Find out all interior points and limit points of $ A =\{(x,y): x^2+2y^2 < 1\}$

I understand this problem graphically, but i'm not quite sure how to prove the answer rigorously using mathematical word.

My try : Let k be an element of A, and suppose t is an element of $\{(x,y):x^2+2y^2 =1\}$, which makes minimum $||t-k||$. Then $N(k,\delta)\subset A$ when $\delta = \||t-k||$.

I suppose this will work, but I want more clear and detailed proof.

Thank you in advance.

  • $\begingroup$ Do you mean $(x,y)\in \mathbb{Q}\times \mathbb{Q}$? $\endgroup$ Apr 5, 2017 at 9:53
  • $\begingroup$ @DietrichBurde oops i forgot it. $(x,y) \in \mathbb R^2$ $\endgroup$
    – user432019
    Apr 5, 2017 at 10:17

1 Answer 1


The easiest way to show that $A$ is open is probably noticing that $A$ is the preimage of the open interval $(1, ∞)$ with respect to the continuous mapping $(x, y) ↦ x^2 + 2y^2$. And similarly, using the preimage of $[1, ∞)$, every boundary point of $A$ lies in $\{(x, y): x^2 + 2y^2 = 1\}$. For the other inclusion $(x, y) ∈ \overline{\{(tx, ty): t ∈ [0, 1)\}} ⊆ \overline{A}$.

  • $\begingroup$ Thanks for the answer, but as a starter, this is somewhat hard to understand for me. Would you please explain more? $\endgroup$
    – user432019
    Apr 5, 2017 at 10:56
  • $\begingroup$ @user432019: In the first part, I claim that every point of $A$ is an interior point of $A$, i.e. $A$ is an open set. And I prove it by using the fact that a mapping is continuous if and only if every preimage of an open set is open… Does it make sence for you? Or which part or notion you don't understand? $\endgroup$
    – user87690
    Apr 5, 2017 at 17:44
  • $\begingroup$ Thanks for additional explanation. Now i started to understand what you mean. I'll try it for myself. $\endgroup$
    – user432019
    Apr 6, 2017 at 3:19

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