 We have seen that if script beta a collection of subsets of x that is script beta is a subset of power set of x, where capital X is any set satisfying the condition that give any x in x that imply there exist a member b from script beta such that x is in b that is we take union of all the members of script beta that is the whole space second b 1, b 2 belongs to script beta and suppose b 1 intersection b 2 is non-empty then for each x in b 1 intersection b 2 that is we have b 1 in our collection script beta, b 2 is in script beta and we x is in the intersection then we can find a b 3 from then for each x in the intersection there exist b 3 from script beta such that x belongs to b 3 which is contained in the intersection. Then we say that such a collection script beta is a basis for a topology on x then using this we define the topology as a subset of x is a member of tau suffix script beta if and only if I mean in case if it is non-empty x is in a implies there exist a member b from script beta such that x belongs to b contained in a that is for a belongs tau beta that is a is open for each x in a will get a basic open set so having this property. So note that script beta itself is a sub collection of tau suffix script beta because x in a we can take b itself equal to a then this condition is satisfied as a trivial example consider x any non-empty set script beta the collection of all singletons x in capital X the question is whether this sub collection this is sub collection of power set of x whether this collection satisfy these two condition namely first condition is yes x take any x in x then that imply there exist a member b which is same as singleton x such that x belongs to b is first condition is satisfied what about the second you take b1 b2 belongs to script beta second b1 intersection b2 should be non-empty only in this case we will have to check whether we can find here b3. So take b1 b2 belongs to script beta and x belongs to b1 intersection b2 then what does it mean because b1 belongs to b1 intersection b2 that means b1 is non-empty b2 is non-empty but all the members of script beta are singletons each member here is singletons hence in this case this would imply b1 hence b1 is singleton x b2 equal to singleton x and in this case b3 and this implies hence this imply there exist b3 which is again the same singleton x such that yes x belongs to b3 which is contained in b1 intersection b2. So hence this collection namely the collection of all singleton set will it is a basis for a topology on x. So what is the topology generated by this collection is nothing but easy to verify this power set of x that is the discrete topology that is you take any a subset of x and x belongs to a then there exist b which is singleton x such that x belongs to b which is contained in a this implies a belongs to our tau suffix script beta hence every set is open. So we get the discrete topology this collection will generate the discrete topology. So these are all trivial examples so one and we already seen that or it is easy to check this collection that is called topology on x induced by script beta is a induced by script beta is in fact a topology on x. This is called topology on x induced by script beta and it is easy to prove that tau is prove that this collection tau prove that tau suffix script beta is a topology on x that is easy to verify. Then now let us see how to as we told in the end how to define how to define interior point of a subset a of x where x is given with a topology where x tau is a topological space. Similarly we define what is a limit point closer of a set boundary point and other related concepts in fact this is one of the basic concept in topology what is called an interior point. So we have a given x tau a topological space so that is given a topological space capital A a subset of x then a point x in a is said to be on is said to be on interior point if as if and only if meaning is under this condition if and only if there exist this meaning there exist at least one open set meaning there exist u belongs to tau psi that x belongs to u in such a case we say that u is a neighbor root of x it is in fact what we say open neighbor root so just we call it neighbor root. So in other words there is a neighbor root u of x such that x belongs to u and that u is contained in a it is not just x is in a is an given subset x is an interior point in there should exist some u which should be an open set containing x that is completely contained in a then we say that x is on interior point then we define interior of a set what do you mean by interior of a in short we just write interior a or a standard notation a circle. So we read as a interior this is defined as the collection of all interior points of a. So a interior equal to set of all x in a that is a mean necessary condition such that just x is an interior point of interior point of a that is for each x is in a interior means we know that there exist u containing open set containing x contained in a we collect the collection of all interior point that is a subset of a by definition seconds by definition a interior is a subset of a well it is prove that in fact a interior is an open set. So you start with any set then collect all the interior points of that set then that set which is a interior is in fact an open set and will prove that it is not only it is an open set it is the largest open set contained inside contained in a. So yeah quickly let us prove this statement so we call it as a theorem we are given a topological space let x, tau be a topological space and capital A be a subset of x then interior will assume that if this interior is either empty that case it is open then even if it is non-empty will prove that it is first then one a interior is an open set first it is an open set second suppose capital B is an open set such that this way such that this open set is contained in a then then our interior is largest open set contained in a then what we have to prove is then this open set is contained in a interior yeah it is in fact it is almost a straight forward result first let us prove that and by definition will assume that if a interior is empty I mean there is no elements there is no interior point then a interior we take it empty which is an open set so let us assume that a interior is non-empty. So take see by the definition it is clear take x is in a interior then there exist an open set so that is u belongs to tau that u depends upon x so we call it then there exist u say suffix x belongs to tau this is just to say that is an open set containing x and that depends upon x there exist ux belongs to tau say that our definition of interior point says there is an open set containing x and completely contained in a correct so in fact we see that now ux is non-empty not only x is in ux then for y belongs to ux for any y in ux y this ux is already contained in a and y is in ux hence there exist an open set ux containing y and contained in a this implies this proves that y is also an interior point of a that means what you take every element of this open set is also on interior point that is we have proved that this open set containing x is contained in a interior that is now take any x then we will get an open set ux then take the union of all those open sets then that gives hence we have we have the following namely a interior itself equal to union of such ux a interior because we have proved that each ux for each x in a interior there exist ux which is a subset of a interior and this imply union of such ux x is in a interior is subset of a interior and conversely if we take any x in a interior that x will belongs to that ux for each x we have at least one open set ux that is contained in ux hence the other way a interior is subset of this union so hence we will have this we will not write everything so that is now a interior is nothing but union of ux where each ux is an open set that is a interior can be written as arbitrary union of numbers of tau our here our index this this a interior is a I mean for each x our index set varies from I mean this is a collection ux x in a interior this is a sub collection of tau that implies this union ux x belongs to a interior is also belongs to tau but that is same as a interior hence a interior is in fact an open set now the second part let us see that suppose capital B is an open set such that it is contained in a interior suppose B is an open set here so already we have we have second part we have capital B an open set and it is a subset of a so what we want to claim prove that capital B is subset of a interior that mean taken so to prove capital B is subset of a interior so take x if B is empty then it is subset of a interior so take if it is non-empty take x is in B but what is B itself a open set so now B is an open set means instead of our u x belongs to B B open B is an open set B is an open set so that B is contained in a that is we got an open set B so that x is in B and contained in a that imply by our definition x belongs to a interior in this case for each x there is an open set that open set is capital B itself so hence we have proved that that is x belongs to B imply x belongs to a interior hence we have this imply capital B is a subset of a interior that is a interior is an open set and it is the largest open set contained in a in fact we can other thing is we can easy to prove that a is if a is an open set open set then a interior equal to a for an open set I mean a interior is a itself in fact a interior already a interior subset of a take x is in a again the u there x is an open set x is in a contained in a so that will give us each point of a is an interior point now we will introduce the other concept then we will come back and see example see what is called a we will define the next concept called limit point or another we also call it or accumulation accumulation point of a subset capital E of a topological space x tau I mean sometimes we just say x it is understood that there is a topology on x here what is the how we define a point or say element x it need not be in a x is in our topological space x a point x in x is said to be a limit point is said to be a limit point of a if and only if it satisfies the following condition namely you give any open set containing x see here like it may be like x need not be in a even it may be x may be in a or x may be the complement say x is here take any open set containing x then that will contain some points of a different from x then we say that that point x is a limit point of a that is if and only if for each here for every open set for each open set you containing x for each open set containing x you intersection from a remove the set x a difference x this set should be non-empty then we say that x is a limit point of the set a like the collection of all interior point there we want at least one open set is completely inside a here for every open set because it is like boundary point mean yeah this we do not say it is boundary point the point may be on the boundary of see like if you say a close interval 0 1 if you take any point either 0 take any neighbor root open set containing mean again we will get a basic open set containing that point open set containing 0 mean it is an interval it will contain points from right side similarly 1 we will get some 1-epsilon 1-epsilon that is a basic open set containing this point which is which is as non-empty intersection with the given set. So, if you collect all the limit points that set we call it derived set of a so for a subset of x we define the set a dash this is our notation for the collection of all limit points of a for a in x we define the a dash or a prime as this a prime equal to all those points in x such that x is a limit point of a then we say that it is a limit point of a this is known as derived set of a then we define we define the closer the closer of whatever inditially we feel closer mean all the points of a and there are points which may not be a but they are limit points under any condition that mean you give any neighbor root of that x that neighbor root will contain some point of a different from x. So, we define the closer of a denoted by closer a or a bar as we write a closer equal to a union all the limit points now let us see that first in fact a closer is a close set like a interior is an open set closer is a close set what is a close set the compliment if we take the compliment that should belongs to tau. So, it is a compliment is an open set and will prove that it is a close set and it is the smallest closed set containing a mean will just indicate the proof okay we will call it again theorem for any a for a subset of x is understood there is a topology on x for a in x first a closer is a closed subset of x thus will prove this so what is here to prove is that is to prove to prove the compliment that we call it from x difference a closer we call it a closer compliment to prove this is an open set such is open set mean like each point should be take a point in that set you will have to prove that that is an interior point so take let x belongs to a closer compliment that is then x is not in a closer then x is not in a closer so x is not in a closer implies this condition mean for x belongs to a closer for every open set in fact we can see that either it is in a then or it is a limit point in either case you take any open set containing x u intersection a is non empty note that is x belongs to an element or say y belongs to a closer if and only if for each open set u containing y u intersection a is non empty what is a closer a closer equal to a union the set of limit point if y is in a already u is also open set containing y so y is in a as well as u so is non empty if y is suppose not in a but still in a closer that mean it is a limit point of a then by the definition limit point mean not only u intersection a is non empty u intersection the smaller set a difference y is non empty in particular this condition is satisfied in fact it is if and only if so now here x is not in a closer so now we will remove this so now x is not in a closer implies there exist on open set say u containing x that is it is a neighbor root of x u containing x such that u intersection a is that condition is not satisfied so u intersection a is empty that is u intersection a is empty this proves the hence u is completely I mean the compliment is contained in a compliment so where we started we started with x is in a closer compliment so but this is not what we have to prove is it is an interior point we have to get an open set containing x yes we have got there exist an open set containing x such that here we have got u intersection a is empty hence u is subset of a compliment but that is not sufficient we have to get an open set in fact the same open set you can see that this is contained in a closer compliment see if we take any x so what we that is to prove that this open set is completely contained in a closer compliment correct so take any y belongs to u u intersection a is empty then what will happen if u intersection see suppose we are getting u completely contained in a closer compliment otherwise it will be u intersection suppose this is intersection a closer is non empty correct so then this will imply this implies we will get a element y in u and y is in a closer but y is in u y is in a closer that means u is a open set containing y that imply u intersection a is non empty but we have u intersection a is empty so hence this cannot happen that is u intersection a closer non empty cannot happen and this proves that hence what we have is u intersection a closer is empty correct so that imply u is contained in a closer compliment so hence the started point is started with on x in a closer compliment and we have proved that there is an open set u containing x that is completely contained in a closer compliment so that imply a closer compliment is an open set so this proves that a closer compliment is an open set and hence a closer is a close set so now we leave it as a second part exercise prove that a closer is already a prove that it is a close set is the smallest closed set containing a just containing a that is what is the meaning is capital B a close set and a is subset of B is a close set containing a that should imply our this is the smallest such close set this is to prove a closer is subset of prove this is to prove so then we will get that a closer is the smallest close set containing a so now let us see some simple examples to have a feeling of what we mean by interior and point and close sets even you take any okay I will take x equal to simple the our set of real numbers with usual topology usual topology on R that is so this we call it usual topology or standard topology that is here all the what are the basic open sets are open interval these are all basic open sets in fact to define all this concept interior even the limit point it is enough to consider only basic open set we can easily prove that x is an interior point if and only if there x is a basic open set if we know that we have a basis like R the collection of all open interval they are all basic open set there x is a basic open set B so that x belongs to B which is contained in A similarly x is a limit point if and only if for each basic open set u containing x u intersection a difference x is non-empty so that yeah you can easily verify those concept see now take symbol okay take a even take anything say q the collection of all all rational numbers so whether this a is non-empty what can you say about a interior by definition it has an interior point if it is non-empty mean x belongs to a interior implies there should exist an open set because there exist an open set u such that x belongs to u containing x and contained in a but x belongs u is an open set this implies there exist by our definition that is how we introduce how now x belongs to u u open in R implies there exist R greater than 0 such that x belongs to x minus R that is what this is a neighbor root open set containing x contained in u and that u is contained in a u is subset of a that mean we got for a given x in our in this case x is in a mean it is a rational number so but we got an R that is radius all the points between x minus R and x plus R this is those y in R such that modulus of y minus x is less than R so this will give all the real numbers between x minus R and x plus R set of all y such that x minus R less than y less than x plus R so but yeah now we can but this set but x minus R x plus R it cannot be contained in a because this is contained only rational here there are in fact infinite number of irrational every interval contains irrational hence this set is empty so at the same time you can see that what is limit the collection of all limit points so if we take any x in R and take any open set u containing x then u intersection q in fact it will contain infinite number of points so any open set mean in particular it a interval this will contain infinite number of rational using the properties of R so we can prove that in this case every point is a limit point hence a closer also a union a prime in that is the whole space R in other words q closer equal to R later we will see that such a set we call it dense we get a subset whose closer is the whole space okay so we will continue in the next lecture.