# interior points and limit points of $\{(x,y): x^2+2y^2 < 1\}$

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.

• Do you mean $(x,y)\in \mathbb{Q}\times \mathbb{Q}$? Apr 5, 2017 at 9:53
• @DietrichBurde oops i forgot it. $(x,y) \in \mathbb R^2$ Apr 5, 2017 at 10:17
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 $(x, y) ∈ \overline{\{(tx, ty): t ∈ [0, 1)\}} ⊆ \overline{A}$.
• @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? Apr 5, 2017 at 17:44