# 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

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You are having difficulty showing that if two sets aren't equal then their complements aren't equal? –  Gerry Myerson Oct 25 '12 at 2:52
A set is open iff its complement is closed. –  Neal Oct 25 '12 at 2:58

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!)

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Thank you so much Brian. Much appreciated. Yes, I was sloppy on the title. Just fixed it. And sorry for the typo. Don't know what I was thinking. In answer to your question, the only open $U$ that is open and $f(U) =F$, where F is closed is if $U = F^c$, right? But why do I not feel that the reasoning is not enough for proving surjection? Sorry, if I am being too picky. $f$ is surjective if for each $F \in M$ there exists at least one $U \in M$ such that $f(U) =F$, right? Doest he reasoning above show that? –  user43901 Oct 25 '12 at 3:33
@user43901: Not to worry: I make my share of them. –  Brian M. Scott Oct 25 '12 at 3:34
@BrianMScott : any inputs on the previous comment? –  user43901 Oct 25 '12 at 19:58
@user43901: Sorry; I think that you edited the later part in after I saw the comment. Yes, $X\setminus F$ is the only possible choice for $U$, and it works: $f(X\setminus F)=F$. Since $F$ was an arbitrary closed set, $f$ is surjective. I don’t know what more you could possibly want: that argument shows that every $F\in K$ is $f(U)$ for some $U\in T$, which is exactly what it means for $f$ to be surjective. –  Brian M. Scott Oct 25 '12 at 20:19

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.

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