How to show the cardinality $0$ is less than the cardinality $1$ I know that to show this I have to show there’s an injection from the $\varnothing$ to $\{\varnothing\}$.
But how can one even define a function on the $\varnothing$?
 A: 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.
A: General idea: in order 2 conditions to be both satisfied, it is already necessary the first one to be satisfied. 
So if non-injectivity requires 2 conditions and if the first one cannot be satisfied, non-injectivity does not hold , and therefore, automatically , injectivity holds. 

Let f be a function from  {  }  to  { {  } }. 
Let me " define" f arbitrarily like this : for all x, f(x) = 2x. 
Now, whatever its " definition" may be, f is a relation, and therefore, in this case, a subset of the cartesian product : 
                 "{  }  cross { {  } }"

that is , the set of all pairs (x,y) such that (1) x belongs to Empty Set and (2)  y belongs to { {  }  }. 
But, since condition (1) cannot be satisfied, a fortiori, there is no pair that satisfies both conditions, so my cartesian product is the Empty set. 
I know that function f, that is a relation,  is a subset of a cartesian product that is empty. But the only subset of the empty set is the empty set itself. So : function f such that f(x)=2x is identical to the Empty set. 
( That would have been true whatever formula I would have used to  "define" f). 
Now, is f injective? 
I know that f is injective just in case : 
IF a pair (a, c) belongs to f ( whatever this pair may be)  , THEN there is no pair ( b, c) belonging to f with element a different from  element b. 
( indeed, if we had at the same time (a, c) and ( b,c) with a different from b, then 2 diffetent elements of the domain would have the same image, and f would not be injective). 
You know that a statement of the form (P --> Q) is false just in case P is true and Q is false. Let's apply this logical fact to our definition of injectivity which is an IF...THEN statement. 
Let's see whether it is possible for f not to be injective. 
For f NOT to be injective , it would be required that : 
(1) there is actually a pair (a, b) belonging to f [ this is the " P is true" part] 
AND 
(2) there is ( at least) a pair (b, c) belonging to f with a different from b  [ this is the "Q is false" part, for, remember, Q was " there is no pair..." ] 
But it is impossible that both consitions hold, since it is already impossible the first one to hold: f being empty, there is no pair (a,c)  that belongs to f, since there is nothing that belongs to f. 
So the proposition " f is NOT injective" is FALSE ( and even necessarily false). 
Consequently, the proposition " f is injective" must be true. 
