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We define $D(A,B) =infd(a, b) : a in A text and bin B$. I know $D$ is not metric. Take $A=1$ and $B=(0, 1)$ then $D(A, B) = 0 $ as $1$ is in closure of $B$ but $Aneq B$.

But I take particular example $A=frac12n + n : n in mathbbN$ and $B=mathbbN$ then my intuition says $D(A, B) =0$ but I am unable to prove this. Can anyone prove this analytically?

Thanks

Let $epsilon > 0$. Then there exists a sufficiently large natural number $N$ in which $epsilon > frac12n$ for every $n geq N$. So, pick any $n geq N$ and it follows that $n + frac12n in A$, $n in B$ and $$|n+ frac12n - n| = frac12n < epsilon.$$ Since $epsilon$ was an arbitrary positive real number, it follows that $D(A, B) = 0.$

For any $frac 12n + n in A$ we will have $n in B$ and $d(frac 12n+n, n)=frac 12n$. So $frac 12n subset d(a,b)$ and... remember. If $K subset M$ then $inf M le inf K$ (remember why?).

So $0le D(A,B) =inf d(a,b)le inf frac 12n= 0$.

That wasn't the most efficient way or direct way, but it was the must intuitive and easiest (IMO).

Doing it directly is pretty easy too.

1) $0$ is a lower bound of $a in A, bin B$

Pf: If $frac 12n + n in A$ and $min B$ then $d(a,b) = |frac 12n + (m-n)|$. As $0< frac 12n < 1$ we know $frac 12n not in mathbb Z$ and $m-n in mathbb Z$ so $d(a,b) =|frac 12n - (m-n)| not in mathbb Z$ so $d(a,b) ne 0$ so $d(a,b) > 0$.

2) If $k > 0$ then $k$ is not a lower bound.

Pf: There exists an $n in mathbb N$ so that $n > frac 12k$ so $frac 12n < k$ and $frac 12n + n in A$ and $n in B$ so $frac 12n = d(frac 12n,n)in d(a,b)$

And thus $inf d(a,b) = 0$.

Let's do a proof by contradiction. Here's $A$ and $B$, as you defined them (50) and (51).

$$ A = left n + frac12n mathop: n in mathbbZ_> 0 right tag50 $$

$$ B = mathbbZ_> 0 tag51 $$

Suppose $D(A, B) = c $ where $c > 0$ . $c$ is an arbitrary constant. The only thing we know about it is that it's a positive real number.

Let's introduce a new constant $c'$ so we know $c'$ is strictly closer to $0$ than it is to $1$ (101 and 102). The choices of $10$ and $4$ are somewhat arbitrary, but I think it makes the proof easier to visualize.

$$ c' = minleft(frac110,; cright) tag101 $$

$$ m = 4 times leftlceilfrac1c'rightrceil tag102 $$

Note that $m$ is an integer by inspection and at least $10$, therefore it is a positive integer and hence (103).

$$ m + frac12m in A tag103 $$

The distance between $m$, which is in $B$ and $m + frac12m$, which is in $A$, is $frac12m$.

$frac12m < c' le c tag104$

However, by hypothesis $c = D(A,B)$, which is defined to be the infimum of the distances between each item in $A$ and each item in $B$ .

Contradiction.

Therefore $D(A, B) le 0$ . However, $D$ is the infimum of a set of non-negative reals, therefore $D(A, B) ge 0$.

Thus, (105) as desired.

$$ D(A, B) = 0 tag105 $$