Assuming P != NP, what is the cardinality of the set of NP-Hard languages?

If P=NP, then every non-trivial language is NP-Hard, so clearly there are uncountably many NP-Hard languages. However it's less clear to me what the cardinality of this set is assuming P != NP.

• Hint: The class NP is clearly countable and a superset of the complement of NP-hard languages. Commented Dec 3, 2019 at 13:28
• That's not true though. Assuming P != NP, there exist languages which are neither NP nor NP-Hard, uncountably many in fact. For example, if P != NP, no unary language is NP-Hard. Take any unary encoding of an undecidable language and it's neither NP nor NP-Hard.
Any upwards-closed non-empty class $$\mathfrak{L}$$ of languages has the cardinality of the continuum, with very few limitations on what kind of reasonable reducibility we are looking at. The reason is if $$A \in \mathfrak{L}$$ and $$B$$ is an arbitrary language, then the language $$A + B = \{0w \mid w \in A\} \cup \{1w \mid w \in B\}$$ satisfies that $$A \leq A + B$$, hence $$A + B \in \mathfrak{L}$$. Since $$A + B = A + C$$ iff $$B = C$$, we see that $$B \mapsto A + B$$ provides an injection from the all languages into $$\mathfrak{L}$$.