prove that the intersection of all open intervals form a sub base on standard topology r

Name: prove that the intersection of all open intervals form a sub base on standard topology r
Uploaded: 2021-10-21
Duration: 2 min 24 s
Channel: Numerade
Description: prove that the intersection of all open intervals form a sub base on standard topology r

Question

Manisha Sarker · Accepted Answer

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&#x27;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.

Ahmad Reda · Answer

In Probleble 24. We have the boreal sigma feed on the rail line, which is given by this definition We&#x27;re absolutely. Here is a sigma feet and I the set of all open interpret all open intelligence first. Let&#x27;s revise the properties of a sigma field. Sigma field means is that if we have set a in sigma field then a compliment or in sigma Phil A and a compliment or in sigma phi. And for a sequence in in sigma field the unions and the intersection or in sigma feet. Take the union for any subsets and the intersection for any subsets we can find them again in sigma field. And finally sigma feed contains the empty set. By using these properties we can prove that be not contains also all closed and half of an interpreter. But before that from this definition with a note is absolutely and excellent is a sigma field. This means peanut in the segment field. As a shortcut to the conclusion that we know is a sigma field then we have been on is a sigma field and we know that we not contains all open intervals. Then if minus infinity to a. Is included, then it&#x27;s compliment is included in better known and this is open and terrible. Then we mean that it&#x27;s in bay. Better note sigma field then its complement is in with a note then minus infinity to a compliment is in beaten. What is the complement of minus infinity? It&#x27;s from a close to infinity. So as if this open and terrible from B to infant E. Is in better node. And we know that it&#x27;s in better note because this is over interval and we know that this is an interval, then we know that it&#x27;s in metal then its complement which is from minus infinity to be is in Britain. Now we have this board is in better note, we have proved that and this birth is in better note then we can use the property that the intersection or in sigma field which means it&#x27;s or it&#x27;s in better. No, the intersection between and the infinity and from minus infinity to be is in better note which means the intersection is from A. To be what A is smaller than be is in better which means that the closed in terrible&#x27;s also is in better No. And to prove that all half open intervals and to prove that all half open in terrible&#x27;s are included in beta nude. We can make use of this conclusion A and B is in better node and of course the open intervals are in beta node which is B. And infinity. And get the intersection between them intersection of A and B and B. And infinite. This just gives us hey to be inclusive and be exclusive. Then it&#x27;s also in beta. No. Then we have proved that all half open intervals are included in beta. Note because the intersection with subsets including better not gives us this, which is included in Britain.

Bryan Lynn · Answer

So we want to prove that the supreme um this set for R. And Q. Such the R. Is less than A. That&#x27;s exactly uh you know for all A. And R. So let&#x27;s just to make things easier, let&#x27;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&#x27;s just given by how we&#x27;re setting up this set. You know what that means? Then the Supreme of this set has to be listening. Hey? So let&#x27;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.

William Semus · Answer

So this question, we have a function F of X, which is x squared minus five X minus 24. And we want to find the intervals where this function is concave up or down as well as determine any x coordinates of inflection points of our function. So let&#x27;s see here in order to analyze can cavity? We have to look at a second derivative. So let&#x27;s get our first derivative. My F prime of X is what this time it&#x27;s two X minus five. And then how about my second derivative? My second derivative is just to now notice that second derivative can never be equal to zero and my second derivative is always positive. That means the original function is always concave up when F double prime is positive. My original function is concave up. And so this is always the case. This time F. This con cave up on the interval from negative infinity to positive infinity. It&#x27;s never concave down. And since F never changes can cavity, there are no inflection points. Notice this graph here is just a parabola. It&#x27;s a quadratic opening up. It&#x27;s always concave up. Never concave down no inflection points. Hopefully, that makes sense. Have a great rest of your day.

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