**Example:** **infinite models rule out certain kinds of proofs**. If you are interested in a property $P$, formulated in the first-order logic of graph theory, and if you can prove the existence of an infinite graph satisfying $P$, then this proves that a conjecture of the form $C(P)$="there does not exist a *finite* graph satisfying $P$" can *never* be proved by reasoning with (0) the axioms of the language of graph theory, (1) your statement $P$, (2) the usual inference rules of first-order logic.

In this sense, awareness of infinite graphs can have a very practical consequence:
you then know that $C$ cannot be proved this way, and---if at all---you will only ever prove it by making essential use of the *finiteness* of the vertex set.

This may appear tautological in this generality, but in practice, it can have real value.

**Example:** there is an (in)famous open problem about *finite* triangle-free graphs, where, for each $n\in\mathbb{N}$,

$P:=P(n):=$"the graph is triangle-free, and is triangle-freeness-preservingly $n$-existentially-closed".

This predicate is easily expressed in the first-order logic of graph theory; the latter property is often abbreviated $n$-e.c. in the literature. Being triangle-freeness-preservingly so means that of course only *those* extension-properties are required to hold which do not immediately create a triangle.

Then $C(P)$ is known to be *false* for each^{0} $n\in4$, in view of explicitly known (and even infinite) families of finite graphs satisfying $P(n)$.
Up to an including $n=3$, many such graphs are known, though they appear not to be completely classified^{1,2}.
But the truth-value of $C(P(4))$ is notoriously open; the most direct approach would be to say "Hey let's go and write $F:=$"the graph is triangle-free", and $E(n):=$the graph is triangle-freeness-preservingly $n$-e.c.", note $P(n)=F\wedge E(n)$, and prove $C(P(4))$ by simply proving via first-order inference-rules that $E(4)\Rightarrow\neg F$." It is not clear a priori that this attempt might not simply work, with the first-order derivations being maybe a bit long, but manageable---however, *this* way ,the mission will be impossible: in a famous paper^{3} Henson proved that there exists a countably-**infinite** triangle-free graph which is triangle-freeness-preservingly $4$-e.c. (and much more than that). Therefore, the (conjectural) $C(P(4))=\mathrm{true}$ cannot be proved in the above straightforward way. Here, infinite graphs prevent you from attempting the impossible, and help you focus on trying other methods of proof. If you would not be aware of the concept of infinite graphs, e.g. by being taught that *all* graphs are finite, this argument would not be available to you.

_{Footnotes.}

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0. Needless to say, for $n=0\in4$, we have $P(0)$ $=$"the graph is triangle-free", and it is evidently false that there does not exist a finite graph satisfying $P(0)$.
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1. G. Cherlin, Two problems on homogeneous structures, revisited. In: Model Theoretic Methods in Finite Combinatorics, M. Grohe and J.A. Makowsky eds.,Contemporary Mathematics, 558, American Mathematical Society, 2011, and also
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2. C. Even-Zohar and N. Linial: Triply Existentially Complete Triangle-Free Graphs. J. of Graph Theory, 78 (4), 2015, 305--317
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3.
C. W. Henson. A family of countable homogeneous graphs, Pacific Journal of Mathematics, 38 (1971) 69–83 }