1. ## 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!

2. 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

3. 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?

4. 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