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Apology for asking highly related but subtly different questions within a very short amount of time.

Let $X$ be a real Banach space, and $A\subset X$ be balanced, convex, and absorbing. Two facts are known.

If $A$ is closed, then it is a neighborhood of $0$.

If $A$ is not closed, then $A$ might have empty interior.

Question. Usually when a set has empty interior but its closure has nonempty interior, the routine thinking is that taking the closure fill in the holes. But in the context of this question, if $A$ is not closed, and contains some holes, these holes can be approached by points from within $A$ from more than 1 directions, and so should(?) lie on some line segment of two points in $A$, and due to convexity, these holes shouldn't be holes at all. So convexity should(?) prohibit $A$ from having holes. If that is true, then why would taking the closure makes $\overline{A}$ a neighborhood?? If $x_n\to 0$ lies completely outside of $A$, then after taking the closure the sequence should still lie outside of $\overline{A}$. No?

Nonetheless the two facts above are true. So there must be something wrong with this intuition. I suspect $A$ can have holes approaching $0$ which are not the convex combination of any points of $A$. But this seems very strange to me. I can't seem to find such an example online or in my textbooks (Rudin/Conway). Thanks in advance.

user760
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  • For your second point, since $A$ is balanced it contains $0$, so it really boils down to if $0$ is in the interior of the closure of $A$ as well. As for your first question, i think that it might require infinite dimensionality of the vector space to have a balanced convex and absorbing set that has empty interior. Maybe this can be modified to fit the bill. – jd27 May 29 '24 at 06:04
  • That closed, balanced, convex, and absorbing sets (so-called barrels) have interior points requires completeness of the normed space. – Jochen May 29 '24 at 09:02
  • @Jochen Sorry I meant Banach space. Will update. I want to see an example of a sequence $x_n\to 0$ in the complement of $A$ eventually being contained in the closure. I've seen such a sequence not contained in $A$, but not being shown they eventually are all "holes" of $A$. At least not in my textbooks. – user760 May 29 '24 at 09:10
  • @Jochen Such as this example: https://math.stackexchange.com/a/3264904/416552. Is it possible to show that sequence eventually are holes? That is, for large $N$, any ball centered at $x_n$, for $n>N$, intersects with $A$? – user760 May 29 '24 at 09:18
  • In the example of Alex Ravsky in the linked question, the closure of $A$ is a neibourhood of $0$ (because $\overline A$ is a barrel in the Banach space $\ell^2$) and hence contains all $x_n$ for $n\ge N$ and some $N\in N$. Hence, every ball around $x_n$ intersects $A$.

    Anyway, I don't understand what you mean by a hole of a (convex) set.

     – Jochen May 29 '24 at 10:35
  • @Jochen Thanks. I want to show every ball around $x_n$, for large $n$, intersects with $A$ without using the fact $\overline{A}$ is a neighborhood of $0$. I want to see those $x_n$ are "enclosed" by $A$ directly. By "holes" I just mean not in $A$ but enclosed by $A$, such as those $x_n$. – user760 May 29 '24 at 10:54
  • @Jochen If using that fact, then there's nothing to ask. But I want a direct construction showing me those $x_n$ really are approached by points in $A$. So, with that linked example, I want to see a concrete sequence $(a_{n,m})$ in $A$ s.t. $a_{n,m}\to x_n$ for large $n$. – user760 May 29 '24 at 10:59
  • @Jochen Just found this answer: https://math.stackexchange.com/a/189459/416552. Now I'm confused. If this link is correct, then does it mean there should be no concrete examples of such an $A$ (denoted as $K$ in that link), and hence no concrete demonstration of what I ask for?? But the example by Alex Ravsky is certainly concrete. Only one of them can be correct, no?? – user760 Jun 01 '24 at 09:00
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    In Alex Ravsky's example, he chooses a maximal linear independent subset of unit length vectors... -- this uses the axiom of choice quite explicitly. – Jochen Jun 03 '24 at 13:12
  • @Jochen Thanks. That makes sense. But I still wonder how to directly prove that sequence $x_n$ eventually lies in the closure of $A$, without citing the fact $\overline{A}$ is a neighborhood of $0$. – user760 Jun 03 '24 at 15:52

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