Why I never came across such explanation under any definition of compactness on the internet? Why do I have to search so long to find such a simple example that finally makes the concept clear? Thank you sir !
I know this is a few months old, but I will just reply here for those that see this and are confused about it. The idea is that if the set is compact, no matter how you many open balls you use to cover the interval, you can take a finite amount of those open balls and still cover the set. For the open interval (0,1), you can do that as well and that will be one example of a way to cover the set with a finite subcover or a finite amount of open balls. But in order for the set to be compact, it has to hold for any cover of the set. And the video shows one example of a cover where you cannot take a finite subcover to cover the interval (0,1). That is why the interval (0,1) is not compact. It has to hold for every cover of the set.
This is just a counter example, Since by definition of compactness every open cover has finite subcover And here we are getting a open cover which does not have finite subcover So this is not compact And for the second one, whenever you take any interval near the endpoint it will overlap with some other interval inside. The moment it overlaps, we will get some finite collection no matter how big it is, it will be finite So closed interval is compact
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Why I never came across such explanation under any definition of compactness on the internet? Why do I have to search so long to find such a simple example that finally makes the concept clear? Thank you sir !
Thanks a lot ! I just wasn't getting this all morning and now it seems all so clear
I don't think so
Is [0,∞) compact manifold with boundary?
sir give me concept of bolzano weietrass theorem of real number
Still I have doubt ..can you help me ?
Thank you for the Insight.
This video is awesome!
why not put the orange intervals near the endpoints for (0,1) like you did for [0,1]??
Same doubt here
I know this is a few months old, but I will just reply here for those that see this and are confused about it.
The idea is that if the set is compact, no matter how you many open balls you use to cover the interval, you can take a finite amount of those open balls and still cover the set.
For the open interval (0,1), you can do that as well and that will be one example of a way to cover the set with a finite subcover or a finite amount of open balls. But in order for the set to be compact, it has to hold for any cover of the set. And the video shows one example of a cover where you cannot take a finite subcover to cover the interval (0,1). That is why the interval (0,1) is not compact. It has to hold for every cover of the set.
This is just a counter example,
Since by definition of compactness every open cover has finite subcover
And here we are getting a open cover which does not have finite subcover
So this is not compact
And for the second one, whenever you take any interval near the endpoint it will overlap with some other interval inside. The moment it overlaps, we will get some finite collection no matter how big it is, it will be finite
So closed interval is compact
same
@@FocusedandChill but the counter example for (0,1) can also be a counter example for [0,1]
Finally this makes sense
The sound is horrible, I
it's painful follow
Hmm i see no issues