Proving that for the posets $\mathcal{A} = ((-1,1),\leq)$ and $\mathcal{B}=((-1,0)\cup(0,1),\leq)$, $\mathcal{A} \ncong \mathcal{B}$

Question

I need to prove that for posets $\mathcal{A} = ((-1,1),\leq)$ and $\mathcal{B}=((-1,0)\cup(0,1),\leq)$, $\mathcal{A} \ncong \mathcal{B}$, where the binary relation is $\leq$.

I don't exactly understand where to begin. How can they be non-isomorphic, when all we did was to split $(-1,1)$ into a union of the intervals $(-1,0)\cup(0,1)$?

Answer 1

Notice that $\mathscr{A}$ has the following property: if $\langle x_n:n\in\Bbb Z^+\rangle$ and $\langle y_n:n\in\Bbb Z^+\rangle$ are sequence in $(-1,1)$ such that

$$x_1<x_2<x_3<\ldots<y_3<y_2<y_1\;,\tag{1}$$

then $\bigcap_{n\in\Bbb Z^+}(x_n,y_n)\ne\varnothing$. This is a property that is perserved by order-isomorphisms, and $\mathscr{B}$, unlike $\mathscr{A}$, does not have it: if $x_n=-\frac1{n+1}$ and $y_n=\frac1{n+1}$ for $n\in\Bbb Z^+$, these numbers satisfy $(1)$, but

$$\bigcap_{n\in\Bbb Z^+}\left(-\frac1{n+1},\frac1{n+1}\right)=\varnothing\;,$$

since $0$ is not in $(-1,0)\cup(0,1)$, the underlying set of the partial order.

I’ll leave it to you to show that the property in question really is preserved by order-isomorphisms.

Answer 2

I'm not sure if this would be helpful as this certainly seems like invoking something too powerful but -

Note that $\mathcal{A}$ and $\mathcal{B}$ are not just posets but in fact, totally ordered. Thus, we can consider the topological spaces $T_\mathcal{A}$ and $T_\mathcal{B}$ obtained by putting the order topology on the sets $(-1, 1)$ and $(-1, 0)\cup(0, 1)$.
Now, if $\mathcal{A}$ and $\mathcal{B}$ were isomorphic, then $T_\mathcal{A}$ and $T_\mathcal{B}$ would be homeomorphic. However, that is not the case as one is connected while the other is not.

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I am quite certain that the above could be converted into a proof just in terms of posets. (Possibly by mimicking the proof as to why intervals are connected.) However, I'm not sure if that would be very simple.