# Is an infinite line the same thing as an infinite circle?

Imagine that you are sitting next to a line that extends infinitely in both directions. Is it possible to distinguish it from an infinite circle?

From my poor understanding of topology, I would guess that it makes a difference if it's a line or a circle: The second is closed, the first isn't.

What they both do to a plane is, that they split it into two parts, left and right or in and out. But is this enough to say that they are obviously the same thing?

EDIT

$\lim_{R\to \infty}$

$\hskip1in$

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So, how do you define an infinite circle? And what exactly is an infinite line? Because I think mine is closed. – benh Mar 11 '14 at 22:48
Concerning your picture, I think that one needs to be careful about infinity here. I assume, that you are talking about points in the euclidean plane $\Bbb R^2$. As a set (we need a set to do topology) the "infinite line" on the left could be defined as a union of line segments of finite length. However, we can't give a similar argument for the circles, because they are not contained in each other when $R$ is growing. To be honest, I can't think of a convincing concept for defining infinite circles in the Euclidean plane. – benh Mar 11 '14 at 23:08
As a corollary is it possible to define straight as the curvature of an infinite circle? – geotheory Mar 12 '14 at 15:12
From Soddy's Kiss Precise "Since zero bend's a dead straight line And concave bends have minus sign," they are equivalent for some purposes – Ross Millikan Mar 12 '14 at 17:25
– Martin Sleziak Mar 13 '14 at 18:25

From a non-rigorous "visual" point of view you wouldn't be able to tell the difference, just like when you look down at your feet you can't tell that the earth is a sphere. The rigorous way to say this mathematically is that a circle and a line are "locally homeomorphic".

Two sets are homeomorphic if there is a continuous bijection between them, with a continuous inverse. If $f:A\to B$ is a continuous bijection, and $f^{-1}:B\to A$ is also continuous, then $U$ is open in $A\iff f(U)$ is open in $B$. This in turn causes $f$ to preserve basically every topological property of $A$ when it maps to $B$, and so we say that $A$ and $B$ are topologically equivalent or homeomorphic. The line and the circle are not homeomorphic. This is easy to see because the line can be disconnected by removing one point and the circle can not.

The circle and the line however are locally homeomorphic. "locally" basically means that the property holds in some sufficently small neighborhood of every point. If you take some small neighborhood of a point on a circle and a small neighborhood of a point on a line, those two sets are homeomorphic. This is why we think that the earth looks like it is a plane, we aren't big enough to see outside some neighborhood on which the earth actually does look like a plane. If you were really tall, you'd be able to see the earth curves. This would correspond to being able to look outside the $\epsilon$-neighborhood where the earth is homeomorphic to the plane.

Note: This answer is for comparing $\mathbb{R}$ to a circle of finite radius. The question as posed doesn't totally make much sense, specifically the bit about the infinite radius circle see the answer from @Kevin Carlson for some details about that.

I'm ignoring the finite line case because if you are located at the endpoint that obviously things get messed up because you can only move in one direction. The answer for a finite line anywhere except at an endpoint is the same as the answer for $\mathbb{R}$

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Notable, though, is that the line is homeomorphic to the circle minus a point. The circle is the real projective line. – Mike Miller Mar 12 '14 at 5:33

Many people have shared your intuition that lines seem like infinite circles, but the problem remains that if a circle is the set of points a fixed distance from a fixed point in the plane, then there are no infinite circles, because there are no points in the plane infinitely far from each other. So your $\lim{R\to\infty}$ does not converge to anything. But still, we'd often like to say that lines and planes are obviously the same thing.

A general solution to such a problem is to simply add as little as possible to our theory to make them the same, and see whether we've ruined everything in the process. As you observe, the only topological difference between a line and a circle is that a circle is compact while a line is not (you say "closed," but that's not a topological property-though it would work in discussing the line and circle as manifolds.)

Indeed, a line becomes a circle as soon as we compactify by adding a single point! So the minimal way to get lines to be "infinite" circles seems to be to add a point "at infinity" to compactify every line. This makes the plane topologically a sphere, and then you can visualize your limit process as a circle contained in one hemisphere expanding until it approaches the equator. In this case it turns out we're very far from having ruined everything-rather, this is a pretty subject called Möbius geometry.

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+1 interesting. Would Möbius geometry also help here: math.stackexchange.com/q/708645/19341 ? – draks ... Mar 11 '14 at 23:47
Well, Möbius geometry happens on the sphere, not the plane, so a circle's radius can't tend to infinity, but remains bounded by the radius of the sphere, and there's no definition of inside or outside. It seems like you've gotten a pretty wide range of answers to various plausible interpretations of that question, what more are you still wanting? – Kevin Carlson Mar 11 '14 at 23:56
right: I thought the more, the better... – draks ... Mar 11 '14 at 23:57
Well, I guess there might be something to add. – Kevin Carlson Mar 11 '14 at 23:59
so, please, go ahead, I'll go to bed... – draks ... Mar 12 '14 at 0:00

One sense in which a line segment and circle convergent to the same line for $R\to\infty$ is the following:

We will consider a vertical line segment going through the origin.
Similarly, let us take the circle going through $(0,0)$ such that the center lies on the $x$-axis. (I.e, the center will be $(R,0)$.) More precisely:

• The line segment for some given $R$ is $L_R=\{0\}\times[-R,R]$.
• The circle for some given $R$ is $C_R=\{(x,y)\in\mathbb R^2; (x-R)^2+y^2=R^2\}$.

Then for $R\to\infty$ both these things obviously converge to the $y$-axis. (Maybe it helps if you draw a picture.)

If you want to replace word obviously in the last sentence by something more rigorous, you can take Kuratowski convergence.

This means that you can ask: What is the set $$L=\{p\in\mathbb R^2; \lim_{R\to\infty} d(p,L_R)=0\}$$ where $d(p,A)$ denotes the distance of the point $p$ and the set $A$.

Similarly, you can ask about $$C=\{p\in\mathbb R^2; \lim_{R\to\infty} d(p,C_R)=0\}.$$

You will find out that both $L$ and $C$ are equal to $y$ axis.

Perhaps the following picture might illustrate how the circles get closer and closer to the vertical line with increasing radius.

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