# Isomorphism to Closed Interval

## Theorem

Let $m, n \in \N$ such that $m < n$.

Then:

$\left|{\left[{m + 1 \,.\,.\, n}\right]}\right| = n - m$

Let $h: \N_{n - m} \to \left[{m + 1 \,.\,.\, n}\right]$ be the mapping defined as:

$\forall x \in \N_{n - m}: h \left({x}\right) = x + m + 1$

Let the orderings on $\left[{m + 1 \,.\,.\, n}\right]$ and $\N_{n - m}$ be those induced by the ordering of $\N$.

Then $h$ a unique order isomorphism.

## Proof

From Unique Isomorphism between Finite Totally Ordered Sets, it suffices to show that $h$ is an order isomorphism.

To this end, remark that, for all $x, y \in \N_{n - m}$:

 $\displaystyle h \left({x}\right)$ $=$ $\displaystyle h \left({y}\right)$ $\quad$ $\quad$ $\displaystyle \iff \ \$ $\displaystyle x + m + 1$ $=$ $\displaystyle y + m + 1$ $\quad$ $\quad$ $\displaystyle \iff \ \$ $\displaystyle x$ $=$ $\displaystyle y$ $\quad$ Natural Number Addition is Cancellable $\quad$

proving $h$ is an injection, and so a bijection, from Equivalence of Mappings between Sets of Same Cardinality.

$x \le y \iff h \left({x}\right) \le h \left({y}\right)$

so $h$ is an order isomorphism.

$\blacksquare$