2

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:

$$\overline{A}=\cap_i \overline{A_i}$$

Arctic Char
  • 16,972
Harton
  • 171
  • Unless I'm misunderstanding something, the standard term for what you called "closed cover" is "closure". – Daniel Fischer Aug 21 '20 at 16:18
  • @DanielFischer Not sure I've made an edit, please check the update, there is an Edit seciont – Harton Aug 21 '20 at 16:35
  • I guess that should be "intersection" instead of "section" (literal translation from your language?), and the symbol $\cap$ instead of $\cup$ (typo). Then that is what usually is called the closure. (I suspect "closed cover" is also a literal translation.) – Daniel Fischer Aug 21 '20 at 16:39
  • @DanielFischer Yes, all correct – Harton Aug 21 '20 at 16:41
  • 1
    The first part is proved here. The question speaks of a closed convex set, but the answers do not assume that. Do you understand these answers? – Daniel Fischer Aug 21 '20 at 16:53
  • @DanielFischer not sure what $C^o$ means, also not quite getting the calculations and not sure.. is p(t) making a line between $C^o$ and $\overline{}$ – Harton Aug 21 '20 at 17:06
  • $C^{\circ}$ is another notation for the interior of $C$. Yes, $p(t)$ gives the line segment from $p$ to $c$. – Daniel Fischer Aug 21 '20 at 17:08
  • For the second part the claim from Daniel Fischer’s link applied to $\overline{C}$ yields $\overline{C}=\overline{\operatorname{int} \overline{C}}=$ (by the first part) $\overline{\operatorname{int} C}$. – Alex Ravsky Aug 22 '20 at 01:15
  • @AlexRavsky theorem fixed, but I am not sure I quite understand the calculations in everything – Harton Aug 23 '20 at 16:54
  • What you don't understand in my arguments? – Alex Ravsky Aug 23 '20 at 16:57
  • First, doesn't the link prove that $\overline{intC}=C$? we want $\overline{intC}=\overline{C}$ . The second top answer does not fit our use case for $\overline{int\overline{C}}=\overline{intC}$, as it's described for topological vector spaces and it's not proved. Question: why is $\overline{C}=\overline{int\overline{C}}$ – Harton Aug 23 '20 at 20:11

1 Answers1

1

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.

zhw.
  • 107,943
  • In the 1st part it's said that there is a sequence for which y is a limit. what is that sequence ? – Harton Aug 27 '20 at 15:33
  • @Harton $(1/n)x +(1-1/n)y, n=1,2,\dots$ will work. – zhw. Aug 27 '20 at 15:48
  • we get $C \subset \overline{intC}$ Do we get $\overline{C} \subset \overline{intC}$ from it and why ? it's obvious that $C \subset \overline{C} $ but why is $\overline{C} \subset \overline{intC}$ – Harton Aug 27 '20 at 16:02
  • $\overline C$ is the smallest closed set containing $C$ and $\overline {\text{int }C}$ is a closed set containing $C.$ Therefore $\overline C \subset \overline {\text{int }C}.$ – zhw. Aug 27 '20 at 16:13
  • I am not sure I understand the sequence Idea ... can I say there is a seq ${y_i}:=(1/n)x+(1-1/n)y$ and limyi =y ? – Harton Aug 27 '20 at 16:14
  • I showed the half open interval $[x,y)$ is contained in the interior of $C.$ The sequence $(1/n)x+(1-1/n)$ lies in $[x,y)$ and converges to $y.$ – zhw. Aug 27 '20 at 16:37
  • That should be $(1/n)x+(1-1/n)y$ above – zhw. Aug 27 '20 at 19:42