Is every function from $\aleph_0 \to \aleph_2$ bounded? If $f$ is a function $f:\aleph_0 \to \aleph_2$, does it mean that the range of f is bounded in $\aleph_2$? Does this hold for all regular cardinals?
 A: Yes.  Recall that given a cardinal $\kappa$ the cofinality of $\kappa$, $\mathrm{cf} ( \kappa )$, is the least cardinal $\mu$ for which there is an unbounded (cofinal) function $\mu \to \kappa$.  Regularity means that $\mathrm{cf} ( \kappa ) = \kappa$, and all successor cardinals are regular.

Added to indoctrinate others into Asaf's anti-Choice programme
Note, also, that there is a fair bit of Choice involved in the above.  Without the Axiom of Choice it is possible that that $\aleph_2$ is singular; in particular it could have cofinality $\omega$, meaning that there is an unbounded function $\aleph_0 \to \aleph_2$.  The answers to this old math.SE question and well as this slightly more recent MO question contain a wealth of information on the broader topic.
A: If $\kappa$ is a regular cardinal, and $\alpha$ is any ordinal less than $\kappa$, then every function $f:\alpha\to\kappa$ is bounded. In particular, any function $f:\omega\to\omega_2$ is bounded.
More generally, all functions $f:\alpha\to\kappa$ will be bounded if $\alpha<\operatorname{cf}\kappa$, the cofinality of $\kappa$.
A: One of the definitions of a cardinal $\kappa$ being regular is that, whenever $\alpha < \kappa$, every function $f : \alpha \to \kappa$ is bounded.
In any case, you can prove this directly, using the fact that a countable union of sets of cardinality $\aleph_1$ has cardinality $\aleph_1$: consider $$\bigcup_{n < \omega} \{ \alpha < \omega_2 : \alpha \le f(n) \}$$
