
Metric Spaces
I was having "trouble" with two problems. Well, i got a solution for the first (a while ago, i hope i can remember it for today), for the second i was not sure how to approach it.
1) Prove that any set in a metric space is an intersection of open sets.
Yes, a basic question, i know. but that's the problem. my professor said i'm thinking way too hard, and there is a very easy, elegant solution to this.
2) Let $\displaystyle A$ be a finite open subset of a metric space $\displaystyle M$. Prove that every point in $\displaystyle A$ is an isolated point of $\displaystyle M$
Definitions and theorems that may come in handy:
Definition: A point $\displaystyle x$ in a metric space is called isolated if the set $\displaystyle \{ x \}$ consisting of $\displaystyle x$ alone is open.
Theorem: A subset of a metric space is open if and only if it is expressible as a union of open balls.
(i think we can use this for the first problem, to prove the claim for singleton points).
Theorem: In a metric space any union of open sets is open.
Theorem: In a metric space a finite intersection of open sets is open.
If any other definitions or theorems are required, i can supply them. just ask. i think this should be enough though
Thanks

I have a solution for problem one. However, it can be regarded as an ugly proof because it is set theory and nothing about metric spaces plus I think it uses the axiom of choice.
First note that any singelton element can be expressed as an intersection of infinitely disks skrinking to that point.
Next (key step) if we have a union of an intersection of sets we can write it as an intersection of union of sets. So for example, $\displaystyle (A\cup B)\cap (C\cup D) = (A\cap C)\cup (A\cap D)\cup (B\cap C)\cup (B\cap D)$ (this is verified by the distribution law of unions and intersections). However, the problem is what happens if we have infinitely sets, possibly uncountably many, does this still work. I think it still works but the problem is that it is not longer a topology question now it is a set theory question, that is the ugliness of my solution because I change one area in math into another area in math.
But now we can prove it. Say $\displaystyle S$ is an nonempty set. Pick any $\displaystyle s\in S$ then we can write $\displaystyle s$ (a singleton) as $\displaystyle A_1 = \bigcap B_i$ where $\displaystyle B_i$ are open disks skrinking to $\displaystyle s$. Pick another point $\displaystyle t\in S$ and do the same idea. Now define $\displaystyle A = \bigcup A_i$ but here it can be an uncountable intersection :eek:, but using the thing that I said above we can write this union of intersections as intersections of union of sets. BUT, the union of open sets is open so we are taking an intersection of open sets. Q.E.D.
(What an ugly construction. (Puke) )

For #2, I am not sure what theorems you have.
I accustomed to seeing that definition used for general topspaces but not metric spaces. For finite sets there is a finite collection of open balls that are pairwise disjoint which covers the set. (i.e. each point is in its own ball and no two balls intersect.) Because A is open M\A is closed. So any point of A is not a limit point of M\A. For any $\displaystyle x \in A$ there must be an open set $\displaystyle O\,,\,x \in O\,\& \,O \cap M\backslash A = \emptyset $. But that means $\displaystyle O\, = \{ x\} $.

My DSL has gone haywire!
Sorry about that DSL goof; I misread your #1 the first time.
It is a theorem that the closure of any set is the intersection of open sets.
PROOF: Suppose that A is a closed set. Let $\displaystyle B(x;r)$ be the ball centered at x with radius r.
Now define $\displaystyle O_n = \bigcup\limits_{x \in A} {B\left( {x;\frac{1}{n}} \right)} $.
It is easy to see that $\displaystyle A \subseteq \bigcap\limits_n {O_n } $
Now suppose that $\displaystyle y \notin A$ then there is a ball $\displaystyle B(y;r) \cap A = \emptyset$.
There is a positive integer N such $\displaystyle \frac{1}{N} < r$.
If $\displaystyle y \in O_N$ then $\displaystyle \left( {\exists a \in A} \right)\left[ {y \in B\left( {a,\frac{1}{N}} \right)} \right]\quad \Rightarrow \quad a \in B(y;r)$ which is a contradiction.
Thus $\displaystyle y \notin \left( {\bigcap\limits_n {O_n } } \right)$ and so $\displaystyle A = \left( {\bigcap\limits_n {O_n } } \right)$.
In the case that A is open just add $\displaystyle O_o = A$ to the collection.