Is the following subset of $\mathbb R^2$, endowed with the usual distance and topology, compact? $A_a = \{(x,y)\mid x^4 + y^8+e^{xy} \leq 4 \}$

Question

Is the following subset of $\mathbb R^2$, endowed with the usual distance and topology, compact?

$$A_a = \{(x,y)\mid x^4 + y^8+e^{xy} \leq 4 \}$$

$x^4 \leq x^4 + y^8+e^{xy} \leq 4$ hence, $|x|^4\leq {4}$ ...hence $x = \sqrt{2}=1.414$

$y^8 \leq x^4 + y^8+e^{xy} \leq 4$ hence, $|y|^8\leq {4}$ ...hence $y = \sqrt{\sqrt{2}}=1.189$

$e^{xy} \leq x^4 + y^8+e^{xy} \leq 4$ hence, $|xy|\leq \ln{4}$ ...hence $xy = 1.386$

However this does not hold?? Am I approaching this correctly?

score 0 · Accepted Answer · answered Aug 08 '20 at 15:08

Let $(x,y)\in A$, If $|x|>1,|y|>1$, $\|(x,y\|^2=x^2+y^2\leq x^4+y^8+e^{xy}\leq 4$ implies $A$ is bounded.

Let $f(x,y)=x^4+x^8+e^{xy}$, $f$ is continuous. Let $z=lim(x_n,y_n), (x_n,y_n)\in A$, $f(z)=lim_nf(x_n,y_n)\leq 4$ implies $z\in A$ and $A$ is closed, therefore compact by Bolzano Weirstrass.

Is the following subset of $\mathbb R^2$, endowed with the usual distance and topology, compact? $A_a = \{(x,y)\mid x^4 + y^8+e^{xy} \leq 4 \}$

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