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(14 points) Let A = [o. 1] u {2+$:new}: Construct an open cover of A in R that does TOt have finite subcover of A_ Give Four reasoning...

Question

(14 points) Let A = [o. 1] u {2+$:new}: Construct an open cover of A in R that does TOt have finite subcover of A_ Give Four reasoning

(14 points) Let A = [o. 1] u {2+$:new}: Construct an open cover of A in R that does TOt have finite subcover of A_ Give Four reasoning



Answers

Show $\sup \{r \in \mathbb{Q} : r < a\}
= a$ for each $a \in \mathbb{R}$.

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

In this problem, we have been asked to prove that the subsets of finite non empty set X are partially ordered by set inclusion. So what we need to do is prove that this relation of set inclusion is a fractional order. So, first of all we need to prove that it is reflexive. This relation will be reflexive because let us consider any subsidy of X. And we can say that he is a subset of E. It is a subset of itself and thus is related to enhance. This relation is reflexive. Next we need to show that it is anti symmetric. Now, for that let us consider any two subsets of X, A and B. And let us assume that he is related to B and B is related to A. So that means that is a subset of B, and B is a subset of A. Now if is included in B and bs included an A. Then the only possible conclusion is that the two sets A and B are equal. So, since this condition implies that is equal to be. Hence we can say that the relation is anti symmetric. And the third condition that we need to show is that it is transitive. So for that let us consider three subsets of X, A, B and C. And let us suppose that is a subset of B, and B is a subset of. See. Now, if A is included in B and bs included in C, then we can conclude that is included in C. For example, if we consider a Venn diagram, this is the set E. This is included in the set B, and that's it is included in the set C. So from this diagram we can see that the set E is included in this set. See So this is what we get. So if it is related to B and B is related to see, we get the day is related to. See. Hence it is transitive. Thus, set inclusion is reflexive, it is anti symmetric, and it is also transitive, thus it is a partial order, and thus we approve that the subsets of a given finite non implicit X are partially ordered by set inclusion.

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.


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