i somehow managed to prove that open does not imply compact for R^1 but it was a very technical and fairly unstandard, it used the idea that open intervals which are subsets of R can be expressed as an "infinite sequence of closed intervals" (the sequences in the intervals say [a_n,b_n] converge to the same limit) in this case monotonically and where forall n [a_n,b_n] share an element with some [a_m,b_m] hence forming an open cover which doesn't have a finite subcover because it'll contradict the monotonic convergence of [a_n,b_n] and hence open sets aren't compact, i'm not sure where to go next with this.
Thanks for the comment. Remember that this is only one direction of Heine-Borel, that compactness implies closed and boundedness. So if we take for example a specific compact subset of R^1 (say [0,1]), we are assuming it is compact and trying to show that it is closed and bounded. So for some point not in [0,1], like x = 1.2, we cover [0,1] with complements of closed balls (which in R^1 is closed intervals) around 1.2, so the complements of [1.1, 1.3], [1.15, 1.25], [1.19, 1.21], and so on, getting infinitely close to 1.2 The complements of those sets are open sets, for example the complement of [1.1, 1.3] is (-infinity, 1.1) U (1.3, +infinity), which is an open set. Notice how [0,1] is a subset of (-infinity, 1.1) U (1.3, +infinity)! So these complements are an open cover of [0,1]. So since we are assuming [0,1] is compact, we can say there is a finite subcover, the infinite process going towards 1.2 stops at a certain point. So we can find a closed ball (a closed interval in R^1) around 1.2 that is NOT in the open cover (which would be exactly the closed interval that the process stopped at). In R^1 we can delete the endpoints of the interval we stopped at, for instance say the process stopped at [1.19, 1.21]. Then we know the open set (1.19, 1.21) is completely in the complement of [0,1] (because [0,1] is covered by those sets which the process stopped). This shows that the point 1.2 is in an open set entirely contained in the complement of [0,1]. Now since this is just for a single point, it is not a proof, but the argument holds if you take an arbitrary point outside of [0,1]. To prove a set is open in R^1 you can show that every point in it is contained in some open interval that is then contained in the set. Also if the complement of a set is open, then the set is closed. So if [0,1] has a complement that is open, [0,1] is closed. That shows that it is closed. Try to figure out why a set like [0,1] is bounded (think about what the definition of bounded in R^1 means, what are some different ways you could define it?) I hope this helps, keep on going with upper math, it is a worthwhile pursuit.
Proof that open balls are closed and bounded. Observe that an open ball centered at the origin with radius 1 is contained in an open ball centered at the origin with radius 2. Therefore, an open ball centered at the origin with radius 1 is contained in the union of a collection of open sets (namely, any collection containing an open ball centered at the origin with radius 2). Therefore, an open ball centered at the origin with radius 1 is contained in a finite number of sets from this collection (just one: the radius 2 ball). By the definition of compactness, this means an open ball centered at the origin with radius 1 is compact. By the Heine-Borel Theorem, this means an open ball centered at the origin with radius 1 is closed and bounded. The same logic holds for open balls centered anywhere, with arbitrary radii. Hence open balls are closed and bounded. QED
Haha! So, lets reread what the definition of compactness is. It has to be every possible open covering, not just a single one. So for every possible open covering of say (0,1), there has to be a finite subcover. Since the union of the collection (0, 1 - 1/n) as n ranges over the positive integers has no finite subcover, the set (0,1) is not compact. Since (0,1) is an open ball in R^1, this contradicts your "proof". Essentially this line "By the definition of compactness, this means an open ball centered at the origin with radius 1 is compact." is false.
Some examples would have been infinitely helpful. Since they were not provided, I made some myself. Take the interval ( 2, 4). There's a lot of ways to cover this with a finite number of open intervals -- e.g., ∪{ ( 1, 3) ( 3, 5)}, etc.. So is it compact? No: We can construct an infinite sequence of open intervals ( 2 + 𝜀, 3), every element of which is needed to cover ( 2, 4). Aha! This is the point of the theorem: To weed out sets that involve an infinitely protracted approach! So what happens if the interval is closed -- [ 2, 4]? Then "it's clear" -- as mathematicians say! -- that no amount of open intervals is enough to cover the lower bound, 2, without overshooting it. As soon as we encounter an open interval that includes 2, it also includes points beyond 2: It's "open ended". And once we get to this interval, we can stop. To be rigorous, we can also consider interior points: There too, the covering has to contain an interval that contains the point, and once we get that interval, we can quit, well shy of infinity. The principle is this: a cover can always involve an infinite number of open sets, but the only case when an infinite number are needed is the case where we approach a point without actually including it. [ 2, 4] is compact, but [ 2, 4] - { 3} is not. Is every closed set compact? No: Here the counter-example is the set of integers.
instead of editing the shit out of the video to correct for the fact that you used m and M contrastively, why didn't you just use different fucking letters in the first place? lOgIc!
Please return king
Great explanation. It would be great if you would make more videos.
Appreciate the amount of hard work put in
i somehow managed to prove that open does not imply compact for R^1 but it was a very technical and fairly unstandard, it used the idea that open intervals which are subsets of R can be expressed as an "infinite sequence of closed intervals" (the sequences in the intervals say [a_n,b_n] converge to the same limit) in this case monotonically and where forall n [a_n,b_n] share an element with some [a_m,b_m]
hence forming an open cover which doesn't have a finite subcover because it'll contradict the monotonic convergence of [a_n,b_n] and hence open sets aren't compact, i'm not sure where to go next with this.
I got this from "Introductory Real Analysis" by Kolmogorov & Fomin, translated by Silverman.
Damn someone that actually knows how to explain things
Why are you not making more video like such?
Great video ,very clear and detailed explanation. Thank you !
Love the video and clarity, any chance you can do one specifically for R1 (just the real number line)
Thanks for the comment. Remember that this is only one direction of Heine-Borel, that compactness implies closed and boundedness. So if we take for example a specific compact subset of R^1 (say [0,1]), we are assuming it is compact and trying to show that it is closed and bounded. So for some point not in [0,1], like x = 1.2, we cover [0,1] with complements of closed balls (which in R^1 is closed intervals) around 1.2, so the complements of [1.1, 1.3], [1.15, 1.25], [1.19, 1.21], and so on, getting infinitely close to 1.2
The complements of those sets are open sets, for example the complement of [1.1, 1.3] is (-infinity, 1.1) U (1.3, +infinity), which is an open set. Notice how [0,1] is a subset of (-infinity, 1.1) U (1.3, +infinity)! So these complements are an open cover of [0,1]. So since we are assuming [0,1] is compact, we can say there is a finite subcover, the infinite process going towards 1.2 stops at a certain point.
So we can find a closed ball (a closed interval in R^1) around 1.2 that is NOT in the open cover (which would be exactly the closed interval that the process stopped at). In R^1 we can delete the endpoints of the interval we stopped at, for instance say the process stopped at [1.19, 1.21]. Then we know the open set (1.19, 1.21) is completely in the complement of [0,1] (because [0,1] is covered by those sets which the process stopped). This shows that the point 1.2 is in an open set entirely contained in the complement of [0,1]. Now since this is just for a single point, it is not a proof, but the argument holds if you take an arbitrary point outside of [0,1].
To prove a set is open in R^1 you can show that every point in it is contained in some open interval that is then contained in the set. Also if the complement of a set is open, then the set is closed. So if [0,1] has a complement that is open, [0,1] is closed.
That shows that it is closed. Try to figure out why a set like [0,1] is bounded (think about what the definition of bounded in R^1 means, what are some different ways you could define it?)
I hope this helps, keep on going with upper math, it is a worthwhile pursuit.
Note: NOT the Hairy Ball theorem!
great video , thanks for your efforts
Proof that open balls are closed and bounded.
Observe that an open ball centered at the origin with radius 1 is contained in an open ball centered at the origin with radius 2.
Therefore, an open ball centered at the origin with radius 1 is contained in the union of a collection of open sets (namely, any collection containing an open ball centered at the origin with radius 2).
Therefore, an open ball centered at the origin with radius 1 is contained in a finite number of sets from this collection (just one: the radius 2 ball).
By the definition of compactness, this means an open ball centered at the origin with radius 1 is compact.
By the Heine-Borel Theorem, this means an open ball centered at the origin with radius 1 is closed and bounded.
The same logic holds for open balls centered anywhere, with arbitrary radii.
Hence open balls are closed and bounded.
QED
Haha! So, lets reread what the definition of compactness is. It has to be every possible open covering, not just a single one. So for every possible open covering of say (0,1), there has to be a finite subcover. Since the union of the collection (0, 1 - 1/n) as n ranges over the positive integers has no finite subcover, the set (0,1) is not compact. Since (0,1) is an open ball in R^1, this contradicts your "proof". Essentially this line "By the definition of compactness, this means an open ball centered at the origin with radius 1 is compact." is false.
13:36 narrator: there was no next time
Great video
Some examples would have been infinitely helpful. Since they were not provided, I made some myself. Take the interval ( 2, 4). There's a lot of ways to cover this with a finite number of open intervals -- e.g., ∪{ ( 1, 3) ( 3, 5)}, etc.. So is it compact? No: We can construct an infinite sequence of open intervals ( 2 + 𝜀, 3), every element of which is needed to cover ( 2, 4). Aha! This is the point of the theorem: To weed out sets that involve an infinitely protracted approach!
So what happens if the interval is closed -- [ 2, 4]? Then "it's clear" -- as mathematicians say! -- that no amount of open intervals is enough to cover the lower bound, 2, without overshooting it. As soon as we encounter an open interval that includes 2, it also includes points beyond 2: It's "open ended". And once we get to this interval, we can stop. To be rigorous, we can also consider interior points: There too, the covering has to contain an interval that contains the point, and once we get that interval, we can quit, well shy of infinity. The principle is this: a cover can always involve an infinite number of open sets, but the only case when an infinite number are needed is the case where we approach a point without actually including it. [ 2, 4] is compact, but [ 2, 4] - { 3} is not.
Is every closed set compact? No: Here the counter-example is the set of integers.
Amazing vid! Why didn’t I come across it earlier...!
instead of editing the shit out of the video to correct for the fact that you used m and M contrastively, why didn't you just use different fucking letters in the first place?
lOgIc!