Let $A$ be a nonempty open connected subset of a (real) topological vector space $X$ such that $$2A-A \subseteq 8A$$ (for instance one could take $A=(-1,2)$).

Question. Is it true that there exists a nonempty open connected set $B\subseteq A$ such that $B$, in addition, is symmetric (i.e., $B=-B$)?

  • $\begingroup$ Have you proved it for $X=\Bbb R$ at least? $\endgroup$ – Adam Chalumeau May 15 at 10:16
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    $\begingroup$ @AdamChalumeau The hypotheses imply $0\in A$. $\endgroup$ – logarithm May 15 at 10:30
  • $\begingroup$ @logarithm Do they? Do you have a proof of this? If so, then $A \cap -A$ is a good choice for $B$. $\endgroup$ – Theo Bendit May 15 at 10:33
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    $\begingroup$ @PaoloLeonetti Those are the only connected open sets of $\mathbb{R}$, and of a general $A$ intersected with a line in a general tvs. $\endgroup$ – logarithm May 15 at 10:49
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    $\begingroup$ @Mirko The example was really supposed to demonstrate why $A \cap (-A)$ may not be connected, even though $0 \in A$ and $A$ is connected. An open example is easy enough too; just add (in the sense of Minkowski sum) an open ball of radius $1/3$ to $A$, and the intersection should still be disconnected (I don't want to compute the exact intersection). However, I agree that $2A - A \subseteq 8A$ (probably) does not hold here. Note that, if $0 \in A$, $A$ is open in a normed linear space, then $A$ always contains an open ball around $0$, which is open, connected, and symmetric. $\endgroup$ – Theo Bendit May 16 at 1:01

I got many ideas from the comments, but I had to verify the details, and to modify and add some elements to convince myself that the answer is yes (as shown below).

So we have $\frac A8\subseteq \frac A4 - \frac A8\subseteq A$ (hence $\frac A8\subseteq A$, also $\frac A{64}\subseteq \frac A8$, and, by induction, $\frac A{8^n}\subseteq A$ for all $n\ge1$). Since $A$ is open, it is not difficult to see that if $C$ is the closure of $A$ then also $\frac A4 - \frac C8=\frac A4 - \frac A8\subseteq A$ (the details are provided in a corollary near the end). If $a\in A$ then the sequence $\frac a{8^n}$ converges to $0$ (I believe even in general TVS (yes, reference provided in a comment below by OP)), so $0\in C$. Hence $\frac A4=\frac A4-0 \subseteq\frac A4 - \frac C8=\frac A4 - \frac A8\subseteq A$. So we have:

(i) $\frac A{32} \subseteq \frac A4$ because $\frac A8\subseteq A$,

(ii) $\frac A{32} \subseteq \frac A8$ because $\frac A4 \subseteq A$, and

(iii) $\frac A4 - \frac A8\subseteq A$.

Using the above we get $\frac A{32} - \frac A{32} \subseteq \frac A4 - \frac A8 \subseteq A$. Thus the set $B=\frac A{32} - \frac A{32}$ works. Clearly it is symmetric, and it is open and connected: It is connected since it is the continuous image of $A\times A$ under the subtraction function (and division by $32$), and the product space $A\times A$ is connected since the factors are.

Here are some details on the condition that $\frac A4\subseteq A$, which was used in the above proof. One way to prove it is without a reference to the closure $C$ of $A$, as follows.

Claim. $\frac A4 \subseteq A$ (or equivalently, $2A\subseteq8A$).

Proof. Take any $a\in A$, we need to show that $2a\in 8A$. Since $A$ is open, there is $n$ such that $a+\frac a{2\cdot8^n}\in A$. Then $2a=2(a+\frac a{2\cdot8^n})-\frac a{8^n}\in 2A-A\subseteq 8A$. This completes the proof of the Claim.

Here is an alternative way to show that $2A\subseteq8A$. Show that $2A-A=2A-C$ (where $A$ is open and $C$ is the closure of $A$). (Then, since $0\in C$ we get that $2A\subseteq2A-C=2A-A\subseteq8A$.)

Lemma. If $U$ is open and $K$ is arbitrary then $U+\overline K= U+K$ (where $\overline K$ is the closure of $K$).

Proof. Pick any $p\in U+\overline K$. Then $p=q+r$ for some $q\in U$ and $r\in\overline K$. Since $U$ is open, there exists a symmetric neighborhood $V$ of $0$ such that $q+V\subseteq U$. Pick $s\in(r+V)\cap K$. Then $v=s-r\in V$ so $-v\in-V=V$ and $q-v\in U$, hence $p=q+r=q-v+r+v=(q-v)+s\in U+K$. Thus $U+\overline K\subseteq U+K$ and $U+\overline K=U+K$.

Corollary. If $A$ is open and $C$ is the closure of $A$ then $2A-C=2A-A$. (So, if, in addition, $2A-A\subseteq8A$ where $A$ is open and nonempty then $2A\subseteq2A-C=2A-A\subseteq8A$, using that $0\in C$ for the first inclusion.)

Proof. Use the above Lemma with $U=2A$ and $K=-A$.

Discussion. So the inclusion $2A\subseteq8A$ was given two different proofs, one direct, and another using the closure $C$ of $A$, along with the above lemma and corollary (providing a quicker approach, at least to me). I did not know (beforehand) if $0\in A$, and did not use it in my proof (though it eventually follows from $0\in\frac A{32}-\frac A{32}=B\subseteq A$), and I do not know if $A\cap(-A)$ must be connected (assuming that $A$ is open and $2A-A\subseteq8A$). An example when $A\cap(-A)$ need not be connected was provided by another user in the comments above, $A$ is the union of the two closed upper semicircles in the plane, of radius $1$ with centers at $(\pm1,0)$, but this $A$ is not open and $2A-A\not\subseteq8A$. (The two semicircles could easily be made open, by ``thickening'' them a little, but it is not clear to me at present if we could also get $2A-A\subseteq8A$, and yet to have $A\cap(-A)$ disconnected). One more comment: If $0\in A$ (where $A$ is open) and if we work in a locally connected space, then the connected component of $A\cap(-A)$ containing $0$ would be open and symmetric (so this component could play the role of $B$). But, we need to assume some extra condition (e.g. $2A-A\subseteq8A$) to show that $0\in A$, and even if we knew that $0\in A$, it might not be immediately clear how $A\cap(-A)$ would possibly help, if the space is not locally connected. (I would be curious to see a proof - if there is one - based on the use of $A\cap(-A)$, showing that either it is connected, or that it contains a connected, open symmetric set.)

  • $\begingroup$ About your question in brackets, the answer is positive, see Theorem 1.15.a in Rudin's "Functional Analysis": let $V$ be any neighborhood of $0$, then $\bigcup_n nV=X$. $\endgroup$ – Paolo Leonetti May 15 at 14:46
  • $\begingroup$ @PaoloLeonetti I think I fixed the details in my proof (with or without the use of the closure $C$ of $A$). So now I know $0\in A$ since $0\in B\subseteq A$. I do not know if $A\cap(-A)$ is connected (and I am curious, but will leave it for now). Clearly $A\cup(-A)$ is connected. $\endgroup$ – Mirko May 15 at 15:29
  • $\begingroup$ The proof depends on the choice of the coefficients, however I find it very nice. $\endgroup$ – Paolo Leonetti May 16 at 14:12

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