As in the title: the existence of which sets is implied by the axioms of $\mathsf{ZF}$? For example one such set would be the empty set whose existence is demanded by the Axiom of the Empty Set.
But are for example all ordinals also present in every model because their existence follows from the axioms? I presume the answer must be no as we can have countable models of $\mathsf{ZF}$ but there are uncountably many ordinals.
Or do ordinals just happen to be present in any standard model because of the structure of the model?
In response to Trevor's last paragraph: Trying to make my question mathematically precise, I would like to ask: apart from $P(x)=$ "$x$ is the empty set", which sets we know intuitively are present in every model. For example, we know there exists an infinite set. But can there be a model where the smallest infinite set is already uncountable so that there are no natural numbers $\omega$? And similarly, can there be a model where there is no set representing the real numbers?
Thank you for your help.
Second edit: After reading Trevor's edit I would like to rephrase my question as follows:
For which sets $s$ that we know, like e.g. empty set, $\omega$, $\mathbb R$, ordinals, does $M$ think that $s$ exists? Trying to write it as a formula: For which $s$ does $M \models \exists x (x = s)$.
Best Answer
As Asaf pointed out, it is not true that $\omega$ is in every model of ZF. It is not even true that every model $(M,E)$ of ZF has a set $X$ such that $(X,E \restriction X \times X)$ is isomorphic to $(\omega, \mathord{\in} \restriction \omega \times \omega)$.
Instead I will answer the question I think you may have meant to ask, namely "what sets are in every transitive model of ZF?"
The sets that are in every transitive model of ZF are the ones that are in $L_\alpha$, the $\alpha^\text{th}$ level of Goedel's constructible universe $L$, where $\alpha$ is least such that $L_\alpha$ satisfies ZF. This is because if $M$ is a transitive model of ZF then $L^M$ is a level of $L$.
By the way, it makes no formal sense to say "an axiom implies $X$" or even "an axiom implies the existence of $X$" where $X$ is a set. What does make sense is to say "an axiom implies the existence of a set $X$ with property $P$." This makes sense because "there exists a set $X$ with property $P$" is a statement. Axioms imply statements, not sets. Of course it should not be against the law to say imprecise things but I think in this case it may be causing you confusion.
EDIT:
ZF proves that the class of all finite ordinals, denoted $\omega$, exists as a set. If $(M, \in)$ is a transitive model of ZF, then $\omega^M$, which is the element of $M$ that the model $(M,\in)$ thinks is the class of all finite ordinals, is equal to $\omega$ itself. Moreover the finite ordinals of $M$ are exactly the finite ordinals.
The class of all ordinals of $M$, denoted $\text{Ord}^M$, is an initial segment of $\text{Ord}$, the class of all ordinals. If $M$ is a set rather than a proper class, then some ordinals will be missing from $\text{Ord}^M$. In this case $\text{Ord}^M$ is itself an ordinal.
By the Loewenhein–Skolem theorem, $M$ may be countable, in which case $M$ does not contain any uncountable ordinals and $\text{Ord}^M$ is a countable ordinal. However, by Cantor's theorem applied in $(M,\in)$, there are many ordinals in $M$ that are uncountable in $M$ even though they are countable in $V$. This (roughly) is known as Skolem's paradox.
Real numbers can be coded as subsets of $\omega$ in an absolute way, so because $\omega^M = \omega$, $\mathbb{R}^M$ is a subset of $\mathbb{R}$. However, equality can fail: for example, if $M$ is countable then $\mathbb{R}^M$ is countable and therefore cannot be equal to $\mathbb{R}$.