Proof of Operations Preserving Convexity in $\mathbb{R}^n$: Intersections, Translates, Scalar Multiples, Sums, Affine Images and Preimages, Closure and Interior

Throughout, a set $K$ is shown to be convex by taking $x,y\in K$ and $t\in[0,1]$ and showing $(1-t)x+ty\in K$, which is what Segments and Convex Subsets of $\mathbb{R}^n$ (i) and Segments and Convex Subsets of $\mathbb{R}^n$ (ii) require; the vector-space laws of $\mathbb{R}^{n}$ and $\mathbb{R}^{m}$ are background facts of the setting and are used without comment.

Proof of (i). For $\emptyset$ there is nothing to check, and for $\mathbb{R}^{n}$ every point $(1-t)x+ty$ lies in $\mathbb{R}^{n}$.

Proof of (ii). Let $x,y\in\bigcap\mathcal{C}$. The point $(1-t)x+ty$ lies in $\mathbb{R}^{n}$, and for each $K\in\mathcal{C}$ we have $x,y\in K$, so $(1-t)x+ty\in K$ by convexity of $K$. Hence $(1-t)x+ty\in\bigcap\mathcal{C}$. (For the empty family, $\bigcap\mathcal{C}=\mathbb{R}^{n}$, and this is (i).)

Proofs of (iii), (iv) and (v). For $x,x',y,y',a\in\mathbb{R}^{n}$ and $\lambda\in\mathbb{R}$ the vector-space laws give

$$(1-t)(a+x)+t(a+x')=a+\bigl((1-t)x+tx'\bigr),\qquad(1-t)(\lambda x)+t(\lambda x')=\lambda\bigl((1-t)x+tx'\bigr),$$ $$(1-t)(x\pm y)+t(x'\pm y')=\bigl((1-t)x+tx'\bigr)\pm\bigl((1-t)y+ty'\bigr).$$

Now take $x,x'\in C$ and $y,y'\in D$. The bracketed points lie in $C$, respectively $D$, by convexity, so each left side, a combination of two points of $a+C$, of $\lambda C$, or of $C\pm D$, lies in that same set.

Proofs of (vi) and (vii). For $x,x'\in\mathbb{R}^{n}$, Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v) and the vector-space laws of $\mathbb{R}^{m}$ give

$$A\bigl((1-t)x+tx'\bigr)+b=(1-t)(Ax+b)+t(Ax'+b).$$

For (vi), take $x,x'\in C$: the right side is a combination of two points of the image, and the left side exhibits it as the image of the point $(1-t)x+tx'$ of $C$. For (vii), take $x,x'$ with $Ax+b,\,Ax'+b\in E$: the right side lies in $E$ by convexity of $E$, so $(1-t)x+tx'$ lies in the preimage.

Proof of (viii). Let $x,y\in\operatorname{cl}C$, put $z=(1-t)x+ty$, and let $r>0$. By Closure of a Subset of a Metric Space (i), the balls $B(x,r)$ and $B(y,r)$ meet $C$; choose $x'\in C\cap B(x,r)$ and $y'\in C\cap B(y,r)$. Then $z'=(1-t)x'+ty'\in C$ by convexity, and $z'-z=(1-t)(x'-x)+t(y'-y)$, so Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (vi) and Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (v) give

$$\lVert z'-z\rVert\le(1-t)\lVert x'-x\rVert+t\lVert y'-y\rVert<(1-t)r+tr=r,$$

the inequality being strict because at least one of the weights $1-t$, $t$ is positive. So $z'\in B(z,r)\cap C$ (Open Ball in a Metric Space (ii), Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (ii)). As $r>0$ was arbitrary, $z\in\operatorname{cl}C$.

Proof of (ix). Let $x,y\in\operatorname{int}C$ and $z=(1-t)x+ty$. By Interior of a Subset of a Metric Space (i) there are radii $r_{1},r_{2}>0$ with $B(x,r_{1})\subseteq C$ and $B(y,r_{2})\subseteq C$; put $r=\min\{r_{1},r_{2}\}$. Let $w\in B(z,r)$ and $h=w-z$. Then $\lVert h\rVert<r$, so $x+h\in B(x,r_{1})\subseteq C$ and $y+h\in B(y,r_{2})\subseteq C$ (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (ii)), and

$$w=(1-t)(x+h)+t(y+h)\in C$$

by convexity of $C$. Hence $B(z,r)\subseteq C$, and $z$ is an interior point of $C$.