Showing there is a bijection from all open subsets to all closed subsets of $M$ (From Pugh's RMA) Let $\mathcal{T}$ be the collection of open subsets of a metric space $M$, and $\mathcal{K}$ be the collection of all closed subsets. Show there is a bijection from $\mathcal{T}$ to $\mathcal{K}$. 
I believe the bijection between $\mathcal{T}$ and $\mathcal{K}$ would be the function $f: \mathcal{T} \rightarrow \mathcal{K}$ that returns the interior of each closed subset; the inverse function would return the closure. 
I attempted to prove this using the Schroeder-Bernstein Thm: there is a bijection between $A$ and $B$ if there are injective functions $f: A \rightarrow B$ and $g: B \rightarrow A$. So I need to show that the closure and interior functions are injective. 
But then I realized the interior function (which takes a closed set $U$ and returns the intersection of all opens contained in $U$) isn't injective-- consider a an interval $[a,b] \in \mathbb{R}$ and consider $[a,b] \cup [p] \in \mathbb{R}$ ($[p]$ is an isolated point). They have the same interiors (I think). 
Where does my approach go wrong? Was it a mistake to choose the interior and closure functions to test for injectivity? How would you find the bijection between $\mathcal{T}$ and $\mathcal{K}$?
 A: A subset is open if and only if its complement is closed. This is everything you need to know to construct the bijection.
A: To explain your exercise, topology can be defined in many ways, common ways being via open sets, closed sets, interior operator, closure operator, neighbourhoods.  These are all equivalent, you'd usually choose one as definition and others as characterizations (i.e. there'd proof that they are equivalent.)
We only know you're given definition of metric (and metric spaces aren't the most general kind of of topological space), interior, closure, open and closed set.
I'd guess you can prove your exercise (but I might be wrong) from following definitions.  If they are not definitions in your material, you might need to prove them to be equal to definitions given in your material.
For any subset $S$ of space $X$.


*

*Set is open only if it's interior of itself.  $S = Int(S)$

*Set is closed only if it's closure of itself. $S = Cl(S)$

*Set is closed if and only if it's complement is open.

*Complement of closure is interior or complement. $X \ Cl(S) = Int (X \ S)$


I think your exercise could well be some combination of above ideas.  Many more are possible.  I've possibly given too much away, but I think mostly it was answered above as well.
I'd like to note that the reason your idea won't work is the above idempotence of closure(interior).  They are adjoint (or Galois connections) to insertions of closed (open) sets to powerset of space, as you've realised, this connection is in general weaker than isomorphism, or bijectivity being enough in this case.
A: Yes, the interior won't get the job done, for the reasons you described.
Fortunately, open and closed sets are much more fundamentally related, and we don't need interiors, or closures, or anything like that.
Open and closed are in some sense inverse properties, right? One defined in terms of the other. 
If you're given a set, what's the most fundamental way to check if it's closed?
A: The complement $L^c$ of an open set $L$ is closed. Thus the map $L \to L^c$ is a bijection between open and closed sets.
