Fax Uncountable Formula with airSlate SignNow

Fax uncountable formula

we're following up with our proof from the last video where we proved that the rational numbers forms a countable set and in this case we're going to prove that the real numbers forms an uncountable set so in other words there is no one-to-one and onto function from the natural numbers into the real numbers okay we're going to do that by way of this thing called the nested interval property which we proved a couple of videos ago and that goes in the following way so let's suppose for all natural numbers k we have a closed interval which we'll call i k and that goes from a k to b k where a k is less than or equal to b k and so in other words this is going to be a non-empty interval and we have this nesting of the eyes so we have i1 contains i2 contains i3 and so on and so forth and we can write this all at once by saying that i n plus 1 is contained within the interval i n and that's true for all natural numbers n and the conclusion of this nested interval property is that the intersection over all of these closed intervals is non-empty okay great so now let's go ahead and look at the proof of the fact that the real numbers is uncountable and we're going to do this by way of contradiction so let's say by way of contradiction suppose that r is countable but this means that there exists a bijection f which goes from n to r great and then we can go ahead and set x i equal to f of i and what that tells us is we have this nice way of listing the elements in r so any in other words we can say r is equal to x1 x2 x3 x4 and so on and so forth where this is the image of one under this bijection that we have and the next one is the image of two and so on and so forth okay good so now what we want to do is construct some closed intervals and we'll do that in an inductive way so for our first step we want to take i1 to be any closed interval not containing x1 so um you could write down maybe a formula for this closed interval if you wanted to so notice it could be the closed interval from x one plus one to x one plus two but we'll just say that it's any closed interval not containing x1 so maybe let's make a picture let's so let's say we have our real number line here and then let's say that x1 is right there then we might as well take i1 to start here remember we call that a1 and end here so there's b1 so in other words this closed interval right there is i1 okay now we want to define i2 so we'll do that in the following way so let's set i2 to be any closed sub interval of i1 not containing uh x2 okay so let's see how that would work so if x2 is outside of this interval i1 then we can just let i1 equal i2 so that's good but let's say x2 was right here so in other words x2 is inside the interval i1 then we would want to take i2 to be something like this so we'd say this one is a2 and this one is b2 and so this interval right here is i2 okay and that's something that's possible to do regardless of where x2 lands so if x2 lands outside of the interval i1 we're okay but if x2 lands inside the interval i1 we're still okay and now we want to continue with this iteratively so we'll just say continue iteratively so in other words we're going to define i n plus 1 as a closed sub interval of i n not containing x in plus one great and so that's our iterative process for constructing all of these intervals okay i'll clean up the board i'll put a summary at the top and then we'll finish the proof so let's see where we are now so by way of contradiction we supposed that the real numbers was countable and we listed all of the elements in the real numbers as follows so we've got x1 x2 and so on and so forth and then from this list we constructed a sequence of nested closed intervals so we let i1 be any closed interval not containing x1 and then inductively we let i n plus 1 be any closed interval of i n not containing x n plus 1. and so now notice that the hypotheses of the nested interval property are satisfied so we've got this nested sequence of closed intervals which is exactly what we need so now what we know is that the intersection of these closed intervals is non-empty so let's go ahead and write that down so now by the nested interval property we know that the intersection over all of these closed intervals is non-empty but then since this intersection is not empty that means there is an element in this intersection but these are all sets of real numbers so what that tells us is that there exists x which is a real number such that x is in this uh intersection of these closed intervals great but then by our assumption that the real numbers is countable we know that this x is one of the numbers on this list so uh just to spell that out we know that x equals x m for sum m in the natural numbers great and so now let's look at these two facts next to each other so let's look at this one versus this one so let's maybe number them one and two so notice that 1 implies that x is in in for all n in the natural numbers so that's what 1 gives us but then 2 gives us something that is contradictory to this and that is by our inductive construction of these i's so 2 implies that x m is not an i sub m plus one but now notice that these are contradictory statements here we have x is an i n for all natural numbers in in other words x is in i m plus one but then on the other hand x is not an i m plus one so that leaves us with a contradiction and what do we contradict we contradict our very first assumption which was that the real numbers was a countable set so we are only left with the possibility that r is not a countable set in other words it's uncountable that's a good place to stop