Interior and closed cover properties of the convex normed subsets Task
Let C be a convex subset of a normed space X with a non-empty interior. Proove that the closure of the interior of C matches the closure of C. Proove that the interior of the closure of C matches the interior of C:
$$int\overline C = int C$$
$$\overline C = \overline{int C}$$
We have to use the following:
Let $x \in \overline{C}$, and $x \notin intC$
From this follows:
$$intC \subset C$$
$$\overline{intC} \subset \overline{C}$$
Attemp:
We can say that intC is open and convex, right?
Theorem: X is Banach space, C - open convex subspace of X, $x_0 \notin C => \exists f \in X^*,f \neq 0$, such that $f(x)< f(x_0) \forall x \in C$
Can we use this theorem to get all the limit points:
$$S = \{x \in \overline{C} : x \neq intC \}$$
Is the following correct:
$$\overline{intC} = intC \cup S$$
From which:
$$\overline C = \overline{int C}$$
Question:
Guess this is not correct, can you please proove these statements and show me the the proof.
Note:
The only potentially good thing I could do is maybe use the corollaries of Hahn-Banach. I think that is what I am assigned to be looking for(But maybe I have to use something else).
EDIT:
What I understand by closed cover:
closed cover of a set А is intersection of all closed sets, containing A:
$$\overline{A}=\cap_i \overline{A_i}$$
 A: Problem 1, $\overline C = \overline{\text{int }C}:$ Let $x \in \text{int }C, y\in C.$ Then there is an open ball $B=B(x,r)\subset C.$ Since $C$ is convex, $tB+(1-t)y\subset C$ for all $t\in [0,1].$ Note that if $t\in (0,1],$ then $tB+(1-t)y$ is an open ball lying in $C.$ Thus the open set
$$U= \bigcup_{t\in (0,1]} tB+(1-t)y \subset C.$$
Because $[x,y)\subset U,$ we see see that $y$ is the limit of a sequence in $U \subset \text{int }C.$ Therefore $\overline C \subset \overline{\text{int }C}.$ Since the reverse containment is obvious, we have the result.
Problem 2, $\text{int }C =\text{int }\overline C:$ Write $\overline C = \text{int }C\cup \partial C.$ Let $x\in\text{int }C, y\in \partial C.$
Claim: For $s>1,$ $x+s(y-x)\notin \overline C.$ (Thus $y$ is the "last point" in $\overline C$ on the ray from $x$ in the direction of $y.$)
Proof: Suppose this is false. Then for some $s>1,$ $z= x+s(y-x)\in \overline C.$ Now $y\in [x,z],$ so there exists $t\in (0,1)$ such that $y=tx+(1-t)z.$ Because $z\in \overline C,$ there exists a sequence $z_n$ in $C$ such that $z_n\to z.$
Choose $B=B(x,r)$ such that $B\subset \text{int }C.$ Then $B_n =tB+(1-t)z_n \subset \text{int }C$ for all $n.$ Note that the center of $B_n$ is $tx+(1-t)z_n$ and its radius is $tr.$ Here's the thing: $y\in B_n$ for large $n.$  To see this, use $y=tx+(1-t)z$ in the inequality $|y-(tx+(1-t)z_n)|<tr.$
Thus $y\in \text{int }C.$ But that's a contradiction, as $y\in \partial C,$ a set disjoint from $\text{int }C.$ This proves the claim.
This implies no open ball centered at $y$ is contained in $\overline C.$ It follows that no point on $\partial C$ can be in $\text{int }\overline C.$ This gives the desired result.
