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By exhibiting an open cover that does not have a finite subcover prove that the following spaces are not compact.•The space (0, 1) ∪ (2, 3) as a subspace...

Question

By exhibiting an open cover that does not have a finite subcover prove that the following spaces are not compact.•The space (0, 1) ∪ (2, 3) as a subspace of R with the standard topology.• The open ball B(0; 1) in R2 with the topology induced by the Euclidean metric. • The interval (0, 1] as a subspace of R with the standard topology.

By exhibiting an open cover that does not have a finite subcover prove that the following spaces are not compact.•The space (0, 1) ∪ (2, 3) as a subspace of R with the standard topology.• The open ball B(0; 1) in R2 with the topology induced by the Euclidean metric. • The interval (0, 1] as a subspace of R with the standard topology.



Answers

Prove that all bounded monotone sequences converge for bounded decreasing sequences.

This question covers topic relating to linear and zebra and the span of uh basis of uh vector space. So you can see here we have the set as a finite and had the route that the the linear combinations of all vectors of S. Is actually there's a set of fee. Okay, so how do we do that? First of all? You refine the uh you can say like we can pick a finite set um the subset of S. And that obviously ideas upset of the vector space V. And because it's the it's an element in sign as actually inside V. So that ai can be rewritten as B J E J. What do you Hj? So E J is um it can be finite or infinite is a basics the basis for V. Okay, so we um so as you can see here um a I can be rewritten as a linear combination of these spaces. Right? And now if we pick any vector, like for example, I pick victor, A and A. Is A and finally linear combination of ai right? So for example, and for I ai right from one, for example, from one to let's say end and maybe uh M. Okay. And so we can re written that some as and for nighttime submission of changing from one to N. B to J. J. And death is just a linear combinations. Uh The finite clinical combination of or the vector E. Uh J. Right? So if you want to write it down precisely, it's just the summation from I from one to M. And information from che from one to end on. For I B J E K. Right? And if you put out the so ehh Okay. So now if you put out each A out put E. J. As you can okay, you can interchange this summation to change this to end and you change it to em. And and for I. B. J. E. J. Right? Okay. So now just uh J equal to one to end of E. J. Um, submission I from 1 to 1 for I beta J. Okay, So this is just a real number, right? So it's a linear combination, so that means I belong to the span of ass. Or so that means the span of S. Is upset of fee.

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

Mhm. We are given a space as that contains polynomial of degree three year. Lower that fulfills a given equivalence Part one asked us to show that this set S. Is a subspace of Polynesia polynomial of degree three. And the solution is quite straightforward since within the definition of the set it only takes polynomial polynomial from Polynomial of degree three. The birth, so all all polynomial and S. They will be in polynomial three. A Parliament of decrease three. Mhm. Uh There does exist the polynomial in S. It's a non empty space as it as we have a table here showing different coefficients for for the different powers of X. In the equation. X cubed plus B, X squared plus C. X plus deep. Okay. Uh so there exists upon them inset therefore some empty set. Therefore it's a proper subspace. And we are done moving on. Part two. We are asked to find a polynomial in S where not all the coefficients are zero. So we opened up a class for all polynomial is in X. We can denote can denote the polynomial as such. Okay. Yeah. Uh I have also noted three times the polo meal here and then the polynomial different differentiated I assume uh listener knows how to differentiate polynomial and I'm going to move on and then I multiplied the differentiated polynomial by X plus one giving us this formula. And finally I am using the equation that was given to us in the definition of set S. Yeah. So we can drive several things from this equation over here. Uh We need this needs to hold true for all access so we need three A minus B uh equals zero. Or in other words we need 38 equals B. Since two B -2 C need to be zero. Include that the equal. See And since it contains three And also contains C -3 d. You conclude that C Must be equal to three D. So you can just pick A # 40. Like one seems to be three times team. So it's three. He needs to be seen. So It's the same as seen was three. And there needs to be three times be. So it's nine. I did the same thing for D equals two. And he goes three. You see that this is true for any real number I just the D equals I. C. Equals three, I. B equals three I. And equals nine i. So our space s can be rewritten as such with it. You can derive any uh any polynomial from s in a very simple manner.

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.


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