Ordering

Let $S$ denote the set containing all the natural numbers that are not divisible by $ 2 $.

And define the binary relation $\ge$ on two natural number $m , n$ , $m \ge n$ if $m = k n$ , for integer k, meaning that $n$ divides $m$ .

How to show that $(S\cup\{0\},\ge)$ is order-isomorphic to $(S,\ge)$ ?

#Algebra

Easy Math Editor

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Comments

It is just for mistake, thanks for the reminder.

A Former Brilliant Member - 4 years, 8 months ago

I edited your post a bit using standard notation for a set and a poset. Can you check and tell me if this is what you wanted to mean? Thanks.

Prasun Biswas - 4 years, 8 months ago

Yes, absolutely.

A Former Brilliant Member - 4 years, 8 months ago

In that case, you can't prove it because they are not order-isomorphic.

Proof. Suppose that they are order-isomorphic. Then, there exists a bijection $f\colon S\cup\{0\}\to S$ such that $x\geq y\iff f(x)\geq f(y)~\forall~x,y\in S\cup\{0\}$ .

Now, since $0$ is divisible by every integer (since $0=0\times a$ for all integers $a$ ), we have $0\geq a~\forall~a\in S\cup\{0\}$ . So, we have $f(0)\geq f(a)~\forall~a\in S\cup\{0\}$ , i.e., $f(0)\geq k~\forall~k\in\textrm{Im}(f)=S$ .

Since $f(0)\in S$ , this means that there must exist some element in $S$ (which would be $f(0)$ ) which is divisible by every element in $S$ . Since $0\in\textrm{Dom}(f)$ , we know that $f(0)$ exists, say $f(0)=m$ .

Now, since $f(0)=m\in S$ , by definition of $S$ , we get that $m+2\in S$ but note that $m$ is not divisible by $m+2$ (obvious), i.e., $m=f(0)\not\geq m+2$ which is a contradiction.

Hence, our assumption is wrong and $f$ doesn't exist.

Hence, we conclude that $(S\cup\{0\},\geq)$ is not order-isomorphic to $(S,\geq)$

An informal argument that one could use to show the same fact is that an order-isomorphism preserves maximum properties and since 0 is a maximum for $(S\cup\{0\},\geq)$ whereas there is no maximum element in $(S,\geq)$ , they cannot be order-isomorphic.

In fact, there's a more general theorem which gives some sufficient conditions for two posets to not be order-isomorphic. You can check it out here.

Prasun Biswas - 4 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$