Does this lemma, which have an easy proof for rational numbers, also hold true for all real numbers?

Question

Recently, I answered this question. A generalized version of what was proved is the following.

Lemma: For $n\in\mathbb{N}$ and $a_1,a_2,\ldots,a_{2n+1}\in \mathbb{Q}$, if for each $1\le i\le 2n+1$, the numbers $\{a_j:1\le j\le 2n+1\wedge j\neq i\}$ can be divided into two sets of size $n$ with equal sum, then all $a_i$ must be equal.

Proof: At first, we limit ourselves to $\mathbb{Z}$. If for some $a_1,a_2,\ldots,a_{2n+1}\in \mathbb{N}$, for all $1\le i\le 2n+1$, the numbers $\{a_j:1\le j\le 2n+1\wedge j\neq i\}$ can be divided into two sets of size $n$ with equal sum, we call $a_1,a_2,\ldots,a_{2n+1}$ a solution.

For any solution $a_1,a_2,\ldots,a_{2n+1}$ and a positive integer $m$, $a_1-m,a_2-m,\ldots,a_{2n+1}-m$ is also a solution. If all the $a_i$ are even, then $\frac12a_1,\frac12a_2,\ldots,\frac12a_{2n+1}$ is a solution as well.

Now, the sum of every $2n$ elements is even, therefore, each $a_i$ has the same parity as the sum of all $a_i$, therefore, all $a_i$ have the same parity.

WLOG, we can assume one of the $a_i$ to be $0$, so all $a_i$ are odd. This means we can divide them all by $2$ and still have a valid solution. However, since one of them is still $0$, they're all even again, so we can divide them all by $2$ again, but then one of them is still $0$ and so on. So all the $a_i$ are finite and divisible by arbitrarily large powers of $2$, so all $a_i$ are equal to $0$.

For $a_1,a_2,\ldots,a_{2n+1}\in\mathbb{Q}$, simply multiply all of the $a_i$ by the product of the denominators to translate the problem to the integers. Q.E.D.

Question: Is the lemma also true for $a_1,a_2,\ldots,a_{2n+1}\in \mathbb{R}$?

Answer 1

HINT:

For the proof you can use the following linear algebra result:

Let $V$ be a vector space over a field $k$ and $\phi_1$, $\ldots$, $\phi_m$, $\psi$ linear functionals on $V$ so that $\phi_1(v)= \cdots = \phi_m(v) = 0$ implies $\psi(v)=0$. Then $\psi$ is a linear combination of the $\phi_i$'s with coefficients in $k$.

Now consider $V=\mathbb{Q}^{2n+1}$, for the $\phi_i$'s all grouping differences, and for $\psi$ one of the $a_s-a_t$.

Answer 2

More generally, it holds in any additive abelian group . . .

Let $\langle{G,+}\rangle$ be an additive abelian group.

Let $X=G^{2n+1}$.

Let $S$ be the set of those elements $x\in X$ which satisfy the specified one-point removal condition.

Let $E = \{x\in X\mid x_1 = \cdots = x_{2n+1}\}$.

Claim:$\;S=E$.

Proof:

If $n=0$, the truth of the claim is immediate, so assume $n \ge 1$.

Clearly we have $E \subseteq S$.

Thus, we only need to show $S \subseteq E$.

Define an equivalence relation $\sim$ on $X$ by $x\sim y$ if $x,y$ are permutations of each other.

Then we have:

• If $x\sim y$, and one of $x,y$ is in $S$, then so is the other.$\\[4pt]$
• If $x,y\in S$, then $x-y\in S$.

Let $x \in S$.

Suppose $x\notin E$.

Permuting the components of $x$, if necessary, we can assume $x_1\ne x_2$.

Let $d=x_2-x_1$.

Then $d\ne 0$.

Let $y$ be the result of swapping the first two components of $x$.

Then $x\sim y$, so $y\in S$, hence $x-y\in S$

But $x-y=(-d,d,0,...,0)$ which fails the one-point removal condition, since the condition can't be satisfied if $-d$ is removed.

Hence we must have $x\in E$.

Thus we have $S \subseteq E$, which completes the proof.