Order Isomorphic Sets are Equivalent

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be ordered sets.

Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be order isomorphic.


Then $S$ and $T$ are equivalent.


Proof

By definition, an order isomorphism is a bijection $\phi$ such that:

$\phi: S \to T$ is order-preserving
$\phi^{-1}: T \to S$ is order-preserving.

So, by definition, there exists a bijection between $S$ and $T$.

The result follows by definition of set equivalence.

$\blacksquare$


Sources