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Let $A$ be a compact convex set in $\mathbb{R}^2$.

What does ''compact convex set'' mean?

What I understand: We have a "bunch" of real points $(x,y)$ in the plane. Any two of them satisfies the fact that a line drawn between them is fully inside this "bunch" of points.

So can this "bunch" be a polygon? (I think it can.)
Also, how does the compact part fit in and what does it mean?
How do I make one in the plane?
Also what is a "compact convex polygon" if it is possible?

This is not me.
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  • Compactness is a very common concept in analysis, and in $\Bbb{R}^2$ specifically, it is equivalent to being closed and bounded. Are you familiar with these terms? – Theo Bendit Nov 23 '21 at 05:26
  • Did you take a real analysis class? This is where one usually meets the notion of a compact subset of $R^n$ for the first time. – Moishe Kohan Nov 23 '21 at 05:26
  • @TheoBendit I have heard of these terms but I do not think I understand them fully. This is the series of definitions I got for closed.

    A set is closed if and only if it contains all of its limit points.

    A point $p$ is a limit point of a set $X$ if every disk centered at $p$ includes a point of $X$ different from $p$.

    A set is said to be bounded if it lies in some circular disk of finite radius.

    This is vaguely understandable.

     – This is not me. Nov 23 '21 at 05:38
  • @MoisheKohan I have not taken such a class. – This is not me. Nov 23 '21 at 05:41

2 Answers2

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A convex set is one where the line between any two points in the set lies completely in the set. Polygons can be convex or concave - for example, a crescent moon is a concave set and you can approximate it with a polygon that stays concave.

A compact set, at least in the real Cartesian plane, is one that is both closed and bounded, which roughly means that (a) it includes its own boundary and (b) it has no points going to infinity. Polygons, assuming you include both the boundary and interior, are compact in $\mathbb{R}^2$.

So yes, a polygon may be both convex and compact, but not every convex compact set is a polygon, and not every polygon is a convex compact set.

ConMan
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  • Thank you. This helps. So a convex hull is a compact convex set? And the compact set just needs to be "contained" in a way? – This is not me. Nov 23 '21 at 05:44
  • The convex hull of a set is the smallest convex set that contains it - imagine covering your shape in a sheet of rubber that grips tight to the bits that stick out and then stretches flat over the concave parts. And yes, compact and bounded are equivalent (in the contexts you're probably dealing with) so as long as you can contain your set inside a ball of some finite size you're good. – ConMan Nov 23 '21 at 22:24
  • @Thisisnotme. The convex hull of a finite set is compact but the convex hull of an infinite set, even if it's bounded, might not be. (For example, consider the set of points strictly inside - but not on the boundary - of a circular disk. Then that set is convex and is therefore its own convex hull, but it's not closed because it doesn't include its boundary, and is therefore not compact.) – Akiva Weinberger Jan 05 '23 at 06:21
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A compact and convex set refers to a set of points that follow the property:

  1. The set is convex, that is, any line that connects any and all the two points chosen from the set lies in the set. Think of the parabola y=x^2. Now choose any 2 points lying inside or on the parabola and try to connect them by drawing a straight line. You will observe that all such lines will lie inside the parabola, which means that the curve is convex.
  2. Compactness refers to the property that the set shall be closed and bounded. The example of parabola from the previous point does not hold true here because it was not a bounded curve. A disk of finite radius for eg, would be a compact and convex curve as it is bounded, has no holes and is convex. Note: Don't confuse the disk with a circle. A circle only refers to the boundary, excluding the interior that is why it is not possible to connect any 2 points without moving the pencil out of it.
tend0pain
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