Show that if I1, I2, …, In is a collection of open intervals...

Key Concepts

Common Intersection of Intervals

The idea of a common intersection refers to finding at least one point that is contained in every interval in the collection. When dealing with intervals on the real line, certain ordering properties allow one to construct a point that lies within each interval, particularly by considering the extremal endpoints of the intervals.

Extremal Endpoints Argument

This technique involves identifying the largest of the left endpoints and the smallest of the right endpoints from the collection of intervals. Using the order properties of real numbers, one can show that if every pair of intervals intersects, the maximum left endpoint is less than the minimum right endpoint, thereby ensuring that the overall intersection is nonempty.

Open Intervals

An open interval is a set of real numbers between two endpoints where the endpoints themselves are not included. These sets have the property that every point inside the interval is an interior point, meaning there is always a small neighborhood around each point contained within the interval. This concept is fundamental in real analysis and topology as it introduces the idea of open sets.

Pairwise Intersection Property

The property that each pair of intervals in a collection has a nonempty intersection means that for any two intervals you choose, there is at least one point that is common to both. This characteristic is crucial in problems dealing with overlapping sets and is the foundation for proving that a larger collective intersection exists.

Transcript

00:01 Hi, this video is going to be a long one.

00:04 So please bear with me.

00:07 In this question we are present with a problem of when we have a real number interval, open intervals, i1 up to in, such that any pair of them, like intersection is not empty, so they have something in common.

00:30 Show that the intersection of every everything is also not empty.

00:37 This is not trivial, it's not intuitive, why that's the case.

00:44 But let's go over this step by step.

00:49 First, the most important thing is to observe when two intervals have their intersection not empty.

01:01 What would that mean? it means not either sorry it means that these two things must happen so the what should i call the left border and the right border of of these two interval must interact in this way so i will call them abcd a must be less than d otherwise if the like go below they're going to have nothing in common right so so this must happen but we can we can also have the other extreme that see go beyond b if that's the case then we have nothing in common as well so we have this tight bow on both both borders of of these intervals that must be uphill at all time okay that's just some thing to know and we will use this later but first let's establish the induction problem so let the statement pfn be that intersection of i1 up to i n is not empty with the property given in the question i will not write it down again but any pair okay i will write down again any pair intersection is not empty okay so basic step we have n equal to 1 obviously it's it is true because it's an interval is not empty now for inductive step suppose that the statement is true for some n now we want to show p of n plus 1 is also true how do we do that i will suppose otherwise and show that it leads to a contradiction.

03:29 So we start off with p of n, right? we have this equation that the intersection of all intervals up to i .n will call it ab.

03:44 It's not empty.

03:46 Now, note that if we, if i label borders of i -i, which is not very good index, sorry, i -i as a -i and b -i.

04:02 It means that this ab, the result of intersection must be the mean and max of their borders.

04:12 You can pause video and think about this, why that's the case.

04:18 But that's what it is.

04:21 Now, once we have that, suppose we add this new newest interval in right because we want the statement pn plus one so we look at this i n plus one and suppose for contradiction that it makes this intersection empty so it make intersection of everything empty what would that mean well first we will have a relationship between this border of ab and and a and plus 1 b and plus 1, right? the relationship would be this.

05:09 So don't be confused just yet.

05:13 We have a, b, right, as the result of the intersection, and we have this newly added interval, a, n plus 1 and b and plus 1.

05:27 Since we know that this two intersections give m33, this set, it means that one of these two things must happen.

05:40 So either, you know, the border is over in some extreme, either b is less than a plus one and the picture will be like this, or the other way around.

05:55 A is greater than b in plus one.

05:59 Okay, this is just a reverse of the property i talked about earlier.

06:06 This joined by m.

06:08 So, so now this joined by all.

06:13 All right.

06:14 What does this has to do? like what does this lead to? well, we also have another properties in this question, which is the relationship between between interval, any interval, i -i -j...

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