I saw this text on Cracking GRE math book.
"HeineBorel theorem: A subset of Rn is compact iff it's both closed and bounded"
Is there any subset which is closed but not bounded?
Closed and bounded
Re: Closed and bounded
The real line is closed but not bounded.

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 Joined: Tue Aug 09, 2011 6:18 pm
Re: Closed and bounded
Slow down. HeineBorel talks about Euclidean spaces, not metric spaces on R^n. Suppose you put a discrete metric on the real line. That is, d(x,y) = 1 if x \neq y, and d(x,x) = 0 . In this strange world, points are open sets. Naturally, any finite subset is going to be compact (this is always true in any topological space). But in fact, finite sets are the only compact sets!
Here, the real line is closed and bounded, but not compact.
Here, the real line is closed and bounded, but not compact.

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: Closed and bounded
Anytime you refer to R^n without specifying the topology, it means the usual topology given by normal euclidean distance
Re: Closed and bounded
Thank you for mentioning this.Topoltergeist wrote:Slow down. HeineBorel talks about Euclidean spaces, not metric spaces on R^n. Suppose you put a discrete metric on the real line. That is, d(x,y) = 1 if x \neq y, and d(x,x) = 0 . In this strange world, points are open sets. Naturally, any finite subset is going to be compact (this is always true in any topological space). But in fact, finite sets are the only compact sets!
Here, the real line is closed and bounded, but not compact.
Re: Closed and bounded
There is also a version of Heine Borel for general metric spaces: A subset of a metric space is compact if and only if it is closed and totally bounded.
The real line with the discrete metric is bounded but not totally bounded, hence not compact.
The real line with the discrete metric is bounded but not totally bounded, hence not compact.
Re: Closed and bounded
The Heine Borel theorem is more involved than appears at first glance. It is kind of fun to play with. Right now, I will only apply it to intervals on the Real Number Line. The tricky part of it is the notion of "compact'. This in turn depends on the notion of an "open cover", basically a set of bounded open ended intervals that cover every point of the interval in question. It does not matter if the covering intervals also cover additional points or if points get covered more than once. Notice that the Heine Borel Theorem sets two requirements for compact. One is that the interval in question must be bounded and thus not go on to infinity. The other is that both ends must be closed. Now, the definition of compactness requires that you can cover the whole interval with a finite number of intervals. This excludes unbounded intervals as there is no way a finite number of bounded intervals can cover something infinitely long, thus compact requires a bounded interval. The next requirement is a good more subtle as it requires the interval to be closed on both ends. It also states that if you cover the interval with any kind of open cover of finite intervals, you never need an infinite number of them. If you do dream up such an infinite cover, you can throw out all but a finite number of them and still cover all the points of your original interval. This is the part that is tricky and needs to be thought through carefully. The requirement that the cover be open is critical. If you use closed covering intervals, you can always force an infinite cover that cannot be reduced to a finite one whether the original interval is open or closed. An example is the interval from 0 to 1 on the Real Number line. If you make your closed intervals 0 to 1/2, 1/2 to 1/4, 1/4 to 1/8 and so on until you cover all the terrain from 0 to 1. You cannot remove any of these intervals without leaving holes in your cover set.The closed cover is inherently, irreducibly infinite. Closed intervals have the very nice property that they can be connected head to tail. They can be easily set up to completely cover a continuous interval whether the ends are closed or not. So, a closed cover is of no use to the Heine Borel Theorem. The situation is rather different if all the intervals are required to be open. These intervals CANNOT be connected head to tail as the connecting points are empty and connect nothing and can cover nothing. An open interval can only cover another point if you use an INTERIOR point of the interval as the end points are empty. This means that when you use open intervals to cover something, they must overlap each other to some finite degree if they are to provide complete coverage of the original interval. These areas of overlap are necessarily of some sort of finite extent and you are not easily going to get an infinite number to fit into a finite bounded interval. This is the first hint that open covers tend to be finite, though of course not necessarily. They do not lend them,selves to infinite covers the way closed intervals do. Now of course you can cover any finite interval with an infinite set of intervals whether they are open or closed or whether the original interval is open or closed. What the definition of "compact" states is that if the original interval is closed, no open cover is INHERENTLY infinite. Any infinite open cover can be reduced to a finite one by throwing out all but a finite number of sub intervals while still being able to cover every point of the original interval. now if you have an OPEN original interval, you can cover it with an irreducible infinite open cover by using a scheme similar to the one I stated above. Use the open interval from 0 to 1 as your original interval. each element of the subcover will look like: 0 to 1/2 (exclusive); 0 to 3/4 (exclusive); 0 to 7/8 (exclusive and so forth. Since the original interval is open, you dont need to cover the end points and can cover all the interior points with an irreducibly infinite open sub cover. To be sure, in the above scheme, since each new element covers the previous ones, you can throw out any finite number of initial covers but you will still have an infinite number of remaining ones. Now the situation is different for a closed interval. Here, you have to cover the end points too as the original interval is closed and includes its end points. This means that you must have open intervals that overlap the end points on both ends and extend both some finite distance within and without the end point. in other words, an interior point of some open interval must lie atop an end point.This creates a very different situation. It effectively means you cannot use the sort of infinitessimal approach I used above. Since intervals must overlap to cover all points, you can generally throw out any pieces, even if infinite in number that are smaller than the degree of overlap. The best way to convince yourself that this is true is to try to come up with some ingenious scheme that covers a closed bounded interval with an infinite number of bounded open intervals as an open cover such that you cannot reduce them to a finite set still covering all the points of the original interval. Just think! If you can overthrow the Heine Borel Theorem by coming up with such a counter example, everlasting mathematical fame will be yours! ^,..,^ go get 'em tiger! Actually, if you try this long enough you will soon be convinced of the futility of such efforts and you will also have a good grasp of the Heine Borel Theorem. The basic idea here is that open intervals are "sloppy" in that you can only get complete coverage by overlapping them with each other and any closed end points. closed covers are much neater because they can be connected end to end through their closed end points and you can always get an irreducible infinite cover regardless of whether your original interval is closed or not. Like I said, the Heine Borel Theorem is lots of fun to play around with and you need to do this to really understand the underlying concepts. Try defeating it!
lotsa luck! JerryBear
lotsa luck! JerryBear