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My question is:

Given the set $A = \{(x, y) \in \mathbb{R}^2: 1 < x^2y^3 < 2\}$, is $A$ an open or closed set and is it also bounded or unbounded?

Even though I am semi-confident in this topic, I'm really not sure how to answer these 2 questions given $A$, so any help will be appreciated.

3 Answers 3

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Let $f(x,y)=x^2y^3$, which is continuous, since it's a polynomial. Then $$A=\{(x,y) : f(x,y) \in (1,2)\}= f^{-1}(1,2)$$

Recall that the preimage of an open set by a continuous function is also open.

Now, for boundedness, try to think if it's possible to find $x,y\in \mathbb R$ so that $x$ is as big as we want, and $x^2y^3 \in (1,2)$. In this case, it's easy to see you can. For example, if we fix $M>0$, and choose

$$x=\sqrt{\frac32M}, \quad y=\frac{1}{\sqrt[3]{M}}$$

then $$f(x,y)=x^2y^3=\frac32M \cdot \frac{1}{M}=\frac32 \in A$$

As we can let $M$ be as big as we want, $A$ is not bounded.

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    Ah I see now, thank you for that. But how would you prove if the set was bounded or unbounded? – 2017-02-20
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Hints:

  1. The inverse image of an open set under a continuous function is open.
  2. For any nonzero $x$, you can find a $y$...
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$f(x,y)=x^2y^3-1, g(x,y)=x^2y^3-2$ are continue, $U=f^{-1}((1,+\infty))$ and $U=g^{-1}((-\infty,2))$ are open and $U\cap V$ is open $x^2n^3=1.5, n>0$ integer, implies $x={1.5\over n^3}$ so $({1.5\over n^3},n)\in A$ and $A$ is not bounded.

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    Ah this makes sense. Does this mean also that $A$ is sequentially compact? – 2017-02-20