# Existence of a symmetric subset $B\subseteq A$ such that $2A-A\subseteq 8A$

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., $$B=-B$$)?

• Have you proved it for $X=\Bbb R$ at least? – Adam Chalumeau May 15 at 10:16
• @AdamChalumeau The hypotheses imply $0\in A$. – logarithm May 15 at 10:30
• @logarithm Do they? Do you have a proof of this? If so, then $A \cap -A$ is a good choice for $B$. – Theo Bendit May 15 at 10:33
• @PaoloLeonetti Those are the only connected open sets of $\mathbb{R}$, and of a general $A$ intersected with a line in a general tvs. – logarithm May 15 at 10:49
• @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. – Theo Bendit May 16 at 1:01

## 1 Answer

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.)

• 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 $\bigcup_n nV=X$. – Paolo Leonetti May 15 at 14:46
• @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. – Mirko May 15 at 15:29
• The proof depends on the choice of the coefficients, however I find it very nice. – Paolo Leonetti May 16 at 14:12