How do I show $\{(x,y) \in R^2 : x^3 + y^2 < 1 \}$

Here is what I have come up with.

We can have $x^3 + y^2 = 0$ by simply setting $x = 0$ and $y = 0$. Hence, it is bounded at $0$. So, now we want to show upper bound is closed as well.

We know for some number $z \in R$ that if $z < 1$ then $z^2 \leq z$. Hence, for some $y,x < 1$ then $y + x \geq x^3 + y^2$.

We want to show that for every $x,y$ there exists some $\epsilon$, s.t. $x + y + \epsilon < 1$. We simply just pick $\epsilon = 1 / (2(x + y))$. Hence, because we can always find a small enough $\epsilon$, the upper bound must be open.

Hence, $[0,1)$, so it is clopen and thus not closed.

  • $\begingroup$ Why does $x^3+y^2=0$ iff $x=0$ and $y=0$? Certainly $x=-1$ and $y=1$ also satisfies the equality. Also, what does "bounded at $0$" mean? This is describing a region in $\mathbb{R}^2$ (two-dimensions), but your language seems to mimic one-dimension. This question appears to be very confused... $\endgroup$ – Michael Burr Jun 2 '16 at 19:03
  • 1
    $\begingroup$ One approach is to use the fact that $f:{\mathbb R}^2 \rightarrow \mathbb R$ defined by $f(x,y)=x^3 + y^2$ is continuous and $(-\infty,1)$ is an open set in $\mathbb R$ along with the knowledge of which sets in ${\mathbb R}^2$ are both open and closed. However, this depends on results you may not yet have covered. $\endgroup$ – Dave L. Renfro Jun 2 '16 at 19:07
  • $\begingroup$ ah, don't know why i wrote that, thanks. $\endgroup$ – Fredrik Jun 2 '16 at 19:07

Let $E=\{(x,y) \in R^2 : x^3 + y^2 < 1 \}$.Then $E=f^{-1}(-\infty,1)$, where $f:\mathbb{R^2}\mapsto \mathbb{R},(x,y)\mapsto x^3+y^2$. Since $f$ is continuous and $(-\infty,1)$ is open, $E$ is open!


Tip : Let $S$ be a subset of a metric space. A set $S$ is closed if

  1. The complement of $S$ is an open set,

  2. $S$ is its own set closure,

  3. Sequences/nets/filters in $S$ that converge do so within $S$,

  4. Every point outside $S$ has a neighborhood disjoint from $S$.

The point-set topological definition of a closed set is a set which contains all of its limit points. Therefore, a closed set $C$ is one for which, whatever point $x$ is picked outside of $C, x$ can always be isolated in some open set which doesn't touch $C$.

I'm sure that the above would help you pick the correct equivalent definition of an open set and reach your solution.


By the way, "clopen" means open and closed.

If you are allowed to use the "closed if and only if it contains its limit points" definition, then the following argument shows it is not closed: consider the points $(0,1-1/n)$. Each lies in your set, but their limit $(0,1)$ does not.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.