 So, let us begin this lecture. So, in the previous lecture, we had studied the notion of a closed subset. We had introduced a notion of closed subset and closure of subset and we also saw in the exercises that the closure of a subset A is the smallest closed subset which contains A. So, let us continue our discussion with closed subsets. So, today we will begin with the definition of a dense subset. So, as always x is a topological space. So, a subset t contained in x is said to be dense, dense in x if for every non-empty u contained in x, we have t intersection u is nonempty. And the proposition, the main proposition is that A be a subset of x, a closure node is closure. So, then so, let a closure have the subspace topology, A contain a closure is dense. So, let us prove this. So, we need to show. So, by the definition of denseness. So, we need to show that if v contained in a closure is a non-empty open subset, then v intersection A is nonempty. So, by the definition of subspace topology. So, by the definition of subspace topology, this said v is equal to u intersection A closure for some open subset u contained in x, for some u open in x. And as v is non-empty, this implies there exists some x in v which is equal to u intersection A closure. So, therefore, u is an open subset of x, u contains x and x is in A closure. So, thus by the definition of closure, this implies that u intersection A is nonempty. Recall the definition of closure x is in A closure if every open subset containing x meets A. So, this implies that u intersection A which we can write as u intersection A closure, intersection A because recall that A is contained in A closure that is trivial and this is equal to b. So, thus v intersection A is non-empty and which shows that A is dense. So, let u be open in x where u has been given the subspace topology. So, then you want to say that then v is open. So, in simple words open subset of an open subset is open in the larger to plus equals plus. So, let us prove this the proof is easy. So, by the definition of subspace topology there is an open subset v tilde in x such that this v is equal to u intersected v tilde. But now u is open in x v tilde is open in x and finite intersection of as both u and v tilde are open in x this implies u intersection v tilde this is equal to v is open in x. So, this completes the proof here. So, and similarly we have a similar result for closed subsets. So, before we prove that result for closed subsets we need this small lemma. Let A contain in x be a subset the closed subsets of A where A is given the subspace topology of A in the subspace topology are precisely of the form z intersection A where z is closed in x. So, let us prove this lemma. So, the open subsets of A are precisely of the form u intersection A where u is open in x and the closed subsets are the complements of open subsets here. So, thus closed subsets of A are precisely of the form A and we remove an open subset. So, now a simple a simple say theoretic check shows that but x minus u is a closed subset of x right. So, thus are precisely z intersection A z is a closed subset. So, this is our z and as a corollary of this lemma let us see the analog of the proposition which says that closed subset of a closed subset is closed in x. So, let z be a closed subset of x and let z 1 be closed. So, then z 1 is closed in x right and the proof of this corollary is very similar. So, by the previous lemma this subset z 1 we can write as some z tilde intersection z where z tilde is a closed subset and as a intersection of closed subsets is closed. So, as intersection of closed subsets is closed this implies z 1 is closed. So, next we want to say that checking whether a subset is closed or not can be checked by restricting our attention to closed subsets. So, precisely what I mean is the following. So, let z 1 z 2 be closed subspaces such that this x we can write as z 1 union z 2. So, a subset z containing x is closed if and only if z intersection z i is closed in z i for i equal to 1 comma. So, this proposition will have a useful corollary which we will see next, but let us prove this proposition first. So, if z is closed then simply because intersection of closed subsets is closed. So, this clearly implies z intersection z i is closed or by the lemma. So, let me just write this implies z intersection z i is closed in z i because this earlier lemma because of this lemma right and. So, let us prove the converse conversely z intersection z i is closed in z i for i equal to 1 comma 2 right. So, then by this corollary because of this corollary z intersection z 1 and z intersection z 2 both are closed in x. So, then by the above corollary let me just keep the corollary here z intersection z 1 and z intersection z 2 are both closed in x and since the finite union of closed subspaces is closed. So, this implies that z which is equal to z union z 1 I am sorry z intersection z 1 union z intersection z 2 right as x is equal to z 1 union z 2 and finite unions of closed subspaces are closed this implies z i is closed right. So, as a corollary of this proposition let us prove the following useful result we have the following useful result continuous maps. Let a and b be subsets of x such that be closed subsets such that x is the union of a and b. So, let a and b have the subspace topology now let f from x to y be a map. So, right now is just a map of sets such that f restricted to a and f restricted to b are continuous. So, then f is continuous. So, what do we mean by f restricted to a and f restricted to b? So, look at this inclusion. So, let us call this i sub a that is f right. So, f restricted to a is simply this f compose this inclusion. So, if f is continuous then since both the inclusions are continuous obviously, the restriction is continuous. So, here this theorem is telling us that if the restriction of f to both a and b is continuous then f is continuous on x right. So, to check that a map is continuous it suffices if we can write our topological space x as a union of 2 closed subspaces 2 or finitely many and such that f restricted to each of these is. So, we will only prove this for the union of 2 closed subspaces but it will be immediately clear that the proof works for finitely. So, to show f is continuous enough to show that f inverse of a closed subspace is closed in x. So, by the previous proposition it is enough to show that f inverse of y prime intersected a and f inverse of y prime intersected b are closed in a and b respectively. But note that f inverse of y prime this is a simple set theoretic check intersected with a is simply f restricted to a inverse image of y prime and similarly f inverse of y prime intersected b is equal to f restricted to b inverse of y prime. So, thus as f restricted to a and f restricted to b are continuous. So, this implies that f inverse f restricted to a inverse is closed in a and similarly f restricted to b inverse of y prime is closed in b. So, this implies that f inverse by prime is closed in x this implies f is continuous. So, as an application of this proposition. So, use this proposition use this theorem use this theorem to show that the maps. So, we have two maps let us say f 1 from r 2 to r f 1 of x comma y is equal to the maximum of x comma y and f 2 of f 2 of x comma y is equal to minimum of x comma y are continuous. So, divide r 2 as a union of two close subspaces and f 1 and f 2 both restricted to each of these subspaces should be continuous that should be trivial and therefore, both these maps are continuous. So, we will end this lecture here.