Notation of rational numbers and reals

Question

I know that $\mathbb Q$ is the rational numbers and $\mathbb R$ is the reals, but what is meant by $ ^\mathbb{Q}\mathbb{R}$ in the question here? In that question he uses the phrase

"I am trying to prove that $ ^\mathbb{Q}\mathbb{R}$~$\mathbb{R}$"

Answer 1

In the theory of categories I have encountered the notation $^IX$ under the name "copower".

So actually: $$^IX=\amalg_{i\in I}X_{i}$$where $X_i$ denotes a copy of $X$ for every $i\in I$ and $^IX$ is a coproduct of these objects.

If this can be applied here (I don't dare to guarantee that) then $^{\mathbb Q}\mathbb R$ can be interpreted as the set: $$\{\langle r,q\rangle\mid r\in\mathbb R,q\in\mathbb Q\}=\mathbb R\times \mathbb Q$$ A disjoint union of sets $\mathbb R$, and for every $q\in\mathbb Q$ we have a set $\mathbb R$.

Answer 2

In the context of cardinality, usually $^\mathbb{Q}\mathbb{R}$ and $\mathbb{R}^\mathbb{Q}$ both mean "the set of functions from $\mathbb Q$ to $\mathbb R$". (Wikipedia uses the second notation; here's an example of the first.) In general, both $^AB$ and $B^A$ can denote the set of functions from $A$ to $B$.

The first notation, $^AB$, has the nice feature that it puts $A$ and $B$ in the same order you say them in.

The second notation is very compatible with many other uses of superscripts. For example, $\mathbb R^2$ can be interpreted as "pairs of elements of $\mathbb R$" or "functions from $2$ to $\mathbb R$", and if $2$ is the set $\{0,1\}$ (as is the convention with ordinals) then these are in bijection with each other.

But sometimes this compatibility is a drawback. We might not want to think of $5^7$ as "functions from $7$ to $5$", even though there are exactly $5^7$ functions from $7=\{0,1,2,3,4,5,6\}$ to $5=\{0,1,2,3,4\}$. On the other hand, $^75$ is not overloaded at all... unless you start talking about tetration.