A bijection between the set of all open sets in $M$ and the set of all closed subsets of $M$, where $M$ is a metric space. Let $T$ be the collection of open subsets of a metric space $M$, and $K$ the collection of closed subsets. Show that there is bijection from $T$ onto $K$.
Here, I was thinking that 
if $f: T\rightarrow K $ is a function such that it maps any open set $U$ to its complement $U^c$, then we are going somewhere. Problem is, I can't figure out how to quite write it out.
Also, how do I show that it is injective and surjective? 
If we take two open sets $U_1$ and $U_2 \in$ T such that $U_1 \neq U_2$, then how do I show that the function $f$ maps these two open sets such that $f(U_1) \neq f(U_2)$?
As far as surjection goes, I am lost how to write it out.
Any help will be appreciated. thanks
 A: Your title is inaccurate: you want to construct a bijection between the set of all open sets in $M$ and the set of all closed subsets of $M$, not from one open set to one closed set. However, your function $f:T\to K:U\mapsto M\setminus U$ is just fine, and I’m not really sure what you’re having trouble with: checking that it’s a bijection between $T$ and $K$ is just a matter of paying attention to the definitions.
For example, to show that $f$ is injective, assume that $f(U)=f(V)$ for some $U,V\in T$ and show that this implies that $U=V$:

Suppose that $U,V\in T$ and $f(U)=f(V)$; then by definition $X\setminus U=X\setminus V$. But then $$U=X\setminus(X\setminus U)=X\setminus(X\setminus V)=V\;,$$ so $f$ is injective.

To show that $f$ is surjective, let $F$ be an arbitrary closed set, and find an open set $U$ such that $f(U)=F$; what is that $U$ going to have to be?
(By the way, the word that you want is complement; a compliment is a very different thing!)
A: In some cases, easier way to prove that a function is bijective is to show that it is invertible. (See Inverse of a Function exists iff Function is bijective.)
In your case you have $f \colon T \to K$ defined by $f(U)=U^c$.
It is relatively easy to see that $g \colon K \to T$ defined by $g(C)=C^c$ is inverse to this function.
Since $f^{-1}$ exists, the function $f$ is bijective.
