 So, in the last two lectures of this course we will prove Reson's Metrization Theorem. So, this theorem tells us that if you put certain conditions on a topological space x then in the topology it actually comes from it. So, first we begin with some pre luminaries. So, lemma let x be a metric space and define a function d prime. So, we want to change the metric on x as follows by d prime of x comma y is equal to minimum of d of x comma y comma 1. So, here d is the metric is the metric on x. So, then d prime is a metric on x and induces the same topology as d. So, let us prove this. So, the check that d prime is a metric. So, is a metric is left as an exercise. So, let us assume that d prime is a metric and the open balls d prime x comma r are the same d x comma r. The open balls are the same when r is strictly resummon. And since open balls of this type form a basis for the topology on x this implies that both generate the same topology. So, I mean d prime is a metric. So, it generates a topology and d is a metric. So, it generates a topology. So, a basis for the topology generated by d is given by open balls of this type and a basis for the topology generated by d prime is given by open balls of this type when r is strictly resummon. So, therefore, both these are the same and therefore, the topology is the same. So, this proves. So, let us take the real line with the standard metric and we apply this construction to this and apply the above. So, then we get that d prime x comma y defined to be minimum of the absolute value of x minus y comma 1 defines a metric on r which induces the standard topology. Now, we will use this metric d prime. So, we will use this to define a metric on r n or index by natural numbers. So, this is the countable product. So, let me write it as x n where each x n is equal to r with the standard topology and this product has the product topology as always. So, we define the following metric. So, define d of x comma y. So, for we take two vectors x and y in this r n and we define this d as supremum among all natural numbers d prime x n comma y n divided by n. So, this supremum obviously exists because each of these d primes is less than equal to 1. So, lemma first we need to show that this actually defines a metric on r n. The function defines a metric clearly this holds this is the second condition for defining metric. This obviously holds this clear from the definition and suppose if x is equal to y then obviously, the d of x comma y is 0 and conversely if this is 0 then this implies that the supremum d prime x n comma y n by n is equal to 0. So, this implies that each of these is less than equal to 0, but that implies that d prime x n comma y n is equal to 0 which implies that x n is equal to this implies. So, we only left to prove the triangle inequality. So, this condition also holds. So, let us prove the triangle inequality. So, let we take three points. So, then we have d prime since d prime is a metric we have the usual triangle inequality and which is divided by n. So, now using supremum of a i plus b i is less than equal to supremum of a i plus supremum of b i. So, using this relation we get. So, when we take supremum over here we get d of x comma y is less than equal to d of x comma z plus d of. So, this implies that d defines it and the main theorem we will prove in this lecture is the following. This is the first step towards proving Ulysses-Meterization Theorem. So, the metric d. So, we have defined a metric on R n. So, this induces a topology on the other hand R n also has the product topology and the content of this theorem is that both these topologies are the same induces the product topology. So, let us prove this theorem. So, let us first show every open set in the product topology is open in the metric topology. So, in order to do this first consider sets of the following sets. So, let we take any point in R space R n define this set to be those y in R n such that sign for j line. So, it is easily checked sets of this type form a basis for the product topology. So, we will show that these sets are also open in the metric topology. So, let y be a point in this set. So, choose some delta positive and which is less than epsilon by k. So, that this ball around y j this is a real number of delta radius delta is contained the ball around x j of radius epsilon. So, this should happen for j line between y and k. So, in other words we have x j here. So, for each coordinate j this is this has length epsilon R y j is a point over here. For each coordinate for each j lying between 1 and k that is because y is over here. So, choose a delta which lies between 0 and epsilon k if we use delta very small then the delta neighborhood for all these y j's will completely lie inside this epsilon neighborhood of x j. So, we claim that the ball in the metric topology of radius delta around y is completely contained inside this set. So, let us check this claim. So, suppose z is a point in this part. So, this implies that supremum. So, this implies that d of. So, let us first write d of y comma z is strictly less than delta, but d recall this was the supremum d prime z n divided by n. So, in particular this implies that d prime by n z n by n is strictly less than delta for all n and in particular also for j for j lying between 1 and k. So, this implies that d prime by j z j is strictly less than j times delta which is less than equal to k times delta which is strictly less than epsilon for all j lying between 1 and k. So, this is the definition of this open set u. So, let us look at this definition. So, this happens for z j. So, this implies that this ball d y comma delta is contained in k epsilon. So, we have this open set we have this basic open set u x bar or rather we have this set yeah u x bar k epsilon we chose a point y bar over here and there is an open neighborhood in the metric topology which is which contains this y and it is completely contained inside this set. So, we can cover this u x bar by open sets of this type and that will show that this u x bar k epsilon is also open in the metric topology right. So, this implies that u x bar k epsilon is open in the metric topology. Now, let us prove the converse yeah. So, now we show that any basic open set v d x epsilon is open in the product topology. So, what we will do is so, we will show that there is a set. So, let me replace this by y we will show that there is a set u x k delta such that k delta is contained in. So, let us try to see why it is sufficient to prove this claim. So, let us start. So, our aim is to show that this open set y epsilon this set is open in the product topology right. So, we can do that if given any point x bar in the set we can find a set u which is open in the product topology and completely contained inside this part. If we can do that then we can cover this we can do that for every x in the set and we can take the union of those. So, that it will show that this b d y epsilon that is open in the product topology. Now, in order to show in order to show this that given any point x there is a open set in the product topology which is completely contained inside this ball. Note that for this x there is an epsilon prime ball right this is b d x comma epsilon prime right which is completely contained inside this larger set. And therefore, we will show that there is a set of the type this type u x bar k delta which is contained in this smaller epsilon this epsilon prime ball right. So, that will prove the assertion. So, let us begin with the proof. So, choose k such that choose k large such that 1 upon k is strictly less than epsilon right. So, then we claim that u x k epsilon by 2 is completely contained inside b d x bar comma epsilon. So, let us prove this. So, if z is a point in this set epsilon by 2 and what this implies is that d prime x j comma z j is strictly less than epsilon by 2 for all j line between 1 and k right. So, this implies that d prime x j comma z j by j is strictly less than 1 upon j epsilon by 2 which is less than equal to epsilon by 2 for all j line between 1 and k. Now, on the other hand if j is strictly greater than k then d prime x j comma z j by j is less than equal to 1 upon j which is because d prime is bounded by 1 this less than 1 upon k which is strictly less than epsilon right. So, therefore, when we take supremum over j d prime x j comma z j by j this is equal to the distance of x bar from z bar this is less than equal to epsilon by 2 which is strictly less than epsilon right. So, this implies. So, we started with this z and we showed that the distance of z from x in the metric topology is less than epsilon. So, which implies that u x comma k comma epsilon by 2 is contained in E. So, this proves that. So, thus we have proved this claim over here. So, we have proved this claim. So, this proves that both the topologies this proves that the metric topology and the product topology. So, this completes the proof here. So, we will end this lecture here.