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Prove every finite subset of R" is closed _ Show by examples that an infinite subset of R" can be closed, open neither closed nor open...

Question

Prove every finite subset of R" is closed _ Show by examples that an infinite subset of R" can be closed, open neither closed nor open

Prove every finite subset of R" is closed _ Show by examples that an infinite subset of R" can be closed, open neither closed nor open



Answers

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

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.

So we want to prove that the supreme um this set for R. And Q. Such the R. Is less than A. That's exactly uh you know for all A. And R. So let's just to make things easier, let's say yes. Is the set okay? R. In Q. Such there are is lessened. Okay then just by the definition here that means for any R. N. S. Our must be less than A. It's just given by how we're setting up this set. You know what that means? Then the Supreme of this set has to be listening. Hey? So let's assume the Supreme A. S. Does not equal A. Mhm. Okay then since the supreme has to be less than able today, that would mean that the supreme um of us is taken less than A. Okay but now by the dense nous of Q. So so this follows this line up. Um But then by the dense nous of Q. Right? By dense. This is just between every two like rational numbers. You can always find I had. Yeah. I mean so in between any two numbers you can find um a number and cute, right? And as long as they are rational. And so this means that there is this some are in Q. Such that yeah is less than r. Is less than a right? The Supreme a mess is less than ours. Okay? But since since this are is less than a than this arm must be in S. Okay. Which tells us that our is less than or equal two. None of this. But this contradicts we had just above right and this is a contradiction occurred when we assumed the supreme was not equal to a. We have a contradiction. So therefore the Supreme one of us is equal today that includes it.

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.

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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