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Problem 3. Let (1.0) be discrete metric space; and {~,} sequence in X Prove that if and only if there exists NeN such that $ Vn > . Problem Let and be two unifor...

Question

Problem 3. Let (1.0) be discrete metric space; and {~,} sequence in X Prove that if and only if there exists NeN such that $ Vn > . Problem Let and be two uniformly equivalent metrics defined on X and {ra} be sequence in X Show that Ir metric in mnetric

Problem 3. Let (1.0) be discrete metric space; and {~,} sequence in X Prove that if and only if there exists NeN such that $ Vn > . Problem Let and be two uniformly equivalent metrics defined on X and {ra} be sequence in X Show that Ir metric in mnetric



Answers

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

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.

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.

In this question here were given. Uh, I am equal to the B N minus. B a minus one. And we're interested in the submission that I am from one option infinity. So notice that they will explain this one when any could you want him to be one minus B zero, then plus B two minus, uh, be one plus the B three months, B two plus the big four minus P five plus up to the PM minus P m minus one. And so, um, and we see them, we can console the B one window. Be one here, be to win the B two B 300 b three. So we're missing the beat three years. So sorry. So let me in. Center doesn't before, uh, be 1234 minus b three here. And then we again this one will be canceled with this one and so on. And here we see that we will cancel this and so on on. We see we have left with only, uh if we consider only up to in the past with some s and for now, then we just stop up to here and then we see that this s and we ego Thio actually end here on me and then we see that this sn it will equal to the We have left with the minus p zero then plus with a p n. And we see that this is here as an we turned the limit here Angus t infinity And then he coaches limit on the industry Infinite. They on the minus B plus B n And we say this one we get echoed you This one will be p zero. So it will be the minus p zero and this leap plus the limit on the PM and goes to infinity. I doesn't implies that the I the submission on the I m converge even only if this limit here also. Uh, I already mitt because you some constant else model that infinity

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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