$ (with the standard convention that $0^0$ is undefined).

Is the complement of the closure, $(\overline{S'})^c$, exactly equal to the following set?

$$ \{(x,y,z) \in \mathbb{R}^3 : |z|<1 \text{ and } ( (x > 1 \text{ and } y > 0) \text{ or } (y > 1 \text{ and } x > 0) ) \} $$

This region excludes the "mixed sign" or "negative/negative" areas where subtraction of powers allows the equation to be solved.

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2 Answers 2

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No. The open set $(0,0.5)^2\times (2,+\infty)$ is disjoint from $S'$.

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To refine the statement covering the complement of the closure of $S'$, we must combine the "Large Base" region with the "Small Base" region (identified by the $(0, 0.5)^2$ counterexample).

$$(\overline{S'})^c⊋\left\{ (x,y,z) \in \mathbb{R}^3 : \underbrace{\left( |z| < 1 \land \big( (x > 1 \land y > 0) \lor (y > 1 \land x > 0) \big) \right)}_{\text{Region A: Large Bases}} \;\lor\; \underbrace{\left( |z| > 2 \land |x| + |y| < 1 \land x \neq 0 \land y \neq 0 \right)}_{\text{Region B: Small Bases}} \right\}$$

The strictness is witnessed by $(5,−0.5,0.1)$.

You can’t honestly turn that inclusion into a nontrivial equality with that kind of simple RHS.

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