 So, welcome to this lecture. So, in the previous lecture we had defined connectedness and today we will continue with that topic and today we are going to define path components of x. I am sorry not the path components the connected components. So, let x be a topological space we put an equivalence relation let us put any relation on the points of x as follows we say that x is equivalent to y if there is a connected subset t contained in x such that x comma y are in t. So, let us check that this is an equivalence relation that this defines an equivalence relation. So, there are three things which we need to check first we need to check that x is equivalent to x for all x and x right. This is clear because by taking t equal to just the singleton set x any singleton set x is connected because for a set to be disconnected it has to have at least two points right because we cannot write the singleton set as a disjoint union of non-empty open subsets. In fact, as a disjoint union of non-empty subsets. So, second point we need to check is if x is equivalent to y then y is equivalent to x right. This is also clear as if t is a connected subset which contains x comma y then it obviously contains y and x again. So, y is equivalent to x 3 if x is equivalent to y and y is equivalent to z then x is equivalent to z right. So, to show this recall that that in the previous lecture we had proved that if t 1 and t 2 are connected subsets of x such that t 1 intersection t 2 is non-empty right then the union t 1 union 2 is not is connected. Now, since x is equivalent to y there exists connected subset t 1 such that x comma y are in t 1 right. Similarly, since y is equivalent to z there is connected subset t 2 such that y comma z belong to t 2 right. So, then y is in the intersection and so it is non-empty and thus t 1 union t 2 is connected and contains x and z. So, thus x is equivalent to z right. So, this shows that this sim is an equivalence relation. Now, every time we have an equivalence relation on a set x we can decompose it into equivalence classes. So, this equivalence relation breaks x into a disjoint union of equivalence of equivalence classes right. So, we can write x as a disjoint union of x i is in some index side. So, and each x i is the equivalence class of some x right and x is equivalent to y if and only if x comma y belong to the same x i right ok. So, we have this following proposition which has let us say 3 parts. So, every connected subspace of x is contained in x i plus some i given any connected subspace of x it is going to be contained in one of the x i's. So, the second part is each x i is connected. So, every connected subspace is contained in some x i and each of these x i's is connected. So, therefore, these two will imply that x i's are maximally connected a maximal connected subsets and the third point is each x i is closed. So, let us prove this proposition. So, let. So, assume t intersection x i is not empty and t intersection x j is not empty. So, we will prove the first point by contradiction. So, let us assume that a connected subspace meets two x i's. So, it meets x i and x j where i is not equal to j right. So, let t comma s belong to t right. So, then by the definition of the equivalence relation. So, then t is equivalent to s right as t is connected and it contains both t and s I am sorry. So, I should have written let t be in t intersection x i and s be in t intersection x j. Then as s comma t are in t this implies s is equivalent to t right, but then this implies that the equivalence class of t is the same which is x i is equal to the equivalence class of s which is x j and this is a contradiction because x i is not equal to x j. So, therefore, given any connected subspace it can be contained only in one x i ok. Next let us prove. So, this proves one next let us prove two. So, let x i be an equivalence class let x be in x i ok. So, we fix we fix this point x ok. Then for any y in x i yeah since x and y both of them are in x i as x is equivalent to y there is a connected subspace t sub y contain in x such that x i sorry no sorry x right. I am just using the definition of the equivalence relation such that x and y belong to t sub y. So, clearly from this part 1 yeah from point 1 it follows that t y is contained in x i. So, thus we can write x i as a union over all these y in x i t sub y right. So, this inclusion is obvious this inclusion is obvious because for each y in x i the subset t sub y that is contained in x i and therefore, when we take a union it is going to be contained in x i and conversely for the other inclusion t sub y for every y in x i t sub y contains y right. So, therefore, this is actually equal right. So, now we have this variant of the earlier lemma we have proved. So, consider the following the following let t i be a family subsets of x and suppose the intersection of all these is non empty right. So, then the union t i is connected. So, the proof of this is very similar to the proof of this lemma that we proved this one. So, let us just try and prove it. So, suppose there exists non empty u and v such that in x both in x such as this union t i is written as the disjoint union intersected with u both these are and they are of course, disjoint open subsets this union. So, now, so let we fix this element a the idea is the same as in the earlier one earlier lemma we fix this element a right. So, now we can intersect both sides with t i. So, this will imply that t i is equal to t i intersected u ok. So, we can fix a and before we and assume a is in this union intersection right let us assume this right. So, now, we intersect this equation with t i on both sides. So, we get t i is equal to t i intersected u disjoint union t i intersected v right. Now, this contains a and so is non empty right and on the other hand t i is connected t i is connected this implies t i which forces that t i intersection v has to be empty right. So, this implies that t i is complain completely contain inside u ok for every i and this happens for every i right. So, this implies that this union t i intersected v is empty which is a contradiction. So, this intermediate lemma tells us that this union is connected when all the t i's are connected and their intersection is non empty right. So, now, we will apply this to our situation. So, applying this to our situation we get that x i we have written as a union of y in x i t y each t y is connected and the intersection of all these t y's it at least contains x right. So, this implies that the intersection is non empty because all the t y's they contain x right. So, this implies that x i is connected. So, this proves to and finally, let us prove 3 as x is connected as x i is connected this implies. So, we had seen this corollary that a is connected in x implies a closure is connected we have proved this earlier. So, this implies that x i closure is connected right, but each every by part 1 part 1 each x i closure is contained in a unique right, but then. So, thus this x a has to be x i because x i is containing x i closure right. Thus x i closure is contained in x i which implies that x i is equal to x i closure. So, this implies that x i is closed. So, this completes the proposition this completes. So, the proposition says that given any topological space x we can break it up into these equivalence classes and each equivalence class is defined by the relation of by that relation that we started lecture with right. And each of these equivalence classes they are connected and they are closed and given any other connected subspace given any connected subspace is contained in one of these equivalence classes. So, let us just definition. So, the equivalence classes. So, as an example let us work this out. So, let q contained in R have the subspace topology. So, what are the what are the connected components? So, we claim that ok. So, before that note that given a in q the subset singleton a is connected right. So, the question is what is the maximum connected subsets which contains this singleton a right and that will be the connected component containing a. Now, we claim that connected component is just this set a right y z. So, basically what we are saying is we have taken written q we can write is at we can write it as a disjoint union of connected components x i. So, now choose each x i is non-empty obviously and just pick any a in this x i and the claim is x i is equal to a right. So, let us see this y z. So, if not suppose b also belongs to x i and b is not equal to a right. So, then let us assume that. So, suppose that b is strictly written a right. So, then we have this a is over here b is over here and we can choose a rational irrational c right choose an irrational c yeah choose an irrational c right. So, then we can write x i. So, c is not in q and x i is in q therefore, c is not in x i. So, then x i does not contain c right and so we can write x i as a disjoint union of open subsets minus infinity comma c disjoint union c comma infinity right, but this contradicts both these are non-empty because this contains a and this contains b right and both these are open and it is a disjoint union right this contradicts the connectedness. So, thus x i is supposed to be a single. So, thus x i is supposed to be a single point right. So, therefore all the connected components are just a in q. So, these are is the decomposition and let us make a remark this obvious x is connected if and only if it has only one ok. So, this remark is obvious right because if x is connected then given any two points x and y the subset t can be taken as equal to x right. So, therefore there will be just one equivalence class in the decomposition and conversely if there is just one equivalence class then that has to be x and therefore, x is going to be connected because each equivalence class is connected. So, we will end this lecture here and in the next lecture we will continue we will introduce the topic of path connectedness.