Proof of Open Boxes: Emptiness, Intersections and Preimages under Concatenation

Proof of (i). A point $x\in R(a,b)$ would satisfy $a_{i}<x_{i}<b_{i}\le a_{i}$ (Open Boxes in ℝⁿ (i)), which is impossible.

Proof of (ii). The points $c$ and $d$ exist as the functions $[n]\to\mathbb{R}$ given by these expressions (Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (ii)). For $i\in[n]$ and $t\in\mathbb{R}$, $\max\{a_{i},a'_{i}\}$ is one of $a_{i}$, $a'_{i}$ and is at least both of them (The Maximum and Minimum of Two Elements of a Total Order (ii), The Maximum and Minimum of Two Elements of a Total Order (iii)), so $c_{i}<t$ if and only if $a_{i}<t$ and $a'_{i}<t$; likewise $t<d_{i}$ if and only if $t<b_{i}$ and $t<b'_{i}$. Hence a point $x\in\mathbb{R}^{n}$ lies in both $R(a,b)$ and $R(a',b')$ exactly when $c_{i}<x_{i}<d_{i}$ for every $i\in[n]$, that is, when $x\in R(c,d)$.

Proof of (iii). For $(\xi,\eta)\in\mathbb{R}^{m}\times\mathbb{R}^{n}$, the point $\iota(\xi,\eta)$ has the component $\xi_{k}$ at $k\in[m]$ and $\eta_{j}$ at $m+j$ for $j\in[n]$ (Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space (ii)); in particular $a_{k}=a'_{k}$, $b_{k}=b'_{k}$, $a_{m+j}=a''_{j}$ and $b_{m+j}=b''_{j}$. Every index in $[m+n]$ is of exactly one of these two kinds (Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space (i)). So $\iota(\xi,\eta)\in R(a,b)$ if and only if $a'_{k}<\xi_{k}<b'_{k}$ for every $k\in[m]$ and $a''_{j}<\eta_{j}<b''_{j}$ for every $j\in[n]$, that is, $\xi\in R(a',b')$ and $\eta\in R(a'',b'')$. By Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii) and Membership in a Cartesian Product, and the Cartesian Product of Two Sets Is a Set (i), this is the asserted identity.