Simple question about posets. Suppose that $X$ is a set, and $\geq$ is an ordering on that set. To determine if  $(X, \geq)$ is a poset, we must reflexivity, transitivity, and anti-symmetry. My question is concerned with this anti-symmetry part. That is, we must check that for each $a, b \in X$, $a \geq b$ and $b \geq a$ implies $a=b$. But isn't this dependent on how $=$ is defined? That is, isn't $=$ defined as $a=b$ if and only if $a \geq b$ and $b \geq a$ to begin with? 
For example, what if we define an order $ \geq$ on $\mathbb{R}^{2}$ by, $(a_{1}, a_{2}) \geq (b_{1}, b_{2})$ if and only if  $(a_{1}, a_{2})$ has SOME element larger than SOME element of $(b_{1}, b_{2})$. Then $(2, 1) \geq (1, 2)$ and $(1, 2) \geq (2, 1)$. Now, these elements are obviously not equal under the traditional pointwise ordering on $\mathbb{R}^{2}$, but they could be seen as equal if we DEFINE equality as $(a_{1}, a_{2}) = (b_{1}, b_{2})$ iff $(a_{1}, a_{2}) \geq (b_{1}, b_{2})$ and $(b_{1}, b_{2}) \geq (a_{1}, a_{2})$? If so, it makes the entire exercise trivial, by why must we refer to some "traditional" ordering? Don't we DEFINE set equality in a way that makes anti-symmetry trivial?
 A: No, $=$ is actual identity, not some notion of equivalence defined by $a \le b \wedge \le a$. The anti-symmetry condition says:if $a, b \in P$, $a \le b$ and $b \le a$ then $a$ really is the same entity as $b$.
An ordered set $\mathsf{P} = (P, \preceq)$ where $\preceq$ is only reflexive and transitive is called a preorder. For preorders, distinct elements $a, b \in P$ can bear the relation $\preceq$ to each other. In that case, the equivalence relation $a \equiv b \iff a \preceq b \wedge b \preceq a$ is nontrivial. If $\mathsf{P}$ is in fact a poset, the equivalence relation is just actual identity.
Equality is identity, it's a primitive in first-order logic, and does not get redefined. If the example you gave worked, you wouldn't be (re)defining equality, you'd be defining an equivalence relation. 
Note that the relation you define is not even a preorder: it's not transitive. Instead of writing $\le$ which is misleading, let's write it as $R$. Then: $(0, 2) R (1, 0)$ and $(1, 0) R (-1, 1)$, but not $(0, 2) R (-1, 1)$.
A: Equality (identity) is essentially defined by the property that $a=b$ implies $\Phi(a)\iff \Phi(b)$ for any predicate $\Phi$. That is, two objects are equal only if they are the same, indistinguishable. 
