0
$\begingroup$

I have already proven the following statement:

Let $(X,d)$ be a metric space and let $B(a;\delta)$ and $B(b;\eta)$ be two open balls about $a$ and $b$ respectively, with $a,b\in X$. Then if $$\delta+\eta\lt d(a,b)$$ it follows that $$B(a;\delta)\cap B(b;\eta)=\varnothing$$

However, I would like to extend this theorem to the union of some arbitrary number of open balls. What do the givens need to be for us to be able to conclude that $$B(a_i,\delta_i)\cap B(a_j,\delta_j)=\varnothing$$ for all $i,j$? Must it be given that $$\delta_i+\delta_j\lt d(a_i,a_j)$$ for each $i,j$, or can I get away with fewer givens than that?

$\endgroup$
  • $\begingroup$ Hi, you mean you want the condition for the intersection of all the balls to be empty, or the pairwise intersection to be empty,etc? $\endgroup$ – gary Jul 15 '17 at 23:47
  • 3
    $\begingroup$ If just a single intersection is empty, then that entire n-fold intersection is empty; so you'd need $| \delta_i - \delta_j| < d(a_i, a_j)$ for just one pair $(i,j)$. $\endgroup$ – Kaj Hansen Jul 15 '17 at 23:49
  • $\begingroup$ @KajHansen Sorry, I miswrote the question. I meant that every pair should be disjoint. $\endgroup$ – Frpzzd Jul 15 '17 at 23:51
1
$\begingroup$

(I). If $Y$ is a subset of $X$ with at least $2$ members and if $\{B_d(y,r_y):y\in Y\}$ is a pair-wise disjoint family, with each $r_y>0,$ then $B_d(y,r_y)\cap (Y$ \ $\{y\})=\phi$ for every $y\in Y$, so for every $y\in Y$ we have $\inf \{d(y,y'): y\ne y'\in Y\}\geq r_y>0.$ So $Y$ must be a discrete sub-space of $X.$

(II). If $d$ is the discrete metric on $X,$ that is if $d(x,x')=1$ whenever $x,x'$ are unequal members of $X,$ then all sub=spaces of X are discrete. And we can let $r_y=1$ for each $y\in Y.$ In this case, when $y,y'$ are unequal members of $Y,$ we have $r_y+r_{y'}=2>1= d(y,y').$

(III). In general, $r+r'> d(y,y')>0$ does not always imply $B_d(y,r)\cap B_d(y',r')\ne \phi$ even when $\sup \{d(y,z): z\in B_d(y,r)\}=r$ and $\sup \{d(y',z'):z'\in B_d(y',r').$ For example with $X=[0,1]\cup [2,3]$ with the usual metric $d(x,x')=|x-x'|$, let $y=1,y'=2,$ and $r=r'=1.$

(IV). On the other hand,for some metric spaces, $B_d(y,r)\cap B_d(y',r')=\phi$ implies $r+r'\geq d(y,y').$ For example, $X=\mathbb R.$ In this example, let $Y=\mathbb Z.$ Let $r_y=1/2$ for all $y\in Y.$ We have $r_y+r_{y+1}=1=d(y,y+1)$ and $B(y,1/2)\cap B(y',1/2)=\phi$ for all distinct $y,y' \in Y$.

(V). A converse to paragraph (I) above: If $Y$ is a discrete sub-space of $X$ with at least $2$ members, then for each $y\in Y$ let $r_y=\frac {1}{3}\inf \{d(y,y'):y\ne y'\in Y\}.$ A little bit of work with the triangle equality will show that $B_d(y,r_y)\cap B_d(y',r_{y'})=\phi$ when $y,y'$ are unequal members of $Y.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.