5

ID1F Ci(10Show that the following: a) Every Compact Metric Spaces is sequentially Compact Spacesb) Every Lindelof Metric Spaces is Separable Spaces...

Question

ID1F Ci(10Show that the following: a) Every Compact Metric Spaces is sequentially Compact Spacesb) Every Lindelof Metric Spaces is Separable Spaces

ID1F Ci(10 Show that the following: a) Every Compact Metric Spaces is sequentially Compact Spaces b) Every Lindelof Metric Spaces is Separable Spaces



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

This problem covers limits. In order to solve part A and B of this problem, we will use the approach of contradiction. Let's look at part A. First. The claim is that if S. N. Is greater than equal to A. For all but finite lee many N than limit, S N is greater than equal to A. Let us assume for a contradiction that this limit Yeah, is less than let us assume. And let's say absalon is equal to a minus S. This is the assumption. By the definition of limits, we can say that. Bye. Definition of limit. We can see that there exists capital N. Substantive small and is greater than capital in then sn minus S is less than epsilon. This is what the definition says. I'll write it down there exists capital and such that Yeah. If and is greater than capital N. Then and send minus S. Yes, less than absolute. So if you open this inequality, you know, you will get minus epsilon Nice between s and minus S taking minus s on both sides. You get s minus absalon as less absalon. Now, you can see here that S plus epsilon is nothing but A. So you will use that factor and I can say that s minus epsilon. Uh it means that essen belongs to s minus absalon. Call my A. What does this mean? It means that S. N. Is less than a, S. N is less than A. But our problems said that our assumption was based that S and will be greater than equal to way, but we got the opposite. It means that for all but finite lee many end S. N. Is less than in this problem, as per the assumption, which is a contradiction. It means that our initial claim was correct that if S. N. Is greater than equal to A. For all but financially many in limit S and will be greater than equal to It similarly will solve part B of the problem. Part B says that if S. N is less than equal to be for all, but finitely many end then limit essen is less than equal to be. Let us assume again for contradiction. We will assume, Oh, that limit essen is greater than B. We're assuming as a contradiction. And let us take absalon as S minus B. Again, by the definition of limit there exists and such that if N is greater than in the sm minus S is less than absolute, just like this. We use their definition of LTD so we get yes and minus S is less than epsilon. Again, opening the inequality minus epsilon, S n minus S rather than absolute, this becomes s minus epsilon. Essen? S less. Absolutely. From here, we can see that S many cepsa loans nothing but B. So I can see here, B essen less than s plus absalon. So from here it is clearly seen that sns greater than B. Or I can say for all. But finitely many N. We have S. And greater than B, which is a contradiction according to our statement. Because we took us and less than equal to be. Hence the claim that if S. N. Is greater than less than equal to be for all. But finitely many end then limit us and less than equal to be was correct. And contradiction proved it false. Let's go to part C. Part C is a direct conclusion from part A and part B from part A. You know that? Yes. Which is the limit limit of essen is greater than equal to A. And from part B. This limit S it's less than equal to me, so S belongs to it to be. That's all. Yeah.

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.


Similar Solved Questions

5 answers
Actual amount of salicylic Calculate your percent yield acid you used 'acetylsalicylic the acid, ciperircenember adjust your theoretical 1 the
actual amount of salicylic Calculate your percent yield acid you used 'acetylsalicylic the acid, ciperircenember adjust your theoretical 1 the...
4 answers
Rebyed Io thot colcge map # 7 ol ihoso Survey subjocts &re pandomly sctectcd #ahoue Srtid 3C0 coleqe Qiad +ics, E016 Icoctted Nut icy Esteted 0 proteson ckrcy thcu colcol Min7 Pman thrct decim HpieS Kosnen in4 idlz Up *IICT Culd Poublyytal 3d Iha - enlrcd # Erotesuon ckscly rualed
rebyed Io thot colcge map # 7 ol ihoso Survey subjocts &re pandomly sctectcd #ahoue Srtid 3C0 coleqe Qiad +ics, E016 Icoctted Nut icy Esteted 0 proteson ckrcy thcu colcol Min7 Pman thrct decim HpieS Kosnen in4 idlz Up *IICT Culd Poublyytal 3d Iha - enlrcd # Erotesuon ckscly rualed...
4 answers
For Ihe reaction Lelow; lhe equilibrium constant Kp 6.4x 10-25. If initially the pressure = 2.7 alm and the pressure of Nzo 6.9 alm with no NO. calculate the prossure of NO at equilibrium. Enter your answer using notation with to s gnificant figures. 02 (g) + 2N,0 (g) 4 NO (g)
For Ihe reaction Lelow; lhe equilibrium constant Kp 6.4x 10-25. If initially the pressure = 2.7 alm and the pressure of Nzo 6.9 alm with no NO. calculate the prossure of NO at equilibrium. Enter your answer using notation with to s gnificant figures. 02 (g) + 2N,0 (g) 4 NO (g)...
5 answers
What is the net charge of the peptide below at physiological pH? 2 B Ala-Arg-Asn-Asp-Glu-Ser-Gly+1 E +2Consider the following amino acid titration curve and select the TRUE statement
What is the net charge of the peptide below at physiological pH? 2 B Ala-Arg-Asn-Asp-Glu-Ser-Gly +1 E +2 Consider the following amino acid titration curve and select the TRUE statement...
5 answers
For the following exercises nrove Identity gwen (sin sin 2tsin(2x) ~2sin(-x) cos (~x)
For the following exercises nrove Identity gwen (sin sin 2t sin(2x) ~2sin(-x) cos (~x)...
5 answers
W1 Funj 0 #encala#W cecatlez It 1 ltue 1
W 1 Funj 0 #encala#W cecatlez It 1 ltue 1...
5 answers
Consider the second order differential equation2y=51' +In(r)_Find the general solution of yt . (b) Then, find y(x) = Y (x) + J,(x) using Method Of Undetermined Coeflicients
Consider the second order differential equation 2y=51' +In(r)_ Find the general solution of yt . (b) Then, find y(x) = Y (x) + J,(x) using Method Of Undetermined Coeflicients...
5 answers
A study of a ! local city is being conducted to find out the average number of people that use public transportation every day: The sample mean should be within 11 people of the mean people using public transportation The standard deviation is 100. What is the sample size necessary for a 90% confidence level?0 152230 2240 14.95
A study of a ! local city is being conducted to find out the average number of people that use public transportation every day: The sample mean should be within 11 people of the mean people using public transportation The standard deviation is 100. What is the sample size necessary for a 90% confide...
5 answers
First WaySecond Way((Pk-Q) + 5) ((-P& Q) + 5) (~(Pk Q) & (Pv Q) )Premise Premise Premise&ERHER E 3EGoal
First Way Second Way ((Pk-Q) + 5) ((-P& Q) + 5) (~(Pk Q) & (Pv Q) ) Premise Premise Premise &ER HER E 3E Goal...
5 answers
The region enclosed by y=x2 andy=x in the first quadrant is rotated about the liney=2. Find the volume of the solid of revolution.
The region enclosed by y=x2 and y=x in the first quadrant is rotated about the line y=2. Find the volume of the solid of revolution....
5 answers
Calculate the Ka of your weak acid using three different points on the titration curve:0% to the equivalence point (meaning before any base was added) 50% to the equivalence point75% to the equivalence point% to the Equivalence PointpHKa value0%50%75%
Calculate the Ka of your weak acid using three different points on the titration curve: 0% to the equivalence point (meaning before any base was added) 50% to the equivalence point 75% to the equivalence point % to the Equivalence Point pH Ka value 0% 50% 75%...
5 answers
Question 2Find the length of the curve Y=1+2+3/2 over the interval 0 <x<1. OA L= (2/27 [10-1o+1] 6L = Vio Ni Oc L=(2/27)[10-1o +2 2 ] OD L=10V1 - 2V2 'OE L= (2/27 [10 /1o - 1]
Question 2 Find the length of the curve Y=1+2+3/2 over the interval 0 <x<1. OA L= (2/27 [10-1o+1] 6L = Vio Ni Oc L=(2/27)[10-1o +2 2 ] OD L=10V1 - 2V2 'OE L= (2/27 [10 /1o - 1]...
5 answers
The one-to-one functions g and are defined as follows_g= {(-6, 7), (3, 0), (5, 3), (6, 8)}h(r) =2x-3Find the following_h"'6)("' .)ts)
The one-to-one functions g and are defined as follows_ g= {(-6, 7), (3, 0), (5, 3), (6, 8)} h(r) =2x-3 Find the following_ h"'6) ("' .)ts)...
5 answers
Find the work done by the force field F(x, Y, 2) = Txi + Tyj + 2k on a particle that moves along the helix r(t) = S cos(t)i + 5 sin(t)j + Stk,0 < t < 21 .
Find the work done by the force field F(x, Y, 2) = Txi + Tyj + 2k on a particle that moves along the helix r(t) = S cos(t)i + 5 sin(t)j + Stk,0 < t < 21 ....
5 answers
A 5.994 g sample of a substance (containing carbon, hydrogen andnitrogen) was burned in oxygen, and the carbon dioxide and waterproduced were carefully collected and weighed. The mass of thecarbon dioxide was 17.67 g, and the mass of the water was 5.428 g.What was the empirical formula of the substance?
A 5.994 g sample of a substance (containing carbon, hydrogen and nitrogen) was burned in oxygen, and the carbon dioxide and water produced were carefully collected and weighed. The mass of the carbon dioxide was 17.67 g, and the mass of the water was 5.428 g. What was the empirical formula of the su...
5 answers
Equal masses of solid Ad liquld are combiued in clostd system and allowed to T equilibrium- plot of temperature v8. timc js sho# below . How dces the specilic heat of the solid compare to that of thc liqquid? Justif; your unswer . (6 points)When the system comes to equilibrium is it all liquid, all solid, or mix of liquid and gull? Iirsltty YoU answer. (6 points)
Equal masses of solid Ad liquld are combiued in clostd system and allowed to T equilibrium- plot of temperature v8. timc js sho# below . How dces the specilic heat of the solid compare to that of thc liqquid? Justif; your unswer . (6 points) When the system comes to equilibrium is it all liquid, all...
5 answers
Ieuasticn 11pesAfunction fx Y) has & critical point atx- 2,Y = 4,ie_at (2,4) The secand partial derivatives of the function ara;in the usual notation; fakxM) = 18x. fxx v) = 2, fMxv) =Usc the sccond derivative test to choose Irom the (ollowing at the point (2,4}Thrro KanLitive nAInWIMlato clatnve mnlmumThryck a edalla pott0 Tha ucond detivaive Icvl /dvud Wr € ArVIO[ Lykind 0t critical [KAnt Ioexochns :1 (2
Ieuasticn 11 pes Afunction fx Y) has & critical point atx- 2,Y = 4,ie_at (2,4) The secand partial derivatives of the function ara;in the usual notation; fakxM) = 18x. fxx v) = 2, fMxv) = Usc the sccond derivative test to choose Irom the (ollowing at the point (2,4} Thrro KanLitive nAInWI Mlato c...
5 answers
4xly" 1 ~Ke + following ' 4xy differential 8x3 equation homogeneous solution Yh
4xly" 1 ~Ke + following ' 4xy differential 8x3 equation homogeneous solution Yh...
5 answers
Flnd the Esolution of the [ 3 Initia value problem,+ 2 2 =5,/ (0) =f() = 1 1,0 <t<
Flnd the Esolution of the [ 3 Initia value problem, + 2 2 =5,/ (0) = f() = 1 1,0 <t<...

-- 0.019937--