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I believe there is no infinite domain where this is true:

$(∀x)(∃y)[x < y < 1]$

It would need to be a domain that increases infinitely, but because of the < 1 that cannot happen. Could someone please explain?

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    Does $<$ have to be a total order? Does $1$ have to be n the domain? (If both is the case, there is *no* such domain as specialization to $x=1$ leads to a contradiction) – 2017-02-13
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    not sure what you mean by total order, but yes 1 has to be in the domain – 2017-02-13

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Assuming $<$ is required to be irreflexive and transitive, then $\exists y(1 < y < 1)$ cannot hold. So $\forall x \exists y (x < y < 1)$ fails with $x = 1$.

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    Does this mean that no infinite domain holds true? – 2017-02-13
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    There is no domain at all (finite or infinite) where this holds if $<$ is given to be an irreflexive and transitive relation on the domain and $1$ is given as an element of the domain. – 2017-02-13
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    If one does not need to be in the domain would there be one that works? – 2017-02-14
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    @user7252850 the fact that you use $1$ in your formula means that it (well, whatever you interpret it as) has to be in the domain! – 2017-02-14
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Rob is absolutely correct, but I would like to address another part of your question, as you seem to be claiming that if you have an infinite domain, you can't keep 'squeezing' more and more number between some $x< 1$ and $1$.

But that is not true: we can keep on squeezing numbers in there. Indeed, there are infinitely many numbers between any number $x < 1$ and $1$.

So, in particular, a claim like:

$\forall x (x < 1 \to \exists y (x < y \land y <1))$

can be satisfied by all the real numbers between $0$ and $1$, of which there are infinitely many. So that statement can have an infinite domain.

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    I haven't gotten into specifics in my discrete math class so I am not sure what < being irreflexive and transitive means – 2017-02-14
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    Irreflexive means that there is no $x$ such that $x < x$. Transitive means that if $x < y$ and $y < z$ then $ x – 2017-02-14