 Let us continue the study of product of cell complexes. Indeed last time we already introduced the definition what is the meaning of locally countable. This includes locally finite also if you countable you replace by finite, you will get locally finite condition also. Alright similar to that. So what is the condition? Every point is contained in a countable sub complex that sub complex must be a neighborhood of the point in the whole space. Okay that is the local, local means this should happen in a neighborhood so that neighborhood has to come here. Okay so if you have such a locally countable CW complex this lemma tells you sorry so that was the that was not definition. Definition was that it is locally countable if every open cell meets only countable many closed cells. Now we are putting the lemma here. Okay so this is what happens. So justifying the definition locally countable. Alright of course if this happens obviously a point will meet only finitely, countable many closed cells that will be also true. Okay so you could have taken this as definition but the definition is this one namely every closed cell meets countable many closed cells or every open cell meets countable many closed cells. Okay every open cell meets countable many open cells is wrong because no open cell meets any other open cell they are rejoined. Okay yeah so let us let us prove this lemma. First of all each X belongs to the interior of a unique cell. So let us say interior of EM and I will state K0 be any finite sub complex containing EM. What you have to do? Are you sure that there is a finite sub complex containing EM? All that you have to do know is that the closure of EM is covered by finitely many closed cells. Okay to take the closure of all that that will be a sub complex which will be finite you need know for this finite lemma okay. You may have to start with a M cell you may have to go down down down down you know to wherever you may start 0 cells whatever EM intersects all those cells you may have to take. So there is a finite sub complex because closure of EM is compact okay. So let us denote that by K0 there is no uniqueness you can choose there is one there is a K0 finite sub complex containing EM therefore it contains the point X also to begin with. Problem is you cannot stop here because this K0 may not be an open cell open may not be a neighborhood of X inside capital X. Now what happens for each closed cell EK which intersects interior of EM you see how many of them are there countably many okay. There is a finite sub complex that contains a closure of EK take any cell K the closure of that is contained inside the finite sub complex. Now I have to do this for countably many of them okay take the union of all these finite sub complexes to get a countable sub complex and that is my K1. So I repeat how I started I started with a point X which is in the interior of EM some simplex some cell okay M dimension cell then I took a sub complex which contains this EM okay. Now the higher dimensional cells their attaching maps may come inside EM in the interior of EM okay. So you have to take for each such EK take a finite sub complex contains that take the union of all these things wherein EK is over all those cells whose boundary intersects EM call that as K1 okay. Now carry on with this procedure what is K1? K1 is countable again okay for each of the top dimension cells in K1 you do another countable sub complex and union take the union of all that and so on you get another larger countable sub complex K2 and so on okay one contained in the other. What is important is is K0 will be closed sub complex of K1 K1 will be closed sub complex of K2 and each of them is countable and K0 contains our point X take K to be the union of KNs countable union of countable family so that is also countable it may be very huge but it is countable okay the claim is that this K contains an open subset in X which is containing the point X so it is a neighbor okay K is a countable sub complex it is very clear. So to see that K is a neighborhood of X what we will do start with the m cells what started with m cells sigma such that X is in the interior of sigma right that interior you call this as U0 and opens subset inside the closed simple X okay it is open in there and actually if it is the m dimensional skeleton there it will be open the U0 is a neighborhood of X in XMM dimension skeleton and is contained in K0 K0 is by definition the finite sub complex of some finite sub complex which contains the cell that is why it is K0 okay so you have a sub complex K0 and this opens subset inside that right now we play the old game let U1 be the extension of U0 to the next skeleton X M plus 1 this we know how to extend right from one skeleton to next skeleton right as in the provision 2.1 so this is going to be again and again used so I have given a lot of emphasis on this one then this U1 will be contained inside K1 why because U1 will take only those simple X those those cells part of those cells which intersect the cell U0 okay and those things are already there inside K1 so U1 is contained inside K1 and U1 is an open subset of X M plus 1 we do not know whether it is open subset of X yet repeat this argument what you will get each time you will get a neighborhood Ui which is a neighborhood of X inside X M plus i the U0 U1 and X M plus i which is an extension of Ui minus 1 okay and each Ui is contained inside Ki I am just using this U0 inside Kmm means X M plus 0 U1 is an extension of U0 U1 is contained is K1 so replacing them by i so that is the inductive step I can do okay for each i greater than or equal to 0 I can do this therefore by theorem or layer of theorem U which is a union of Ui's obviously intersects with each X M plus i it is open there so it must be open X belongs to that and U will be contained inside K because Ui is contained inside Ki and K is the union of all the Ki's okay so many of these ideas must have been clear to you but you must now practice yourself writing down such a proof this proof is just balanced correctly we can now give a number of instances when the product topology coincides with weak topology that was our aim okay start with any two CW complexes the product topology coincides with the CW topology on X cross Y in the following instances this is not a if and only if theorem mind you X and Y are finite then it is okay this was the right from the beginning we knew X is finite or Y is finite then also it is okay X or Y is locally compact or equivalently locally finite X and Y have countably many cells both of them may be infinite so this is something new the last one is X and Y are both locally countable so this is the best thing that we could say so far you will see that to make this is not true we will have a counter example also but there may other directions wherein you can you can try to do that thing but if X and Y are both of them are locally countable even if one of them is not locally countable okay a particular case may be there but we will give a counter example counter example that there are one for some case for everything that is not true it is not any friend on this theorem yeah one two and three they are already seen how when we have seen X is X or Y is finite also you can see because if Y is finite it is compact this will be always compactly generated compactly generated and compact product we know it is compactly generated similarly X or Y is locally compact one of them is locally compact another is compactly generated it is also you have seen earlier okay all the three of them we have seen earlier the first thing which we have not yet seen is X and Y have countably many cells and X and Y are locally countable these two things could not have been seen earlier this is special case for CW complex okay so 4 and 5 needs to be proved okay to prove 4 so let W be an open subset of this CW complex product take a point X naught Y naught inside W you should produce a neighborhood U of X naught and a neighborhood V of Y naught such that U cross V is contained inside W if you have done this for every point that would mean that X cross that this W is now open in X cross Y okay that is the product policy what we have started if the view is open inside the CW topology okay so how do we how do you use the the hypothesis in 4 so says X and Y are countably many cells this is what you have to use so write X as union of KNs and Y as union of LNs what are KN and LN they are increasing union but they are finite sub complexes now I cannot take skeletons and so on now a countable complex can always written as union of finite complexes sub complexes how you enumerate enumerate all the cells okay take the first cell it may not be a sub complex but it is contained in a finite sub complex to take that test K naught now put second one the union K naught and the second one must be contained in another finite sub complex okay like this Ki you have constructed take the i plus one cell which you have enumerated put that one also together that may not be sub complex but that will be contained in a larger sub complex which is finite and so on okay so you can always write a countable complex as a increasing union of finite sub complex so write it write them like that okay and assume that X naught and Y naught are the the first one namely K naught and L naught the CW topology on X and similarly C-duplicology in Y you can just go to these KNs okay actually if you take KN that is right then it is similar to our earlier argument if you take the the induced topology co-induced topology with respect to this family KN it will be finer than the compact uh generated thing but then each KN is again contained either a finitely many covered by finitely many compact things therefore it is equal okay the first thing to note that CW topology on X similarly on Y is equal to the co-induced topology by these families therefore to say something is open or closed inside X I can intersect it with KN and see that whether it is open or closed inside K since KN and LN are compact if you just take KN cross LN and the topology and the CW topology on that it is the same thing as KN cross LN by the first case therefore if you take W intersection KN cross LN okay this is a neighborhood of X naught Y naught in KN cross LN why because if I put a W here it is a neighborhood but they they are the same here so it is a neighborhood of KN cross LN now this KN cross LN in the standard product topology since W is a neighborhood of X inside K inside X cross Y with the with the usual topology with the with the CW topology W intersect KN cross LN must be a neighborhood inside KN cross LN W but I do not have to write W that is all I am telling you okay inductively we shall construct sets UNVN inside KN and LN such that UN intersection KN minus 1 is UN minus 1, VN intersection LN minus 1 is VN minus 1 and our starting point X naught Y naught is in each of the UNVN contained in W intersection KN cross LN okay then we put U equal to UN, V equal to VN and as before we can appear to theorem 2 point here to conclude that U and V are open subsets of X and Y respectively so that inductive step has to be done again okay for this we have to use one more small topology so first of all we shall construct UNVN with an additional property which is stronger property and that will help us to carry out the inductive property carry out the inductive step what is the additional property we will do that let us say UN bar and VN bar are compact so we are in the construction we are going to make such a each UN and VN okay such that their closures are compact so these UNVN are compact neighborhoods okay and UN bar itself does not go out of W intersection KNW so that we keep extending okay so this is this is a clever way of trying to do the extension we should try to do extension you will see that you better assume this one so that means in the earlier you have to have constructed it in that way otherwise you cannot assume that right okay so when N equal to 0 what is it U0 V0 where are they they must be inside K0 and L0 right X0 Y0 belongs to U0 V0 U0 V0 bar contained is a U0 V0 U0 bar cross V0 bar in W of intersection K0 such that U0 bar V0 bar is compact why you can do this where are the X0 and Y0 they will be in some cells right they will be interior of some cell right that cell is an open set its closure is compact right so this is happening inside K0 and L0 are both compact subsets I mean finite CW complexes so that is why this is possible in other words all that I am telling you here is that a compact order of space is locally compact also therefore I can do that X0 belongs to U0 U0 bar is compact contained inside W intersection K0 remember this W intersection K0 L0 we have already noticed we have we have we have just first noticed that it is open neighborhood of X0 Y0 and KN cross LN for each N so I am using that here the first thing is that they are compact subsets so I can choose this one to be compact subsets construction for N equal to 0 is over now assume this is done for nth stage okay satisfying these two conditions okay then using compactness of UN bar and VN bar you can now find open sets say UN prime and VN prime in KN plus 1 LN plus 1 so set UN prime is contained in a UN plus 1 prime VN prime contain VN plus 1 prime the closures of product is inside W intersection KN plus 1 so this is a version of Walman's theorem for compact subsets I have stated it in exercise okay I will just tell you what is this exercise it is there below suppose you have a neighborhood of K cross L where K and L are compact subsets inside X cross Y okay then you can get a U and a V inside X and Y so said K is contained inside U L is contained inside V U cross V itself is contained in the given neighborhood W I repeat suppose you start with the product space X Y arbitrary product space is okay you can assume they are host drop spaces whatever K and L are subsets compact subsets K cross L is contained in an open set then you can choose open subsets U V containing K and L respectively such that U cross V itself is contained inside the given open set okay so this is what we are going to use this is what we have used used here UN prime UN plus 1 prime and VN plus 1 prime will be in K and L plus N cross L okay now use 2.1 repeatedly if needed you can extend U N V N to open subsets U N plus 1 V N plus 1 in K N plus 1 and L N plus 1 respectively such that U N plus 1 bar is inside U N plus 1 prime okay the bar of the extended thing is inside this one V N plus 1 bar is in V N plus 1 prime the point here is in the in choosing this one U N plus 1 bar prime intersection with K N may not be equal to U N but now we can make that this was this was the gist of this proposition right we can extend that extension means intersection with this one must be the given open set okay so here U U N prime and V N prime you know there are some arbitrary open subsets larger than the original ones but when you intersect with the old one okay say U N prime intersect with K N may not be equal to V N prime may be larger that will cross problem so this extension when you say it is precisely equal to okay and that is needed into the final contribution once you have that your problem is over so what we have done I will just recall what we have done is that if you have countable both is on 4th one both extend where countable many cells then the product topology coincides with the suitable topology the fifth one is easier if you use the fourth one now okay so given a point X naught Y naught in X cross Y by the previous lemma the lemma which we started with today there are countable sub complexes K comma L respectively in X and Y right and K and L are neighborhoods of X naught and Y naught in X and Y respectively to fix X naught belonging to A contained inside K why not belonging to B contained inside Y such that A and B are open sets so that is possible because they are neighborhoods okay now essentially what we have we are trying to do is to reduce this problem you know problem 5 to problem 4 for the both of them are countable case okay now let W be yet we don't have that picture yet because the whole thing has to be inside X cross Y right not that K and L that is fine but you have to prove that this is finally the neighborhood is in X cross Y right so you have to that W be open in X cross Y Y the C double topology and take X naught belong to right okay then W intersection K cross L W this will be open in K cross L W but this is equal to L by 4 that is the point therefore there are open sets Q in K and V in L such that U cross V inside W but then A intersection U is open inside U okay therefore open inside X see A and B are open inside you have got A A here is at K and B is at Y they are open subsets okay because K and L is K and L it should be this should be L K and L are countable sub complexes which are neighborhoods okay therefore A intersection U is open in U therefore open in X similarly B intersection V is open in B so it is open in Y therefore X naught Y naught which is belongs to A cross U A intersection U cross B intersection V that would be inside W okay so 4 actually helps to solve this problem here is an example due to Docker which tells you that if both X and Y are not locally countable then product topology need not coincide with the CW topology okay so for this we start with the following notation let this N denote the set of natural numbers and L denote the set of all functions free from N to N consider the real vector space R power N which is the direct sum of copies of R as many copies as as natural numbers and R power L the direct sum of copies of R as many as capital this curly L memory indexed by functions from N to N both of them you give the compactly generated topology okay so that we are now going to construct a subspace of this R power N that would be called X and N is subspace of R power L that will be some sort of Y automatically they will be having compactly generated topology and they will be just one dimensional CW complexes when we take the product of X with Y the product topology will not coincide with the CW topology so that is the idea okay so let us choose these standard basis elements for R capital N and R curly N namely V N and U phi indexed by N be the natural number and phi is functions on the natural numbers to natural numbers now put X equal to all the line segments emanating from 0 and ending at the end point V N so we can write them as R times V N where R is real number between 0 and 1 so look at all of them they will all be incident at the origin of R power N okay and the end point will be the vector V N okay look at this complex this is a one-dimensional complex okay consisting of just edges all of them having single point common and indexed by the these vectors V N themselves so of course R and W indicate the weak topology the topology generated by finite subspaces which is same thing as compactly generated topology similarly Y will be R U phi phi since R capital L and 0 less than R less than R to N so it is exact similar only the number of edges here to be very huge then both X and Y are CW complexes which are one-dimensional okay and you can think of them as one point union of one point union of number of edges okay the claim is as as told you the CW topology on X cross Y is strictly finer than the product topology so in order to do that what we are going to do is we will display a subset a subset which is discrete inside the the CW topology but it will have 0 as a limit point the 0 in the origin as a limit point in the product topology so that space is precisely this P consisting of points P N comma phi indexed by N as well as phi what are they the first coordinate is V N divided by phi N second coordinate is U phi divided by phi N phi N and you know for all N phi N is some natural number so I can divide out by that so this will be one of the points on the edge cross edge okay so this will be one of the point in the two-dimensional cell one cell here cross the other cell here okay as you take N and phi different okay N prime and phi prime that will be in a different two cell so all these points are interior in inside exactly one cell each point belongs to the interior of exactly one two-dimensional cell 0 comma V N cross 0 comma U phi okay therefore by our old lemma the set P is a discrete closed subset of the the the topology namely the CW topology however all that we have to show is now the origin 0 in R N cross R L is in the closure of P in the product topology what is the meaning of that take any open subset around 0 show that it intersects P okay so an open subset around 0 will consist in the product topology it consists of a smaller open subset of the form U cross V where U is a neighborhood of 0 in R N and V is a neighborhood of 0 in R L okay once you have such a thing automatically for each N and phi each line segments emanating from 0 we will have some R N such that and R N and S phi numbers depending upon N and phi such that the entire line segment lambda times V N where lambda varies from 0 to R N is contained inside you similarly here lambda ring from 0 to S phi lambda times U phi that will be contained inside this will be true for every N and every phi okay therefore now I would like to give you a sequence of points which converge to to the origin so consider psi from N to N given by psi N equal to maximum of N or look at 1 by R N take its integral part integral part of 1 by R N so these two are some natural numbers now okay add one more take the maximum and add one more that is my psi N then psi N will tend to infinity as N tends to infinity because this is bigger than N hence we can choose the little m such that this psi M is bigger than any number 1 by S psi okay 1 by R N is there the integral part of that plus 1 is bigger than that one so choose psi M bigger than 1 by psi automatically this one it follows that 1 by psi M into V psi M 1 by psi U psi U psi this will be inside P as well as inside U cross V because this number whatever you have taken should be less than R N and that second part should be less than S psi so that is what if you if you take psi M like this 1 by psi will be less than S psi here okay so that completes the proof of the counter example thank you