Why is a circle not a convex set?

Why is a circle not a convex set?

My attempt: I was googling it, but I didn't find the correct answer, as I didn't understand the explanation why a circle is not a convex set as defined here:

As shown in diagram 1, points A and B and points C and D lie inside the circle, thus I think the circle must be a convex set, as clearly shown in figure 1.

Why isn't the circle a convex set?

• @Arthur, this text seems rather shallow in content (apart from not being worded too well). Doesn't it basically say that the only convex curves are a straight lines? – Imago Nov 12 '17 at 13:04
• "Circle" means just the curve/loop. "Disk" usually means the circular area inside. Your can't connect 2 points on the circle without going through the interior. – jdods Nov 12 '17 at 13:06
• The points inside the circle (e.g. the intersection of the lines AB and CD) do not belong to the circle. – md2perpe Nov 12 '17 at 20:06
• A set that is not convex is called concave? What? – Carsten S Nov 12 '17 at 23:58

They mean just the edge of the circle, without the interior. That is not a convex set.

The confusion arises from the notation. A circle, in the second part of the sentence, is defined as just the points on the boundary, excluding the interior. From your picture you can see that the line connecting A and B has A and B on the boundary, but the other points in the segment lie in the interior of the circle. Therefore, if we define a circle as the set of points on the boundary, it is not convex.

On the other hand, if a circle is defined as the set of points in the boundary plus those on the interior, it is a convex set.

The sentence in your book is misleading since it did not define what a circle is, and it seems to be using both interchangeably, adding more confusion.

• Did you check out that book to see if it really doesn't define anywhere what a circle is? I mean, usually that definition is in one place, not everywhere where a circle occurs, and it is quite unlikely that this is the very first place where the book mentions a circle. BTW, in mathematics a circle is always defined as the line. – celtschk Nov 12 '17 at 21:27
• True, but that seems what the OP is confused about – user Nov 12 '17 at 22:59
• But I don't think the book is to blame about it. – celtschk Nov 13 '17 at 8:32
• I will edit the answer later to reflect that – user Nov 13 '17 at 12:44
• If the book were consistent about defining a circle to be the boundary of a disk, it wouldn't say "the interior of a circle is convex" except to mean "the interior of a circle is the empty set, and the empty set is trivially convex". It would say 'the interior of a disk is convex". So I think the book is to blame. – Misha Lavrov Nov 13 '17 at 16:48

The circle is given by the black curved line. Note that the white interior is not part of the circle.

Now consider the intersection point of the line AB and the line CD. This intersection point clearly lies on the line AB between the points A and B. The points A and B are on the circle, so if the circle were convex, then the intersection point would need to be on the circle as well. But it obviously is not; it in inside the circle. Therefore the circle is not convex.

On the other side, the inside of the circle is convex, as if you chose any two points inside the circle, then the full line segment connecting them is also inside the circle.

A set $A$ is called convex if for each pair $x,y\in A$ holds $$(1-t)x+ty\in A\hspace{1cm} \forall t\in[0,1].$$ If you have a ball like $$B_n=\{x\in\mathbb R^n~:~\|x\|\leq 1\}$$ you can prove that it is convex. But the sphere $$S^{n-1}=\{x\in\mathbb R^n~:~\|a\|=1\}$$ is not convex. Consider some $x\in S^{n-1}$ and you get $-x\in S^{n-1}$ but $\left(1-\frac12\right)x+\frac12(-x)=0\notin S^{n-1}$. You might thought about the disk $$B_2=\{(x,y)\in\mathbb R^2~:~x^2+y^2\leq 1\}$$ but the circle $$S^1=\{(x,y)\in\mathbb R^2~:~x^2+y^2=1\}$$ is not convex.

• I see. But is there any advantage for excluding the sphere from the set of convex sets? – Ooker Nov 13 '17 at 7:13
• We don't "exclude" the sphere to be convex, but it doesn't satisfy the definition to be convex. We can't decide which set is convex and which isn't convex. – Mundron Schmidt Nov 13 '17 at 8:32
• Sorry, what I mean is why do we choose a definition that makes the circle unsatisfied with it? Can't we have a better definition that can graph our intuition that the circle is convex? – Ooker Nov 16 '17 at 10:07
• Or geometrically, how to see the circle isn't a convex set? – Ooker Nov 16 '17 at 10:08
• A definition has to be useful and the definition I gave is in fact very useful for a lot of theorems. You can define a property $A$ such that each convex set and also the circle has the property $A$. But this property is not the same as convexity. Hence if you have the theorem which needs convex sets, in general, you can't use sets with property $A$.You can see geometrically if a set is convex. By definition, a set is convex iff for every two points in the set the line between the two points belongs to the set, which is wrong for the circle. – Mundron Schmidt Nov 16 '17 at 10:44

This is a confusion caused by people using language differently. When doing mathematics, we must be quite precise in assigning meanings to different terms, which is why we take so much trouble to define them. And the point of a definition is, once we've understood what somebody means by a term, we won't make the mistake of thinking they mean something else!

So here's how we use certain common English words in ordinary ("Euclidean") geometry:

• A circle is a closed curve in space, formed by the set of all points an equal distance from a fixed point called the centre of the circle.
• A disk (or disc) is the set of all points contained inside (*) a circle, including its centre.

Note that a disk does not necessarily include the circle itself! (When studying topology - which is rather like geometry would be if distances weren't fixed or important - we distinguish between a closed disk, which includes the boundary or outside edge, and an open disk, which doesn't. And we often call a disk a ball - think of a squishy rubber ball.)

We also define a convex set to be one that includes every point that lies on a line segment joining any two of its points. So you can see that the inside (or interior) of a disk is a convex set, since even in a squishy rubber ball, any one point that lies between two others of the interior must be in the interior, no matter how much we squish it. But the outside (or boundary) of a disk is not a convex set, since there are points between any two points on the circle that don't also lie on the boundary, that is, they lie inside the circle, in the interior of the open disk. For example, as pointed out in the answer by celtschk, the intersection of those two chords AB and CD in your diagram is such an interior point that lies between points of the circle.

(*) Now I haven't defined what I mean by point, inside or a line segment. But in the case of a circle in Euclidean geometry, the ordinary notion of "inside" and "outside" works well enough; a point is as small a dot as we can possibly draw in a picture; and a line segment is that part of a straight line that connects two points. More generally, we can't define everything! Even in mathematics, where we try to be as precise as we can, we sometimes have to say "Enough!" and just accept certain ideas as basic, undefinable notions. We usually take a geometric point as one of those basic notions.

Because the chord is not placed on the circle.

• @ MIchael Rozenberg here it say that every segment joining the two point on the circle contain point that are not on the circle......im not getting this line as from diagram it is clearly contain the point that are on the circle,,,,,,,,,pliz explain this what its exactly mean to say ? – lomber Nov 12 '17 at 13:16
• @lomber lego Let $\Phi$ be our circle and $AB$ be a chord. Thus, $\{A,B\}\subset\Phi$. If $\Phi$ is a convex set then for all $C\in AB$, where $AB$ is a segment, we need $C\in\Phi$, which is wrong. – Michael Rozenberg Nov 12 '17 at 13:22
• how is this saying anything different from the text the OP included in her question? If she was confused by that, is this going to clear up the confusion? No, as evidenced by the subsequent comment. I never thought I'd see an answer as lazy and useless as this one get upvotes... – Nick Pavlov Nov 12 '17 at 13:30
• @Michael Rozenberg still not getting...suppose i take A=(1,1) and B =(-1,-1) now using the distance formula i got $AB^{2}$ = 8,,,now how can i show that C ∈ AB? – lomber Nov 12 '17 at 13:30
• @lomber lego In your case $C\in AB$ it's equivalent to $C\left(\frac{1-t}{1+t},\frac{1-t}{1+t}\right)$, where $t\geq0$. You need $t>0$ only, of course. – Michael Rozenberg Nov 12 '17 at 13:45