# Math Help - metric space

1. ## metric space

In a metric space $(S,d)$, let $A$ and $B$ be disjoint closed subsets of $S$. Prove that $\exists$ disjoint open sets $U$ and $V$ of $S$ such that $A \subset U$ and $B \subset V$.

2. Let $r=\inf_{x\in{A},y\in{B}}d(x,y)$, then $r>0$ since they are disjoint and closed (I'll leave it to you to show that this is true). Then for each point $x\in{A}$ create the neighborhood $N_{\frac{r}{3}}(x)$ (neighborhood about x with radius r/3), similarly create $N_{\frac{r}{3}}(y)$ for $y\in{B}$.

Finally take $U=\bigcup_{x\in{A}}N_{\frac{r}{3}}(x)$ and $V=\bigcup_{y\in{B}}N_{\frac{r}{3}}(y)$.

3. Originally Posted by putnam120
Let $r=\inf_{x\in{A},y\in{B}}d(x,y)$, then $r>0$ since they are disjoint and closed
This is not true: Take the graph of $f(x)= \frac{1}{x^2}$ in $\mathbb{R} ^2$ it is like two hyperbolas that get arbitrarily close as you approach $0$ (they both go to infinity). It is clearly closed (each 'hyperbola' is also closed), and the two 'hyperbolas' are disjoint but $\inf =0$. for this to work either $A$ or $B$ must be compact

4. I don't think that $f(x)=\frac{1}{x^2}$ is closed. There is no point so that $f(x)=0$. Or is there some other definition of closed that you are using?

5. Originally Posted by putnam120
I don't think that $f(x)=\frac{1}{x^2}$ is closed. There is no point so that $f(x)=0$. Or is there some other definition of closed that you are using?
I'm saying the graph of $f$ is (topologically) closed as a subset of $\mathbb{R} ^2$ (it's easy to check the complement is open).

6. Originally Posted by Kat-M
In a metric space $(S,d)$, let $A$ and $B$ be disjoint closed subsets of $S$. Prove that $\exists$ disjoint open sets $U$ and $V$ of $S$ such that $A \subset U$ and $B \subset V$.
For each point $x\in S$, define $d(x,A) = \inf\{d(x,s):s\in A\}$. Then $d(x,A) = 0\ \Longleftrightarrow\ x\in A$. (That is because A is closed. In fact, that condition is equivalent to A being closed.) Similarly, define $d(x,B) = \inf\{d(x,s):s\in B\}$. Again, this will be 0 if and only if $x\in B$.

Now let $U = \{x\in S: d(x,A) and $V = \{x\in S: d(x,A)>d(x,B)\}$. All you have to do now is to show that U and V are open.

7. ## closed set

Originally Posted by Opalg
For each point $x\in S$, define $d(x,A) = \inf\{d(x,s):s\in A\}$. Then $d(x,A) = 0\ \Longleftrightarrow\ x\in A$. (That is because A is closed. In fact, that condition is equivalent to A being closed.)
so if $x \in S$ is an elemnt of $B$, $d(x,A) > 0$ since $x \notin A$. Is that correct? if so, doesnt putnam120's argument work?

8. Originally Posted by Kat-M
so if $x \in S$ is an elemnt of $B$, $d(x,A) > 0$ since $x \notin A$. Is that correct? if so, doesnt putnam120's argument work?
That is correct: if $x\in B$ then $d(x,A) > 0$. But putnam120's argument doesn't work for the reason pointed out by Jose27. The distance from B to A can be zero, as shown by the example of the two components of the graph of y=1/x^2. But the distance from any individual point in B to A must be strictly positive.

9. Originally Posted by Kat-M
so if $x \in S$ is an elemnt of $B$, $d(x,A) > 0$ since $x \notin A$. Is that correct? if so, doesnt putnam120's argument work?
My argument doesn't work because I took the $\inf$ over all pairs $x\in{A}$ and $y\in{B}$. But Opalg's argument you look at a particular $x\in{S}$. To see that $d(x,A)>0$ if $x\notin{A}$ just use the limit point definition of a closed set.

I have a question for Jose27: How are you able to see that my original argument would work if one of A or B were compact?

10. Originally Posted by putnam120
My argument doesn't work because I took the $\inf$ over all pairs $x\in{A}$ and $y\in{B}$. But Opalg's argument you look at a particular $x\in{S}$. To see that $d(x,A)>0$ if $x\notin{A}$ just use the limit point definition of a closed set.

I have a question for Jose27: How are you able to see that my original argument would work if one of A or B were compact?
Let me get into this nice discussion. If A is compact $d(\cdot,B):A\to R^+$, $x\mapsto d(x,B)$ is a continuous function, hence the infimum is attained (it is a minimum I mean). It can't be 0, if this were the case x would belong to $B$ since $B$ is assumed to be closed, contradicting the assumed disjointness.