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Let $ be 6 nahemply open subset of R Hhat is boundel fro m below _ shoW thart inf $ exists but that mins does Kot exis}....

Question

Let $ be 6 nahemply open subset of R Hhat is boundel fro m below _ shoW thart inf $ exists but that mins does Kot exis}.

Let $ be 6 nahemply open subset of R Hhat is boundel fro m below _ shoW thart inf $ exists but that mins does Kot exis}.



Answers

Prove that if $a>0$, then there exists $n\in\mathbb{N}$ such that $\frac{1}{n}< a < n$.
Thank you!

In this problem, we are given to subsets of our name, the S. And T. And we have the property that given us and T. S. Is going to be less than or equal to T. For every S. And S. And T. N. T. So we're first supposed to show that S. Is bounded from above. So in order to do that, we just pick a tea from tea. So it's like T. One B. From T. Then for every S. And S. S. Is less than or equal to D. One us founded above. Similarly for bounded below. For S. Let her T. Sorry, let S. One B. And S. Then for every T. And T. S. One is less than or equal to T. Us founded. Hello. We're now supposed to show approve that's the supreme um of S. Is less than the infamous um of T. So let's do this by contradiction. So suppose this is not the case. That's a pretty mom of S strictly greater than inform um of T. And this implies there exists some S. In S. Such that the S. Is greater than the infamous um of T. Furthermore there exists a T. Into such that the in form um of T. Is less than or equal to T. It's less than S. Because otherwise if that wasn't the case S. Would be the in form. Um Right. And we have picked that S. Is greater than the infamous. Um Yeah so that means there exists T. And S. Such. That T. Is less than S. But this is a contradiction because all asses are less than three. Quality us. The in form um O. T. Is less than or equal to the supreme um of us. So then we're supposed to come up with a couple of examples where we have S. Intersect T. Is not the empty set. And we can simply put this as well. S. B. 0 to 1 and that T be 1 to 2. All right. Every element of S. Is less than a record of every element of T. And the intersection it's not the empty set. We want. Now an example where the supreme um of S. Equals the in form um of tea but we want s intersex T. To be the empty set. So how can we do this? Well we can take our other example and just open it up. Sp 0 to 1. The open interval and let T. V. One two. The closed interval open interval. So the in form um of T. Is one, the supreme um of S. Is one and those are equal.

Okay. So the question is as follows, where we look at the status, which we say is the smallest sigma algebra, What is a signal of zero generated by the sigma algebra in in our right, generated by the set are Come on R Plus one. Where are is a rational number. So this is the problem statement should be on our Okay, so moreover, we need to show that S is the collection of moral subsets of our So we need to say that S is the borough subsets. Okay, well the barrel uh, the barrels sat on our is you can think of it as the smallest. Take my algebra generated by uh let's say a basis for the topology of our and a basis for the topology of our is just as I said, A B where am br elements of the reels. Okay, so the topology, the standard topology and our is generated by such sets. And if we can show that S contains uh a set of the form, A comma, B where A and B are real numbers, then that's sufficient to demonstrate. That. Is is uh wow the moral sigma algebra on our and the way that we do this is what we grab one of these. Uh the goal here is to make that said a B. Right. Where A and B are real numbers. You're seeing or from Sets of the form are comma are plus one where are is irrational. Uh taking these sets to these such sets to be an element of sigma algebra. Okay then the way that we do this is let me make some space for these. Oops. So let's look at this set A Excuse me? A comma eight plus one. Right. And uh let me let me do something. A sub N. Coma A sub N plus one. Where S A Ben is a series of rational numbers, right? Such that he saw Ben converges to a rational number. A uh sorry, a real number A From the right. Okay. Then if we take the union over end of all. A seven come on in some men Plus one, guess what? We're going to get this set A comma a Plus one. And this set will be an element of S. And it's important to note this because A Now is a real number. Right? And we want to do something similar. Right? Now we look at the set B. Seven come up beasts up and plus one where of course be summoned once again is rational, right? And we say that Visa ban converges to be from the left and B is a real number. Then it turns out that the complement of this set complement of this set, which will be let's look at let me copy this. I don't have to rewrite it. Once again, the complement of this set will be the same from negative infinity coma be suburban Union be. So then plus one comma infinity. Right? And this set will be in S because see my algebra. Czar close with respect to compliments. And now if we grab this set and we intersected with the set that we found down here, so let me grab it and copy they grabbed the set right, Win or sickbed. We'd are set here. It's what we're going to get. We're making the assumption. Obviously that is less than strictly less than the So keep that in mind. Well, if we do these, we're going to get this set a comma B. So then closed. And if we take the union over end of these sets, we will get a comma B, which is the goal that we were trying to achieve. So the sigma algebra, in conclusion, the sigma algebra generated by this set contains sets of this, the following form and all sets of the following form. And we know that all such sets for my basis for the topology of our therefore the city. S. Is the world sigma algebra owner

Yeah. So in this problem we need to sure that any bounded decreasing sequence of real numbers converges. So we are given about a decreasing sequence. We named that A. N. And we need to show that A. N. Can gorgeous to what limit? We don't care about this. Okay, so let's start with the proof, send the sequence in is bounded in particular. It does bounded below that is the set A. And such that and comes from natural numbers is a bounded subset five. Have we known that are has completeness property also known as and you'll be property that is least upper bound property. And this is equal into G. L. B. Property also known as greatest lower bound property, basically this means, but if any subset of R is bounded below, then it must have an infamy. Um So in fema of the scent and coming from and coming from national numbers is some real number. Let's call this L. So our claim is like the sequence N. Can we just to L now, since L is the greatest No, it bound after said of elements of sequences of the terms of sequences. So for any absolutely. Didn't see role as minus epsilon, which is strictly less than N. I'm sorry, L plus epsilon, which is strictly created. An L cannot be by lower bound. So there exists some. And in particular there exists some term A N. That is less than L plus of silence. But since in is a decreasing sequence we have a N less than equal to capital and for every and greater than or equal to capital. And so in particular we have and less than L plus a silent for every angry to an end no sense. And is can feel them. We have less than equal to a N. For every n natural numbers. In particular, l minus epsilon, which is strictly less than L, would also be less than a N for every angrier than capital. And so we have this inequality right here and this inequality right here and we combined them to find that for every epsilon created in zero, there exists a national number and such side for every end. After that comes after that natural number added minus upside and is less than a. In which is less than L. Plus of silent. That means A. In the sequence A in Canada. Just to L. This completes option for this term.

Okay. So I need to prove that A. Is greater than zero and there exists and such that one over less than a. This and then so we're going to use the property where A. Is greater than zero and B. Is greater than zero. Then for some integer N. We have that in times A. This period can be yes, it's a comedian happen. So first we'll say let's let A equal one and be equals A. Then there this property must exist and then such that and one times one. Which is it was a be greater than a. Okay in other words this this um and one description A. So now and say well let's let a little A. And this baby is going to equal on it. Okay, so by the same property there must be some other some different yeah such that A. It's greater than one over and okay. And this comes from did kind of skip a step. Um that's derived from if we'd written it out as and to a. Being greater than one the bible side. Yeah to it this thing so now we're just gonna let N. N. So the N. B. The maximum of these two introduce then what we have is that end it's greater than or equal to N. One and and it's greater than or equal to and to the only what happens to be more such ad A. Such that which that A. Is less than and one is less than or equal to then. Okay that's going to follow up the first one and yeah one over N. It's just an equal to one over and two less than A. That follows from that second round. We did okay so we just got to combine these two pieces, plug them together we get that A. Is greater than one over N. But less than and Okay and so and that is our solution.


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