Exercise 1.1 ( )

Proof.

(a)

Such property only depends on the group structure of $X$: Each $x$ in $X$ has an opposite $x$. Let $x^{\prime }$ be any opposite of $x$, so that $x-x=0=x+x^{\prime }$. Thus, $x+x-x=x+x+x^{\prime }$, which is equivalent to $x=x^{\prime }$. So is established the uniqueness of $x$. It is now clear that $x+z=y$ $z=x+y$, which asserts both the existence and the uniqueness of $z$.

(b)

Remark that $$\begin{array}{lll}\hfill 0\cdot x& =(0+0)\cdot x=0\cdot x+0\cdot x& \hfill \text{(1)}\\ \hfill & =(0+0)\cdot x=0+0\cdot x& \hfill \text{(2)}\end{array}$$

then conclude from (a) that $0\cdot x=0$. So,

$$\begin{array}{lll}\hfill 0=0\cdot x=(1-1)\cdot x& =x+(1)\cdot x\Rightarrow 1\cdot x=x.& \hfill \text{(3)}\end{array}$$

Finally,

$$\begin{array}{lll}\hfill \alpha \cdot 0\stackrel{(3)}{=}\alpha \cdot (x+(1\cdot x))=\alpha \cdot x+\alpha \cdot (1)\cdot x=(\alpha -\alpha )\cdot x=0\cdot x=0,& & \hfill \text{(4) }\end{array}$$

which proves (b).

(c)

Remark that $$\begin{array}{lll}\hfill 2x=(1+1)x=x+x& & \hfill \text{(5)}\end{array}$$

for every $x$ in $X$, and so conclude that

$$\begin{array}{lll}\hfill 2A=\{2x:x\in A\}=\{x+x:x\in A\}\subset \{x+y:(x,y)\in A^2\}=A+A& & \hfill \text{(6) }\end{array}$$

for all subsets $A$ of $X$; which proves (c).

(d)

If $A$ is convex, then $$\begin{array}{lll}\hfill A\subset \frac{s}{s+t}A+\frac{t}{s+t}A\subset A;& & \hfill \text{(7)}\end{array}$$

which is

$$\begin{array}{lll}\hfill \mathit{sA}+\mathit{tA}=(s+t)A.& & \hfill \text{(8)}\end{array}$$

Conversely, the special case $s+t=1$ is

$$\begin{array}{lll}\hfill \mathit{sA}+(1-s)A=A.& & \hfill \text{(9)}\end{array}$$

The latter extends to $s=0$, since

$$\begin{array}{lll}\hfill 0A+A\stackrel{(b)}{=}\{0\}+A=A.& & \hfill \text{(10)}\end{array}$$

The extension to $s=1$ is analogously established (or simply use the fact that $\text{+}$ is commutative!). So ends the proof.

(e)

Let $A$ range over $B$ a collection of balanced subsets, so that $$\begin{array}{lll}\hfill \alpha \bigcap B\subset \mathit{\alpha A}\subset A\subset \bigcup B& & \hfill \text{(11)}\end{array}$$

for all scalars $\alpha$ of magnitude $\le 1$. The inclusion $\alpha \bigcap <!--\; FUNCTION\; APPLICATION\; -->B\subset A$ establishes the first part. Now remark that

$$\begin{array}{lll}\hfill \mathit{\alpha A}\subset \bigcup B& & \hfill \text{(12)}\end{array}$$

implies

$$\begin{array}{lll}\hfill \alpha \bigcup B\subset \bigcup B;& & \hfill \text{(13)}\end{array}$$

which achieves the proof.

(f)

Let $A$ range over $C$ a collection of convex subsets, so that $$\begin{array}{lll}\hfill (s+t)\bigcap C\subset s\bigcap C+t\bigcap C\subset \mathit{sA}+\mathit{tA}\stackrel{(d)}{\subset }(s+t)A& & \hfill \text{(14)}\end{array}$$

for all positives scalars $s$, $t$. Inclusions at both extremities force

$$\begin{array}{lll}\hfill s\bigcap C+t\bigcap C=(s+t)\bigcap C.& & \hfill \text{(15)}\end{array}$$

We now conclude from (d) that the intersection of $C$ is convex. So ends the proof.

(g)

Skip all trivial cases $\Gamma =\varnothing$, $\{\varnothing \}$, $\{\{x\}\}$, $\{\varnothing$, $\{x\}\}$ then pick $x_1,x_2$ in $\bigcup <!--\; FUNCTION\; APPLICATION\; -->\Gamma$, so that each $x_i$ ( $i=1,2)$ lies in some $C_i\in \Gamma$. Since $\Gamma$ is totally ordered by set inclusion, we henceforth assume without loss of generality that $C_1$ is a subset of $C_2$. So, $x_1,x_2$ are now elements of the convex set $C_2$. Every convex combination of our $x_i$’s is then in $C_2\subset \bigcup <!--\; FUNCTION\; APPLICATION\; -->\Gamma$. Hence (g).

(h)

Simply remark that $$\begin{array}{lll}\hfill s(A+B)+t(A+B)=\mathit{sA}+\mathit{tA}+\mathit{sB}+\mathit{tB}=(s+t)(A+B)& & \hfill \text{(16)}\end{array}$$

for all positive scalars $s$ and $t$, then conclude from (d) that $A+B$ is convex.

(i)

Given any $\alpha$ from the closed unit disc, $$\begin{array}{lll}\hfill \alpha (A+B)=\mathit{\alpha A}+\mathit{\alpha B}\subset A+B.& & \hfill \text{(17)}\end{array}$$

There is no more to prove: $A+B$ is balanced.

(j)

Our proof will be based on the following lemma,

If $S$is nonempty, then each of the following three properties

(i)

$S$is a vector subspace of $X$;

(ii)

$S$is convex balanced such that $S+S=S$;

(iii)

$S$is convex balanced such that $\mathit{\lambda S}=S(\lambda >0)$

implies the other two.

To prove the lemma, let $S$ run through all nonempty subsets of $X$. First, assume that (i) holds: Clearly, every $S$ is convex balanced. Moreover, $S+S\subset S$. Conversely, $S=S+\{0\}\subset S+S$; which establishes (ii). Next, assume (only) (ii): A proof by induction shows that

$$\begin{array}{lll}\hfill \mathit{nS}=(n-1)S+S=S+S=S(n=1,2,3,\dots )& & \hfill \text{(18)}\end{array}$$

with the help of (b) and (d). Pick $\lambda >0$ then choose $n$ so large that $1<\mathit{n\lambda }<n^2$. Thus,

$$\begin{array}{lll}\hfill \mathit{nS}\stackrel{(18)}{\subset }S\subset n\mathit{\lambda S}\subset n^{2}S,& & \hfill \text{(19)}\end{array}$$

since $S$ is balanced. For instance, set $n=\lceil 1∕\lambda \rceil +\lceil \lambda \rceil$. Dividing the latter inclusions by $n$ shows that

$$\begin{array}{lll}\hfill S\subset \mathit{\lambda S}\subset \mathit{nS}\stackrel{(18)}{\subset }S,& & \hfill \text{(20)}\end{array}$$

which is (iii). Finally, dropping (ii) in favor of (iii) leads to

$$\begin{array}{lll}\hfill \mathit{\alpha S}+\mathit{\beta S}\stackrel{(a)}{=}|\alpha |S+|\beta |S\stackrel{(d)}{=}(|\alpha |+|\beta |)S\stackrel{(\mathit{iii})}{=}S(|\alpha |+|\beta |>0);& & \hfill \text{(21) }\end{array}$$

where the equality at the left holds as $S$ is balanced. Moreover (under the sole assumption that $S$ is balanced), this extends to $|\alpha |+|\beta |=0$, as follows,

$$\begin{array}{lll}\hfill \mathit{\alpha S}+\mathit{\beta S}=0S+0S\stackrel{(b)}{=}\{0\}\stackrel{(b)}{=}0S\subset S.& & \hfill \text{(22)}\end{array}$$

Hence (i), which achieves the lemma’s proof. We will now offer a straightforward proof of (j).

Let $V$ be a collection of vector spaces of $X$, of intersection $I$ and union $U$. First, remark that every member of $V$ is convex balanced: So is $I$ (combine (e) with (f)). Next, let $Y$ range over $V$, so that

$$\begin{array}{lll}\hfill I+I\subset Y+Y\subset Y;& & \hfill \text{(23)}\end{array}$$

which yields

$$\begin{array}{lll}\hfill I+I=I& & \hfill \text{(24)}\end{array}$$

(the fact that $I=I+\{0\}\subset I+I$ was tacitely used). It now follows from the lemma’s (ii) $\Rightarrow$ (i) that $I$ is a vector subspace of $X$. Now temporarily assume that $S$ is totally ordered by set inclusion: Combining (e) with (g) establishes that $U$ is convex balanced. To show that $U$ is more specifically a vector subspace, we first remark that such total order implies that either $Z\subset Y$ or $Y\subset Z$, as $Z$ ranges over $V$. A straightforward consequence is that

$$\begin{array}{lll}\hfill Y\subset Y+Z\subset Y\cup Z.& & \hfill \text{(25)}\end{array}$$

Another one is that $Y\cup Z$ ranges over $V$ as well. Combined with the latter inclusions, this leads to

$$\begin{array}{lll}\hfill U\subset U+U\subset U.& & \hfill \text{(26)}\end{array}$$

It then follows from the lemma’s (ii) $\Rightarrow$ (i) that $U$ is a vector subspace of $X$. Finally, let $A,B$ run through all vector subspaces of $X$: Combining (h) with (i) proves that $A+B$ is convex balanced as well. Furthermore,

$$\begin{array}{lll}\hfill A+B\stackrel{(i)\Rightarrow (\mathit{ii})}{=}(A+A)+(B+B)=(A+B)+(A+B),& & \hfill \text{(27)}\end{array}$$

where the equality at the right holds as $X$ is an abelian group. We now conclude from (ii) that any $A+B$ is a vector subspace of $X$. So ends the proof.