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Consider the sequence $ \displaystyle{ A_n = \{ (x,y) \in \mathbb{R} ^2 : (x-n)^2 + y^2 \leq n^2 \} ; \quad n \in \mathbb{N} }$ of circles.

Find the limit $ \displaystyle { \lim_{n \to \infty} A_n }$.

The only thing I can see that the sequence $ \displaystyle{ (A_n)}$ is increasing with drawing the circles, but I can't prove it.

Any ideas?

Thank's in advance!

edit: I made a typo. It is $n^2$ the right side. Sorry for the confusion.

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5 Answers 5

up vote 3 down vote accepted

If you replace $n$ with $n^{2}$, this is a sequence of circles that at each stage is increased in radius by $1$ and then shifted to the right by $1$. I claim that $\displaystyle\lim_{n \to \infty} A_{n}$ is actually the right-half plane with the origin, i.e.

$$ \{(x,y) \in \mathbb{R}^{2}: x > 0 \} \cup \{0,0\}$$

It is easy to see that

$$ \{(x,y) \in \mathbb{R}^{2}: x <0 \} \cup \{(0,y): y \neq 0\} $$

is not in $\displaystyle\lim_{n \to \infty} A_{n}$. Take $(x,y)$ with positive $x$ coordinate. We wish to find an $N$ large enough so that $(x,y) \in A_{n}$ for $n \geq N$. Without loss of generality, we assume $y$ is positive.

Since the ball is given by

$$(x-n)^2 + y^{2} \leq n^2$$

we can solve for $y$, to get

$$y \leq \sqrt{n^{2} - (x-n)^{2}}$$

since $x > 0$, $$n^{2} - (x-n)^{2} > 0$$

Moreover, its derivative is $2(n-x)$, so that if $n > x$,

$$\sqrt{n^{2} - (x-n)^{2}}$$

is increasing. Thus, taking $N$ larger than $x$ such that

$$y \leq \sqrt{N^{2} - (x-N)^{2}}$$

holds, we see that $(x,y) \in A_{n}$ for $n \geq N$.

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1  
Wouldn't the limit be $\{(x,y) \in \mathbb{R}^{2}: x > 0 \} \; \cup \; \{(0,0)\}$? I'm assuming, by the way, that the OP intended limit to refer to equality of $\liminf$ and $\limsup$ of a sequence of sets, with no topological considerations being involved. –  Dave L. Renfro Mar 28 '12 at 19:02
    
Why is not : $y \leq \sqrt{n^{2} - (x-n)^{2}}$ ? –  passenger Mar 28 '12 at 19:05
1  
You can actually see that via a simple geometric argument. Suppose that $(x,y)$ is in $A_{n}$, so that the distance of $(x,y)$ from $(n,0)$ is less than $n$. We want to show that the distance from $(x,y)$ to $(n+1,0)$ is less than $n+1$. By the triangle inequality, the distance from $(x,y)$ to $(n+1,0)$ is less than the sum of the distance from $(x,y)$ to $(n,0)$ and from $(n,0)$ to $(n+1,0)$, which is $n+1$. This shows that $A_{n} \subset A_{n+1}$. To see a strict containment, continued... –  Isaac Solomon Mar 28 '12 at 19:28
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To see that the containment is strict, take $(n,n)$. This point is exactly a distance of $n$ away from $(n,0)$. However, its distance from $(n+1,0)$ is $\sqrt{n^{2} + 1} < \sqrt{(n+1)^{2}} = n+1$. Thus, we can take $(n,n+\epsilon)$. If $\epsilon$ is sufficiently small (not THAT small, really), the distance from $(n,n+\epsilon)$ to $(n+1,0)$ is still less than $n+1$. However, the distance from $(n,n+\epsilon)$ to $(n,0)$ is greater than $n$, so that $(n,n+\epsilon) \in A_{n+1} \setminus A_{n}$. –  Isaac Solomon Mar 28 '12 at 19:33
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@IsaacSolomon: Thank's once again! Your comments are really clear and helpfpul! –  passenger Mar 28 '12 at 21:13

$A_n$ is a circle centred at $(n,0)$ with radius $\sqrt{n}$. Since $n -\sqrt{n} \to +\infty$, there isn't a finite limit; the circles march off toward infinity along the positive $x$ axis.

Perhaps you meant the right side to be $n^2$ instead of $n$. That would make things more interesting, as all points on the $y$ axis would be limit points of this sequence of circles.

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Yes it is $ n^2 $ the right side. I edit it. –  passenger Mar 28 '12 at 18:49

These circles are centered at $(n,0)$, so the center is not converging. It's moving toward $(\infty,0)$. The radius of the $n$th circle is $n$, so the radius is diverging as well, but left-most point of the circle is always $(0,0)$.

Take any $(x,y)$ in the right half-plane. Consider the line segment from the origin to this $(x,y)$. Its perpendicular bisector will intersect the $x$-axis at some point, say $(X,0)$. Now for all $n>X$, $(x,y)$ is within $A_n$. So every point in the right half-plane is in the limit of $\left\{A_n\right\}$.

Considering a sequence of $(x_n,y_n)$ in the right half-plane that converges to $(0,y)$ on the $y$-axis, every point on the $y$-axis is also in the limit of $\left\{A_n\right\}$.

So $\lim A_n=\left\{(x,y)\mid x\geq0\right\}$.

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The circles are centered at $(n,0)$ with a radius of $n$. Considering this you can decide for every point in the x-y-plane whether it will be in $A_n$ for large n. ($A_n \subset A_{n+1}$)

  1. Any negative x: can never be inside the circle as every point with negative $x$ value is further from $(n,0)$ than $n$.

  2. Any point $(x,y)$ with positive x is included in a large enough circle that still touches $(0,0)$ and has its center on the x-axis.

The limit thus is $\{(x,y)|x\ge 0\}$.

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Can you think of some points that belong to none of the circles, such as $(-1,-1)$? It should be pretty clear that a certain half-plane of points will not belong to any of the circles. (Can you prove this?) Consider points on the boundary of this half-plane, points in the interior of this half-plane, etc. and prove what seems evident from a picture. For some of these results, you can argue by making use of the Pythagorean Theorem applied to right triangles with a vertex at the origin and legs along the positive $x$- and $y$-axes. For other results, sufficiently large values of $n$ will be useful. For example, can you come up with a value of $N$ such that $n>N$ implies the circle associated with $n$ contains the point $\left(10^{100},10^{1000}\right)$?

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