My Question is:

With D = {$x = (x_1, . . . , x_d) ∈ \mathbb{R}^d: 5 ≤ {x^4}_j ≤ 6, j = 1, . . . , d$} Determine if the set D is open or closed and also if its bounded or unbounded.

The way this set is presented on paper has got me confused and cant seem to determine the 2 questions ive asked. Any help will be appreciated.

  • 1
    $\begingroup$ This doesn't seem to have anything to do with complex numbers. $\endgroup$ – Fimpellizieri Feb 21 '17 at 3:45
  • $\begingroup$ Please stop tagging your questions with complex-analysis and several-complex-analysis. $\endgroup$ – mrf Feb 21 '17 at 8:02

$D$ is the set of points $x=(x_1,\dots, x_d)$ in $\mathbb{R}^d$ such that each coordiante $x_i$ of $x$ satisfies

$$5\leq (x_i)^4\leq 6,$$

that is, the fourth power of each coordinate lies between $5$ and $6$.

$D$ is both closed and bounded.

It is closed because the inequalities in its definition are 'less than or equal' (rather than simply 'less than'). More precisely, let $F:\mathbb{R}^d\longrightarrow \mathbb{R}^d$ be given by $F(x_1,\dots,x_d)=({x_1}^4,\dots,{x_d}^4)$. Then $F$ is continuous and $D=F^{-1}(S)$, where $S={[5,6]}^d$. In otherwords, $D$ is the preimage of a closed set by a continuous function, and hence is closed.

It is bounded because each coordinate is bounded. Indeed, it's easy to see that every $x_i$ satisfies $-\sqrt[4]{6}\leq x_i\leq \sqrt[4]{6}$.


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