Intuition behind order isomorphisms

Question

So I've tried to teach myself some discrete math and topology and I've recently come to a grinding halt on the topic of order relations and specifically order isomorphisms. I'm finding it difficult to grasp how isomorphic sets are equal and the definition of the function itself.

Is there any intuition and basic examples of $2$ isomorphic sets as I can't quite understand what exactly it means for 2 sets to be order isomorphic and what an isomorphism is. Looking on Wikipedia I came across this definition

Formally, given two posets

Formally, given two posets $(S,\leq _{S})$ and $(T,\leq _{T})$, an order isomorphism from ${\displaystyle (S,\leq _{S})}$ to ${\displaystyle (T,\leq _{T})}$ is a bijective function ${\displaystyle f}$ from ${\displaystyle S}$ to ${\displaystyle T}$ with the property that, for every ${\displaystyle x}$ and ${\displaystyle y}$ in ${\displaystyle S}$, $x\leq y $ if and only if $f(x)\leq f(y)$. That is, it is a bijective order-embedding.

So my confusion is what exactly does it mean when they are isomorphic how is it that they can be essentially said to be the same is there any simple example of such a function that obeys this rule that would give some intuition?.

Furthermore I can't understand why the function need be bijective doesn't the definition of the sets being partially ordered imply infectivity? If not are order isomorphisms not applicable to strict total orders?.

Thanks in advance.

Answer 1

Look at the set $S = \{0, 1, 2\}$ with partial order given by $0 \le 0$, $1 \le 1$, $2 \le 2$, $0 \le 1$, $0 \le 2$, and nothing else. You might represent this partially ordered set with a V, where the bottom of the V is $0$, and the two heads of the V are $1$ and $2$.

Now look at another set $T = \{3, 4, 5\}$ with partial order given by $3 \le 3$, $4 \le 4$, $5 \le 5$, $3 \le 4$, $3 \le 5$, and nothing else. Again, you can represent this with a V, where the bottom is $3$, and the heads are $4$ and $5$.

You may complain that these two partially ordered sets are really "the same", just that the names of the elements are different. In particular, if we map $x \in S$ to $x+3 \in T$, then we seem to be able to identify (= make identical) the two partially ordered sets.

How do we formalize this notion of being "the same up to renaming of elements"?

Well, renaming elements is just specifying a bijection, because a bijection gives you a one-to-one correspondence between elements of one set and elements of the other set.

But it cannot be just any bijection. It has to preserve the partial order. For example, you wouldn't consider the function mapping $x \in S$ to $5-x \in T$ to be an appropriate renaming, because $0$ and $1$ are mapped to $5$ and $4$ respectively, and $0 \le 1$, but $5 \not\le 4$.

So, "preserving the partial order" means that $x \le y$ if and only if $f(x) \le f(y)$, where $f$ is our bijection.

Here's a little exercise if you made it to the end: can you show that there are exactly $2$ order-isomorphisms from $S$ to $T$?

Answer 2

If there is an order isomorphism between two posets, a good picture to have in mind is that they are the same poset, just with differently labelled vertices. If $S$ and $T$ are order isomorphic with isomorphism $f:S\to T$, this means that for every $s\in S$, we an "relabel it" with $f(s)$. What we get when we do this is exactly a copy of $T$. What this means is that $s_1 \leq s_2 \iff f(s_1)\leq f(s_2)$, so all the order relationships that exist in $S$ also exist in $T$. Picture $S$ and $T$ as being directed graphs. Them being order isomorphic means that the graphs are the same shape.

To illustrate why the map needs to be bijective, consider for example the posets $\mathbb{Z}\to \mathbb{Z} \amalg \mathbb{Z}$ (where $\amalg$ is the disjoint union. The order on $Z$ is the standard order, and the order on $\mathbb{Z} \amalg \mathbb{Z}$ is such that elements are comparable only when they come from the same copy of $\mathbb{Z}$. This is essentially just taking two copies of the poset $\mathbb{Z}$ and putting them next to each other. Then the map $f:\mathbb{Z}\to \mathbb{Z} \amalg \mathbb{Z}$ which just sends $\mathbb{Z}$ to the first factor. Then $f$ satisfies $a \leq b \iff f(a)\leq f(b)$ for all $a$ and $b$, but clearly we don't want to consider these two to be the same poset.