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# 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 $$(S,\ge)$$ ?

Note by L Km
9 months, 3 weeks ago

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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. · 9 months, 2 weeks ago

Yes, absolutely. · 9 months, 2 weeks ago

It is just for mistake, thanks for the reminder. · 9 months, 3 weeks ago