How to show the cardinality $0$ is less than the cardinality $1$

A function, $f:X\to Y$ is a subset of a cartesian product, $X\times Y$ where each element $a \in X$ is "represented" precisely once.

$h:\emptyset \to \{\emptyset\} \subseteq \emptyset \times \{\emptyset\} = \{(x,y)|x \in \emptyset, y \in \{\emptyset\}\}$. But as $\emptyset$ is empty $x \in \emptyset$ is impossible so so $\emptyset \times \{\emptyset\} = \{(x,y)|x \in \emptyset, y \in \{\emptyset\}\} = \emptyset$.

So $h= \emptyset$. Is $\emptyset$ injective? What does that mean?

$f\subseteq X\times Y$ is injective if for any $b\in Y$ there is at most one $(x,b)\in f$.

So for any $b \in \{0\}$ how many $(x, b) \in h = \emptyset$. Well, zero as $h$ is empty. So is there at most one? Well, there sure as heck aren't more than one!

....

The emptyset is always an empty function from $\emptyset \to $ any set $X$ and it is vacuously injective.

Vacuous statements are, admittedly, irritating and confusing and you have my empathy. If it helps, you can put a mental close in most definitions as "An $x$ is GLOOP if when it exists GOBBLETY occurs. And when it doesn't exist it is GLOOP by default. That's kind of cheating but it helps.

The old stumbling block is "Every element in the emptyset is red". That is true. Every element of the emptyset, all zero of them, are red because there aren't any elements. But if we say "Every element in the emptyset, if it exists, is red" that's very clear. Unfortunately it's inaccurate because the element doesn't exist.... well, it's still technically accurate to say it is red.... sigh... it gets easier.

Althernatively, Every element is GLOOP $\iff $ there is no element that is not GLOOP. It's very clear that the RHS is always true for an emptyset as there no elements at all.