Does L contain the real numbers? On the one hand, there is no shortage of exposition about real numbers within the constructible hierarchy $L$, for example this thread. On the other hand you have the model $L(\mathbb{R})$, which would seem redundant if $L$ already contained the real numbers (or I guess to be precise, a set order-isomorphic to the real numbers).
After puzzling over this a bit, what I think I understand is: $L$ doesn't actually contain all the real numbers (the entire set $\mathbb{R}$ as it's normally thought of in analysis, etc.), but rather $L$ contains only what one might call the "constructible" real numbers.
Is this right, or is there some other subtlety that I'm missing?
 A: Well, it is consistent that $L$ contains all the real numbers, e.g. if we assume $V=L$ holds true in our universe. In fact, if we force over $L$ by a forcing that does not add new reals (e.g. adding a branch to a Suslin tree; shooting a club into some stationary subset of $\omega_1$; adding some Cohen subset to a regular uncountable cardinal $\kappa$), then all the reals in that universe will also be in $L$.
But at the same time, if we add a Cohen real to $L$, then we very clearly have new reals that are not in $L$. If we collapse $\omega_1$ to be countable, then the set of reals in $L$ is itself countable, so clearly some new reals have been added.
Large cardinal hypotheses can also tell us something about the real numbers and $L$. If $0^\#$ exists, which follows from the existence of a measurable cardinal, then we can actually show that the reals in $L$ only form a countable set. In fact, $0^\#$ is a kind of tipping point in "how close the universe is to being $L$". So sufficiently large large cardinal axioms imply that the universe is quite far from $L$, and this starts already with the real numbers.
We often work in models like $L(\Bbb R)$ when studying descriptive set theory, and these are taken in models that are either generic extensions of $L$, e.g. the Solovay model, or under the assumption of large cardinals, e.g. infinitely many Woodin cardinals and then some.
On the other hand, we can prove that if there is a perfect set of reals which are all in $L$, then all the reals are in $L$.
So, which is it? Are all the reals in $L$ or not? Well, that depends on your universe.
