Finding inf

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March 26th 2011, 05:41 PM
Connected

Finding inf

Let $A=\left\{\dfrac1n:n\in\mathbb N\right\},$ $B=(-1,0]$ and $d(x,y)=|x-y|,$ then find $d(A,B)=\underset{\begin{smallmatrix} a\in A \\ b\in B \end{smallmatrix}}{\mathop{\inf }}\,d(a,b)=\underset{\begin{smallmatrix} a\in A \\ b\in B \end{smallmatrix}}{\mathop{\inf }}\,\left| \dfrac{1}{n}-b \right|.$

I know it's zero, but don't know exactly how to prove it.

Thanks for the help!
March 26th 2011, 06:47 PM
tonio

Quote:

Originally Posted by Connected

Let $A=\left\{\dfrac1n:n\in\mathbb N\right\},$ $B=(-1,0]$ and $d(x,y)=|x-y|,$ then find $d(A,B)=\underset{\begin{smallmatrix} a\in A \\ b\in B \end{smallmatrix}}{\mathop{\inf }}\,d(a,b)=\underset{\begin{smallmatrix} a\in A \\ b\in B \end{smallmatrix}}{\mathop{\inf }}\,\left| \dfrac{1}{n}-b \right|.$

I know it's zero, but don't know exactly how to prove it.

Thanks for the help!

Since $\displaystyle{0\in B\mbox{ and } \frac{1}{n}\xrightarrow [n\to\infty]{}0$ , you have that

$\displaystyle{\underset{\begin{smallmatrix} a\in A \\ b\in B \end{smallmatrix}}{\mathop{\inf }}\,d(a,b)=\underset{ b\in B}{\mathhop{\inf}} \lim\limits_{n\to\infty}\left|\frac{1}{n}-b\right|$ .

Tonio
March 26th 2011, 07:04 PM
Connected

Quote:

Originally Posted by tonio

Since $\displaystyle{0\in B\mbox{ and } \frac{1}{n}\xrightarrow [n\to\infty]{}0$ , you have that

$\displaystyle{\underset{\begin{smallmatrix} a\in A \\ b\in B \end{smallmatrix}}{\mathop{\inf }}\,d(a,b)=\underset{ b\in B}{\mathhop{\inf}} \lim\limits_{n\to\infty}\left|\frac{1}{n}-b\right|$ .

Tonio

Okay, this seems to be a property or something, which is exactly?
March 26th 2011, 07:08 PM
tonio

Quote:

Originally Posted by Connected

Okay, this seems to be a property or something, which is exactly?

No...it just follows from the definition. You could as well choose an arbitrary $\epsilon > 0$ and prove that

there exist $a\in A\,,\,\,b\in B\,\,s.t.\,\,d(a,b)\geq\epsilon$ , and again by definition 0 is the infimum.

Tonio