Isomorphism to Closed Interval

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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.


By Ordering on Natural Numbers is Compatible with Addition and Natural Number Addition is Cancellable for Ordering, it follows that:

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

so $h$ is an order isomorphism.

$\blacksquare$


Sources