Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Assume a circle with radius $R$ in a plane. Let $R$ go to infinity.

What is larger: The inside or the outside of the circle?

EDIT

My naive way of thinking about "largeness" was just to compare $\displaystyle\lim_{R \to\infty} \pi R^2$ with the rest of the plane. I didn't think of where the origin of circle would be or any other things like theories. If this is wrong, don't blame a layman...

share|improve this question

closed as unclear what you're asking by Andres Caicedo, M Turgeon, Pedro Tamaroff, TMM, Dilip Sarwate Mar 12 at 1:59

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

7  
Define "larger". –  Asaf Karagila Mar 11 at 22:51
2  
Even after you define larger, this doesn't seem to make sense after taking the limit. Can you formalize the question in predicate calculus? –  Git Gud Mar 11 at 22:52
1  
@AsafKaragila How would the choice of theory affect the answer? –  draks ... Mar 11 at 23:13
1  
@draks... I don't understand the 'no, I can't' part in relation to my comment. A possible way to better 'formalize' the question is to take the definition of $\lim$ I give in my answer in the link above. According to that definition, whatever center you take, the limit is $\mathbb R^2$ and the 'outside part' is the empty set. So now you're comparing $\mathbb R^2$ with $\varnothing$. Cardinality wise the outside part is smaller. Lebesgue measure wise the outside part is smaller. Order theoretically wise (with the natural order) the outside part is smaller. –  Git Gud Mar 11 at 23:33
2  
@Dror I very much disagree that the question was clear, but if you think so, please formalize it in predicate calculus and $\sf ZFC$. –  Git Gud Mar 11 at 23:34

6 Answers 6

up vote 3 down vote accepted

If you keep the circle tangent to the "y" axis as the circle gets larger, in the limit about 1/2 the points are inside and 1/2 are outside of the circel.

If you move the center of circle away from the orgin faster than the radius increases, all the points are outside of the circle.

share|improve this answer
    
This is actually a better answer than mine. By making the fixed point of the tangent of the circle the vertical line y = x_0, you can actually get to any proportion of area you want! This alone should show that the question is ill-posed. –  Sumedh Joshi Mar 11 at 23:56

For all finite R, the inside is smaller. For an infinite R, there is no circle.

share|improve this answer
    
limit is a process. There is a circle all the time. –  Dror Mar 11 at 23:01
1  
Infinity is not a number, and the radius of a circle needs to be a number. The limit of the process you're describing does not exist. –  Sumedh Joshi Mar 11 at 23:02
    
heh. the limit is the process. Who said $\infty$ is a number? –  Dror Mar 11 at 23:03

After your edit, I understand that you want to do the following: look at the circle $B_R=\{x\in {\bf R}^2\mid d(x,0)\leq R\}$ of radius $R$ on the plane, centered at $0$, and compare the area $A(B_R)$ (usually called Lebesgue measure and variously denoted by $\mu,\lambda$ or $\lambda_2$, but I will call it $A$ for your convenience) with that of its complement $A({\bf R}^2\setminus B_R)$.

  1. As you have noticed, $A(B_R)=\pi R^2$.
  2. On the other hand, $A({\bf R}^2\setminus B_R)=\infty-\pi R^2=\infty$.
  3. So as $R\to \infty$, both $A(B_R)$ and $A({\bf R}\setminus B_R)$ tend to infinity (the latter somewhat vacuously): $$\lim_{R\to \infty} A(B_R)=\lim_{R\to\infty}A({\bf R}^2\setminus B_R)=\infty$$
  4. On the other hand if we look at the union $B_\infty=\bigcup_{R>0} B_R$, then $B_\infty={\bf R}^2$, while its complement is empty, so $A(B_\infty)=\infty$ and $A({\bf R}^2\setminus B_\infty)=0$.
  5. So what do these two last points tell us? This is a property of area, that it is not continuous from above. More precisely, if $A_n$ is a decreasing sequence of sets for which it makes sense to talk about area, then the limit of areas is not always the area of the intersection.
  6. On the other hand, one can show that area is continuous from below, that is, if $A_n$ is an increasing sequence of nice sets, then the limit of their areas is the same as the area of their union (which is always a nice set in this case). It also is continuous from above, assuming that $A_n$ have finite area, so it was no coincidence that continuity failed on sets of infinite area.

What the answer to your question is depends on how exactly do you want to compare the areas. Most obvious choices are the one given in points 3. and 4., but there are others, as Asaf hinted.

To understand more formally the concepts I mentioned above, you'd have to get acquainted with elementary measure theory. Such an elementary introduction should be found in any good introductory textbook to analysis or probability theory.

share|improve this answer
    
thanks (+1) for plenty of points of view... –  draks ... Mar 11 at 23:55

It completely depends on where the center of the circle is while it gets larger. If the center stays at $(0,0)$, then every point in the plane is inside the circle for sufficiently large $R$. That is, $$\limsup_{R\rightarrow \infty}C_R=\mathbb{R}^2.$$ If the center is at $(-R,0)$, then $$\limsup_{R\rightarrow\infty}C_R=\{(x,y)\;|\;x< 0\}\cup \{(0,0)\}:$$ the limit is the left half-plane plus the origin. If the center is at $(-2R,0)$, then every point is outside the circle for sufficiently large $R$: $$\limsup_{R\rightarrow\infty}C_R=\emptyset.$$ More or less however you quantify the "size" of a set, the inside of the circle is respectively larger, the same size, and smaller than its outside in these three situations.

share|improve this answer
    
hard to imagine that the choice of the centre matters that much. In my mind it's always the same circle... –  draks ... Mar 11 at 23:53

One common aspect of the various limit processes that have been described so far is that one has to take the limit of some quantity associated with each finite circle-it's not clear how to take the limit of the circles themselves if we insist on drawing our circles in the plane. But the plane plus a point at infinity is a sphere (think about the point at infinity at the north pole-hopefully I can get away with a bit of geography lingo here to simplify notation,) and this permits us make this problem finite, or compact. So for instance on the sphere the sequence of circles centered at the origin with radius increasing without bound converges to the north pole; a sequence running off to the right more quickly than its radius increases also converges to the north pole; a sequence centered at $(0,r)$ for radius $r$ converges to the prime meridian; and so on.

The answer still depends on a definition of "inside" and "outside" for circles on the sphere. Probably the most direct generalization of the same idea from the plane is that a point is inside a circle on the sphere iff it's at a lower "latitude," i.e. height, than every point of the circle. That makes the entire plane inside the limit of our first two example cases, but nothing inside the limit of our third, since the prime meridian includes the south pole. On the other hand, there's nothing outside the prime meridian either, since it includes the north pole! So this reformulation of the problem suggests that you need the ability to say "neither" in certain cases to answer completely.

By playing around with the third example it's easy to get sequences of circles which converge to any circle including the north pole. (The north pole has to be included since the radius went off to infinity.) So other than the latitude lines which include the south pole as well, these all have an "inside" but an empty "outside." This suggests that whenever the limit converges, the answer to your question is either "the inside is larger" or "neither is larger," and the former is in some sense more common. The fact that it's hard to come up with a natural sense in which the outside could remain larger, even though it's infinite at every time, is perhaps another reflection of the discontinuity of area from above that was mentioned in another answer.

share|improve this answer

Let's ignore the question about the center in relation to the radius for a moment.

I think It's obvious that by "inside"/"outside" we mean $|x|<R$ / $|x|>R$ .

In terms of Lebesgue measure, the inside of the circle has measure $\infty$ when taking the limit $R \to \infty$ (And I mean by that: As large as we please).

In terms of cardinality: the inside of the circle had the cardinality of $\mathbb{R}^2$ when $R$ was still finite.

So, If given the center is always arranged in a way that prohibits points from the plain from being contained in it, then you can deduce that both the inside and the outside have the same Lebesgue measure and the same cardinality.

But, if the center is fixed, or arranged in a way that does not prohibit some point from the plain to be contained in it for all $R$, then the "outside" is actually the empty set, and is strictly lesser in cardinality and measure.

share|improve this answer
    
Quite the opposite of user1457's answer. Interesting... –  draks ... Mar 11 at 23:31

Not the answer you're looking for? Browse other questions tagged or ask your own question.