# metric space

• Aug 14th 2009, 04:43 PM
Kat-M
metric space
In a metric space $\displaystyle (S,d)$, let $\displaystyle A$ and $\displaystyle B$ be disjoint closed subsets of$\displaystyle S$. Prove that $\displaystyle \exists$ disjoint open sets $\displaystyle U$ and $\displaystyle V$ of $\displaystyle S$ such that $\displaystyle A \subset U$ and $\displaystyle B \subset V$.
• Aug 14th 2009, 05:15 PM
putnam120
Let $\displaystyle r=\inf_{x\in{A},y\in{B}}d(x,y)$, then $\displaystyle r>0$ since they are disjoint and closed (I'll leave it to you to show that this is true). Then for each point $\displaystyle x\in{A}$ create the neighborhood $\displaystyle N_{\frac{r}{3}}(x)$ (neighborhood about x with radius r/3), similarly create $\displaystyle N_{\frac{r}{3}}(y)$ for $\displaystyle y\in{B}$.

Finally take $\displaystyle U=\bigcup_{x\in{A}}N_{\frac{r}{3}}(x)$ and $\displaystyle V=\bigcup_{y\in{B}}N_{\frac{r}{3}}(y)$.
• Aug 14th 2009, 05:29 PM
Jose27
Quote:

Originally Posted by putnam120
Let $\displaystyle r=\inf_{x\in{A},y\in{B}}d(x,y)$, then $\displaystyle r>0$ since they are disjoint and closed

This is not true: Take the graph of $\displaystyle f(x)= \frac{1}{x^2}$ in $\displaystyle \mathbb{R} ^2$ it is like two hyperbolas that get arbitrarily close as you approach $\displaystyle 0$ (they both go to infinity). It is clearly closed (each 'hyperbola' is also closed), and the two 'hyperbolas' are disjoint but $\displaystyle \inf =0$. for this to work either $\displaystyle A$ or $\displaystyle B$ must be compact
• Aug 14th 2009, 07:47 PM
putnam120
I don't think that $\displaystyle f(x)=\frac{1}{x^2}$ is closed. There is no point so that $\displaystyle f(x)=0$. Or is there some other definition of closed that you are using?
• Aug 14th 2009, 08:00 PM
Jose27
Quote:

Originally Posted by putnam120
I don't think that $\displaystyle f(x)=\frac{1}{x^2}$ is closed. There is no point so that $\displaystyle f(x)=0$. Or is there some other definition of closed that you are using?

I'm saying the graph of $\displaystyle f$ is (topologically) closed as a subset of $\displaystyle \mathbb{R} ^2$ (it's easy to check the complement is open).
• Aug 15th 2009, 12:03 AM
Opalg
Quote:

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

For each point $\displaystyle x\in S$, define $\displaystyle d(x,A) = \inf\{d(x,s):s\in A\}$. Then $\displaystyle d(x,A) = 0\ \Longleftrightarrow\ x\in A$. (That is because A is closed. In fact, that condition is equivalent to A being closed.) Similarly, define $\displaystyle d(x,B) = \inf\{d(x,s):s\in B\}$. Again, this will be 0 if and only if $\displaystyle x\in B$.

Now let $\displaystyle U = \{x\in S: d(x,A)<d(x,B)\}$ and $\displaystyle V = \{x\in S: d(x,A)>d(x,B)\}$. All you have to do now is to show that U and V are open.
• Aug 15th 2009, 11:24 AM
Kat-M
closed set
Quote:

Originally Posted by Opalg
For each point $\displaystyle x\in S$, define $\displaystyle d(x,A) = \inf\{d(x,s):s\in A\}$. Then $\displaystyle d(x,A) = 0\ \Longleftrightarrow\ x\in A$. (That is because A is closed. In fact, that condition is equivalent to A being closed.)

so if $\displaystyle x \in S$ is an elemnt of $\displaystyle B$, $\displaystyle d(x,A) > 0$ since $\displaystyle x \notin A$. Is that correct? if so, doesnt putnam120's argument work?
• Aug 15th 2009, 11:41 AM
Opalg
Quote:

Originally Posted by Kat-M
so if $\displaystyle x \in S$ is an elemnt of $\displaystyle B$, $\displaystyle d(x,A) > 0$ since $\displaystyle x \notin A$. Is that correct? if so, doesnt putnam120's argument work?

That is correct: if $\displaystyle x\in B$ then $\displaystyle d(x,A) > 0$. 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.
• Aug 15th 2009, 12:39 PM
putnam120
Quote:

Originally Posted by Kat-M
so if $\displaystyle x \in S$ is an elemnt of $\displaystyle B$, $\displaystyle d(x,A) > 0$ since $\displaystyle x \notin A$. Is that correct? if so, doesnt putnam120's argument work?

My argument doesn't work because I took the $\displaystyle \inf$ over all pairs $\displaystyle x\in{A}$ and $\displaystyle y\in{B}$. But Opalg's argument you look at a particular $\displaystyle x\in{S}$. To see that $\displaystyle d(x,A)>0$ if $\displaystyle x\notin{A}$ 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?
• Aug 15th 2009, 02:53 PM
Enrique2
Quote:

Originally Posted by putnam120
My argument doesn't work because I took the $\displaystyle \inf$ over all pairs $\displaystyle x\in{A}$ and $\displaystyle y\in{B}$. But Opalg's argument you look at a particular $\displaystyle x\in{S}$. To see that $\displaystyle d(x,A)>0$ if $\displaystyle x\notin{A}$ 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 $\displaystyle d(\cdot,B):A\to R^+$, $\displaystyle x\mapsto d(x,B)$ 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 $\displaystyle B$ since $\displaystyle B$ is assumed to be closed, contradicting the assumed disjointness.