Is not first order axiomatizable? I have the following question: ¿Being a: connected graph, Archimedean field, characteristic field 0,  well order, is not definable first order?, if the answer were true we would have to show that there is no first order $\varphi$  statement (in proper language) such that $\mathfrak{A}⊨\varphi$ if and only if $\mathfrak{A}$ is connected graph, characteristic field 0, etc. I need help to prove or disprove this question.
 A: The case of well-orders is a good example of this this kind of argument usually runs. Let $LO$ be the theory of linear orders, $(c_i)_{i\in\omega}$ be a new set of constants and for each $i\in\omega$ let $\varphi_i$ be the sentence in $c_i<\ldots <c_{1}<c_0$ in the expanded language.
If there were a sentence $\psi$ that a structure satisfied exactly when it's a well-order, then $(\omega,<)$, for example, would satisfy it; but we can also interpret any finite number of the $\varphi_i$ in $(\omega,<)$, which means that $LO\cup\{\psi\}\cup\{\varphi_i:i\in\omega\}$ is satisfiable, so there's a model $M$ of this theory.
But notice that $M$ cannot be well-founded on account of there being an infinite strictly decreasing sequence through the ordering. Yet it must satisfy $\psi$, a contradiction; so there is no such $\psi$.
The other examples work similarly; you use compactness to show that there is a structure in which your special property (well-foundedness, connectedness, being archimedean) fails, but which still satisfies all the same sentences as some structure which does satisfy that property.
A: In addition to the compactness technique mentioned in the other answer, we can use Ultraproducts or Games to show that two structures satisfy all the same first order sentences. Then by arguing that one of the structures is well ordered/connected/characteristic 0/etc. while the other isn't, we get the claim.
Let's start with ultraproducts. The important theorem in this area is that, given some sequence of models $\mathfrak{A}_n$, we can find a model $\mathfrak{A}$ (their ultraproduct) so that $\mathfrak{A} \models \varphi$ if and only if "many" $\mathfrak{A}_n \models \varphi$.
Here "many" is measured by some ultrafilter, but for most cases of interest we get

*

*If all but finitely many $\mathfrak{A}_n \models \varphi$, then the ultraproduct $\mathfrak{A} \models \varphi$ too.


*If $\mathfrak{A} \models \varphi$, then infinitely many $\mathfrak{A}_n \models \varphi$.

So, for instance, if we want to show that "characteristic 0" isn't first order in the language of rings. We might proceed as follows:
Let $\mathfrak{A}_p$ be the field $\mathbb{Z}/p$ for every prime $p$. We can write down a sentence $\varphi_q$ which says "I am not characteristic $q$", and all but finitely many of the $\mathfrak{A}_p$ will satisfy $\varphi_q$ (indeed, all but $p = q$ will satisfy it).
Then when we ultraproduct the $\mathfrak{A}_p$ together, the resulting model $\mathfrak{A}$ cannot be of characteristic $p$ for any $p$. It must be a field, though, since all but finitely many (indeed all) of the $\mathfrak{A}_p$ satisfy the field axioms. So $\mathfrak{A}$ is a field of characteristic $0$.
Now if any $\varphi$ individually picked out the fields of characteristic $0$, every $\mathfrak{A}_p$ would model $\lnot \varphi$. This means $\mathfrak{A}$ would have to model $\lnot \varphi$ too. A contradiction.

Another useful technique is that of Ehrenfeucht-Fraisse Games. Loosely, Alyss really likes some model $\mathfrak{A}$. Bob likes Alyss, but only has access to some model $\mathfrak{B}$. He (perhaps misguidedly) has decided to trick Alyss into thinking he has access to model $\mathfrak{A}$. She decides to test him with the following game:

*

*Alyss names a number $n$.

*Alyss names either an element of $\mathfrak{A}$ or $\mathfrak{B}$

*Bob names an element of whichever model Alyss didn't choose.

*Alyss again names an element of either model

*Bob responds.

*Play continues for $n$ rounds, and Bob wins if and only if the $n$ elements of $\mathfrak{A}$ is the same (finite) structure as the $n$ elements of $\mathfrak{B}$.

If Bob can successfully trick Alyss with this game, then (theorem) $\mathfrak{A}$ and $\mathfrak{B}$ satisfy all the same first order sentences.
Ok, why is this useful? Let's look at two graphs:

*

*$A$ an infinite path (think $\mathbb{Z}$ where $n$ is adjacent to $n \pm 1$)

*$B$ the disjoint union of two infinite paths.

I claim Bob can successfully win against Alyss in the above game. Why?
Whenever Alyss chooses some element of $A$, then Bob names the corresponding element in the first component of $B$. Whenever Alyss chooses some element of the first component of $B$, then Bob can again be honest and pick the corresponding element in $A$.
If, however, Alyss chooses some element of the second component of $B$, then Bob has to lie. But Bob knows how long the game will last. So he picks some element of $A$ that's really far away from all of the other elements that have been named so far. In particular he picks a big enough number that Alyss doesn't have enough turns left to connect that element to anything played before.
Since Bob has a winning strategy, $A$ and $B$ are elementary equivalent. But since $A$ is connected and $B$ isn't, we find there can be no first order formula defining connectedness.

I hope this helps ^_^
