🔶09 - Bound Set and Bolzano Weierstrass Theorem | Upper Bounds, Lower Bounds and Limit Points

[@KhadijaKamil-el3js]

Please, for the last question, why is the set closed

[@baklash5676]

Can you have recap sessions where you just solve different types of questions please

[@baklash5676]

@SkanCity Academy

[@SkanCityAcademy_SirJohn]

Okay, will consider that

[@baklash5676]

@@SkanCityAcademy_SirJohn appreciate it🙏

[@emmanuelarthur3081]

Great video. But I didn't quite get the last example .I understand why the set SB has a greatest lower bound of 0.9, but can you please explain to me why the set SA has a least upper bound element of 1.1.Since the set is approaching infinity must not the greatest element be infinity

[@SkanCityAcademy_SirJohn]

The set is approaching 1, notice that all the values are getting closer to 1

[@emmanuelarthur3081]

@@SkanCityAcademy_SirJohn That is for the set Sb right? Because it seems the set Sa continues from 1.1.Does Sa also approach 1?

[@SkanCityAcademy_SirJohn]

​@@emmanuelarthur3081 No its for the entire set, the entire set approaches 1, having 1.1, 1.101 and all that doesn't mean its approaching infinity, you realize that those values are more closer to one than they are to infinity, i only separated the who set into two different sets so that we can easily identify the g.l.b and the l.u.b​ @Emmanuel Arthur No its for the entire set, the entire set approaches 1, having 1.1, 1.101 and all that doesn't mean its approaching infinity, you realize that those values are more closer to one than they are to infinity, i only separated the who set into two different sets so that we can easily identify the g.l.b and the l.u.b

[@emmanuelarthur3081]

@@SkanCityAcademy_SirJohn I think I get it now .The sequence is in two folds, the first one has the 1.1 reducing towards 1 and the second has the 0.9 increasing towards one. So they both approach 1.Thank you

[@SkanCityAcademy_SirJohn]

all the values in the set are such that they are very close to one, all the values cluster around 1, so the limit is definitely 1 and not infinity, even though it looks like it is more than 1 and moving on and on, it is still closer to 1.

[@samueladdai8500]

Please can you make videos on sequences

[@SkanCityAcademy_SirJohn]

Yes, I can, but in the meantime, I'd advise you to know how you can relate this to sequence because it is doable as you await for videos on sequence.

[@teenamorrison5326]

The limit point is where the set approaches right? and infinity is like saying the set never ends, am I right? If so, how do the sets have a limit point but still called infinite?

[@SkanCityAcademy_SirJohn]

yes, but you need to understand that not all infinite sets have a limit point. the condition is, for an infinite set to have a limit point, its needs to be bounded, in the sense that it should have a lb and an ub. Now one's the infinite set is bounded, it means we have an idea of the values of the extremes, and hence we can know where the sequence or set is approaching even though it may not be specified in the question. So not every infinite set, but a bounded infinite set.

[@SkanCityAcademy_SirJohn]

the idea of infinite set is just to tell us that the last value is not given or made known in the set...

[@teenamorrison5326]

@@SkanCityAcademy_SirJohn thank you very much

[@SkanCityAcademy_SirJohn]

You're welcome Teena

[@evanstwedeer593]

Also, why is that 1 is not a limit point of set S? Neighborhoods of 1 intersection S has many points but you didn't include it. You just took 0

[@SkanCityAcademy_SirJohn]

1 can't be part. Look at the set carefully. 1, 1/2,.... are single elements in S. It is different from an interval like (1, infinity) so delta N of 1 intersect S contains only 1. Hence 1 is an isolated point but not a limit point.

[@kelvinkwawafful4649]

In the last video why is the limit 1 and not 0

[@SkanCityAcademy_SirJohn]

It's 1 because all the members of the set are converging at 1. Eg 0.9, 0.99, 0.999, 1.001, 1.01, 1.1 etc All these numbers are much closer to 1 than they are to zero

[@asaredurowaadoris195]

pls wat if the interval given is an open interval like (0,4) , will it have upper and lower bound

[@SkanCityAcademy_SirJohn]

yes, it will have. the idea is once you have a real number at both ends, regardless of the type of interval it is bounded, eg (0,4) and [1,5] are all bounded above and below. but (-inf, +inf) is not bounded, because we don't know its end values, and (0, +inf) is half bounded. Hope you are cool with the explanation?

[@asaredurowaadoris195]

Thank you

[@SkanCityAcademy_SirJohn]

You are welcome

[@evanstwedeer593]

Please why is the l.u.b. for set P not 1?🤔

[@SkanCityAcademy_SirJohn]

it cant be 1, because, the upper bounds are the numbers >= the largest number in the set which is 2, numbers >= 2 are: 2,3,4,5 etc. the least of these numbers is 2 hence l.u.b of P = 2

[@Novatron47]

can you do videos on electrical machines

[@SkanCityAcademy_SirJohn]

Due to certain situations, I'm not doing machines this sem. Sorry about that

[@baklash5676]

@@SkanCityAcademy_SirJohn what about transformers or thermodynamics

[@SkanCityAcademy_SirJohn]

Sorry I'm not doing that one too this sem. Sorry

[@albertkoomson7740]

👍🏽

[@KhadijaKamil-el3js]

Please, for the last question, why is the set closed

[@KhadijaKamil-el3js]

Please, for the last question, why is the set closed

[@SkanCityAcademy_SirJohn]

The set is closed because it contains it's limit point, which in this case is 1