# Is the following correct? (concerns sets and convergence)

Let A = {(x,y) in R^2 | x^2 + y^2 <= 81}
Let B = {((x,y) in R^2 | (x-10)^2 + (y-10)^2 <= 1}

then here "A intersection B" is the empty set.
Then let x_n be the sequence (0,10-(2/n)) which is a sequence in A and y_n be the sequence (10/n,10) which is a sequence in B.
would |x_n - y_n| tend to 0 in this case?

jgens
Gold Member
Why do you think xn is a sequence in A? It is not.

Is it not? Isn't a sequence in A that converges to a point in A compliment?

jgens
Gold Member
A sequence in A is a sequence all of whose values are in A. Almost all of the values of xn are not in A.

Is it not? Isn't a sequence in A that converges to a point in A compliment?
A is a disk with radius 9. The sequence $x_n$ converges to (0,10). So the sequence is not in A.

But can't a sequence x_n $\subseteq$ S converge to a point in S compliment?

But can't a sequence x_n $\subseteq$ S converge to a point in S compliment?
Not if S is closed, no.

Oh right I see. Thank you.

Can there exist a pair of non-empty closed sets A,B $\subseteq$ R^2 with A $\cap$ B equals the empty set with a pair of sequences x_n $\subseteq$ A and y_n $\subseteq$ B such that ||x_n - y_n|| -> 0.

I was trying to find a specific example and hence my original thought was my first attempt.

Yes, such a thing exists. Take

$$A=\{(x,0)~\vert~x\in \mathbb{R}\}$$

and

$$B=\{(x,1/x)~\vert~x\in \mathbb{R}\}$$

(Note the distance between these two sets is 0). Can you find suitable sequences in these sets?

Could you explain why the distance between the two sets is 0 please. I thought it was because d|x-y| = (x,0) - (x,1/x) = (0, -1/x) but wasn't sure how this means the distance is 0.

Also I'm finding it hard to grasp the concept of sequences in 2 dimension in the sense of how to represent them but my guess would be,
x_n = (3n, 0) and y_n = (3n, 1/3n) ?

The distance between two sets is usually defined as the greater lower bound of the set of distances between 2 points in the sets. In symbols:
$$d(A,B) = \inf \{d(a,b) : a \in A, b \in B\} \, .$$
Take a look at this: http://en.wikipedia.org/wiki/Distance#Distances_between_sets_and_between_a_point_and_a_set

So you were right that $d(a,b) = |a-b| = |(x,0) - (x, 1/x)| = |(0,-1/x)| = 1/|x|$. But now you need to find the greatest lower bound of the set $\{d(a,b) : a \in A, b \in B\}$. Do you see why it will be zero?

Is it because as x gets smaller and tends to -infinity but you take the modulus and so u get limn-> infinity 1/x = 0?

Also were my choices of sequences okay? That is the main issue I am stuck with? I need x_n in A and y_n in B such that ||x_n - y_n|| tends to 0.
I think x_n=(3n,0) and y_n=(3n/1/3n) would work right?

Yes, that choice of sequence is a good one!