Prove that if a subset of $\mathbb{R}^2 $ is convex closed and has non empty interior they are equal to the closure of their interior

Question

This question was asked here 5 years ago Here but there was a typo in the body that resulted in the question not being answered. There was a comment that linked to another different question but I didn't get the solution for the one that I am stuck on

The question is as follows:

Suppose that $A \subset \mathbb{R}^2$ is convex, closed and has nonempty interior. Prove that A is the closure of it's interior

I drew a picture and was able to clearly able to see that by convexity and closure if $x \in \text{int} A$ there is a line segment that connects it to every point in the closure of A but I'm not sure if that's relevant or if I'm on the right track. Can someone give me a nudge in the right direction?

Answer 1

Let $A\subset {\mathbb R}^2$ be a convex, closed set with nonempty interior. Let $y\in A$ and let $x \in {\rm int}\,A$. We will show that $y$ is in the closure of $\rm{int}\, A$.

Let the ball of radius $\epsilon$ centered at $x$ be in $A$, $B(x, \epsilon)\subset A$. Such $\epsilon > 0$ exists since $x$ in the interior.

Now consider $z_t = (1-t)y + tx$ for $t\in (0,1)$. We will show show that $z_t\in {\rm int}\, A$ and thus conclude that $y$ is in the closure of the interior of $A$.

To see that $z_t\in {\rm int}\,A$, note that $B(z_t, \epsilon t)\subset A$. Why? Any arbitrary point in $B(z_t, \epsilon t)$ can be written as $z_t + \epsilon t v$ where $v$ is a vector with magnitude smaller than one. Thus, $$ z_t + \epsilon t v = (1-t) y + t (x + \epsilon v) $$

Since $x+\epsilon v \in B(x, \epsilon) \subset A$, convexity implies that $z_t + \epsilon t v \in A$. So $B(z_t, \epsilon t)\subset A$ and thus $z_t$ is in the interior of $A$ for all $t\in(0,1)$. Thus $y$ (which was an arbitrary point in $A$) is in the closure of the interior of $A$.