Prove that ρ is a partial order on A.

Question

Let $a_1, a_2,\ldots$ be a sequence of real numbers such that $a_i \ne a_j$ when $i \ne j$, and let $A = \{a_1, a_2,\ldots\}$. Define a relation $\rho$ on $A$ as follows. For all $a_i, a_j\in A$, $a_i\mathrel{\rho}a_j$ if and only if $i \le j$ and $a_i\le a_j$. Prove that $\rho$ is a partial order on $A$.

Answer 1

HINT: This time you have to show that $\rho$ is reflexive, antisymmetric, and transitive.

Reflexivity: If $a_i\in A$, then it’s certainly true that $i\le i$ and $a_i\le a_i$, so $a_i\mathrel{\rho}a_i$. Thus, $\mathrel{\rho}$ is reflexive.

Antisymmetry: You need to show that if $a_i,a_j\in A$, $a_i\mathrel{\rho}a_j$, and $a_j\mathrel{\rho}a_i$, then $a_i=a_j$. Note that if $a_i\mathrel{\rho}a_j$, then $i\le j$ and $a_i\le a_j$. In similar fashion translate the hypothesis that $a_j\mathrel{\rho}a_i$ into statements not involving $\rho$, and see if you can see why it must be true that $a_i=a_j$.

Transitivity: You need to show that if $a_i,a_j,a_k\in A$, $a_i\mathrel{\rho}a_j$, and $a_j\mathrel{\rho}a_k$, then $a_i\mathrel{\rho}a_k$. Here again you should translate the hypotheses that $a_i\mathrel{\rho}a_j$ and $a_j\mathrel{\rho}a_k$ into statements that don’t involve $\rho$. For instance, the first one means that $i\le j$ and $a_i\le a_j$. If you put everything together, you should be able to deduce something about the relationship between $i$ and $k$ and the relationship between $a_i$ and $a_k$. Then translate the desired conclusion, $a_i\mathrel{\rho}a_k$, into statements that don’t involve $\rho$, to see just what it is that you’re really trying to deduce from the hypotheses.