Lecture 4: examples of topological spaces, coarser and finer topologies, bases and closed sets
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1 Lecture 4: examples of topological spaces, coarser and finer topologies, bases and closed sets Saul Glasman 14 September 2016 Let s give the definition of an open subset of R. Definition 1. Let U R. We say U is open if for every r U, r is contained in an open interval which is contained in U; that is, for every r in U, there are positive numbers ɛ 0, ɛ 1 > 0 such that (r ɛ 0, r + ɛ 1 ) U. Proposition 2. With this definition, the open subsets of R form a topology on R. We ll deduce this result as a special case of a more general theorem. In order to state this theorem, it ll be helpful to introduce the idea of a basis for a topology. As we ll see, when defining a topology, it s often annoying to explicitly describe all the open sets of the topology; instead, it s convenient to describe building blocks for the topology. This is what bases are for. Definition 3. Let X be a set. A basis for a topology on X is a collection of subsets of X B P(X) 1 such that 1. U B U = X; that is, every x X belongs to at least one element of the basis. 2. If U 1, U 2 B and x U 1 U 2, then there s some V B with Let s give a couple of examples here. x V, V U 1 U 2. 1 Recall that P(X) is notation for the power set of X, the set of all subsets of X. 1
2 Example 4. The set of open intervals in R (including the empty interval) is a basis for a topology on R. Why is this a basis, according to the definition above? Well, let s note that if B satisfies the first condition and is closed under finite intersections - that is, a finite intersection of elements of B is in B - then it automatically satisfies the second condition. Lemma 5. If B is a collection of subsets of X such that U = X U B and such that B is closed under finite intersections, then B is a basis for a topology on X. Proof. We need to show that whenever U 1, U 2 B and x U 1 U 2, we can find V with x V and V U 1 U 2. But U 1 U 2 B, so we can just take V = U 1 U 2. The set B R of open intervals in R satisfies this condition: the intersection of open intervals is always an open interval (possibly empty). So it s a basis. Example 6. An open disc in R 2 is the collection of points less than distance r away from a fixed point (x 0, y 0 ): D = {(x, y) R 2 (x x 0 ) 2 + (y y 0 ) 2 < r 2.} We claim that set of open discs forms a basis for a topology on R 2. I won t give a rigorous proof of this, but I ll give an illustrative diagram. Note that unlike open intervals in R, the intersection of two open discs is not an open disc. Example 7. An open rectangle in R 2 is what you think it is: a set of the form R = {{(x, y) R 2 x 0 < x < x 1, y 0 < y < y 1 } for some x 0, x 1, y 0, y 1 in R. This is also a basis for a topology on R 2, since the intersection of open rectangles is an open rectangle. Example 8. For any set X, the set is a basis for a topology on X. B d isc = {{x} x X} This is clear: the union of the points of X is X, and the intersection of one-point sets is either a one-point set or empty. We ll soon see that both of the bases for topologies on R 2 I ve mentioned generate the same topology - the so-called product topology, which we ll get into in general soon - and B disc generates the discrete topology on X. But I haven t yet told you how to build a topology from a basis, so let s do that now. 2
3 Lemma 9. Let X be a set and let B be a basis for a topology on X. Let U X be a subset. Then the following two conditions on U are equivalent: 1. U is a union of elements of B. 2. For each x U, there is an element V B such that x V and V U. Proof. First we show that 1 implies 2. Suppose U = i I V i where V i B for all i. Then for each x U, x V i for some i. In checking 2, we can take V = V i. Now we show that 2 implies 1. For each x U, let V x be an element of B with x V x, V x U. Then U = V x. x U If U satisfies either of these equivalent conditions, we call it B-open. Now by the definition I gave earlier, an open subset of R is one which is B R -open, where B R is the set of open intervals. Warning 10. If anyone s taken linear algebra, this is a good time to remark that the usage of the word basis here is quite different from the linear algebra usage. In linear algebra, any vector can be written uniquely as a linear combination of basis vectors, but in topology, it s usually possible to write an open set as a union of basis sets in many different ways. Theorem 11. If B is a basis for a topology, the collection T B of B-open sets is a topology. We call it the topology generated by B. Proof. We have to check the axioms for a topology. 1. is B-open, since it s the union of no elements of B. 2. X is the union of all elements of B, so it s B-open. 3. Let (U i ) i I be a collection of B-open sets. We need to prove that the union U = U i i I is B-open. For each x U, x U i for some i; since U i is B-open, there is V B such that x V, V U i. But V U too, so U is B-open. 4. For each finite collection U 1, U 2,, U n 3
4 of B-open sets, we have to show that is B-open. U = n i=1 This is a good place for a little digression about a useful way of checking the finite intersection condition. If you want to check that some class T is closed under finite intersections, it s enough to show that the intersection of two elements of T is in T. Indeed, if U 1, U 2, U 3 are elements of T, then U n U 1 U 2 U 3 = ((U 1 U 2 ) U 3 ) But since U 1 U 2 T and U 3 T, so is U 1 U 2 U 3. I could do the same thing if I had n-sets: just iteratively intersect them one at a time. U 1 U 2 U 3 = ((( (U 1 U 2 ) U 3 )...) U n ). Incidentally, in case you don t know what a proof by induction is, this is one. So let s check that if U 1 and U 2 are B-open sets, U 1 U 2 is B-open. Condition 2 of the definition of a basis was designed just for this. If x U 1 U 2, then there are sets V 1, V 2 B with Therefore, there s a set V B with This shows that U 1 U 2 is B-open. x V 1 U 1, x V 2 U 2. x V V 1 V 2 U 1 U 2. In particular, this shows that the topology on R, as defined at the beginning of this lecture, is actually a topology. Not bad. For the rest of this lecture, I d like to change tack a bit and talk about some smaller topics concerning topologies. First, let s establish some notation for relating different topologies on the same set. Definition 12. Let T, T be topologies on a set X. We say that T is at least as fine as T if T T ; that is, whenever a set is open in the topology T, it s open in the topology T. Obviously, if T is at least as fine as T and T and T are not equal, we ll say that T is finer than T. We ll also say that T is coarser than T. 4
5 I like the metaphor that Munkres uses here, with truckloads of gravel (p. 77), although it makes a topological space seem more disconnected than it really is. I prefer to say that a finer topology is higher resolution than a coarser topology: if we think of open sets as being areas of similar color on a monitor or something, then a finer topology really does give us higher resolution pictures. Having a finer topology isn t always desirable: the very finest topology on any set X is the discrete topology on X, which is the topology in which all sets are open (think of a monitor which is always displaying random static). The coarsest possible topology on X is the indiscrete topology on X, which has as few open sets as possible: only and X are open (think of a monitor which can only display a solid field of black or white). Interesting topologies are balanced between these two extremes. To wrap up today, let s talk about one more example of a topology. Definition 13. Let X be a set. The finite complement topology T fc on X is the following class of subsets: U T fc if and only if U = or X \ U is finite. In order to prove that T fc is a topology, it ll be convenient to introduce closed sets in a topological space. The definition is simple: Definition 14. Let X be a topological space. A set Z X is called closed if its complement Z \ X is open. We ll talk a lot more about closed sets later, but for now you should think of closed sets as sets which are in sharp focus : they have no fuzzy edges. Let s have some examples of closed sets. Example 15. Since X \ = X and X \ X =, and X are always closed. Example 16. In R, the closed interval [a, b] = {r R a r b} is a closed set. To show this, we need to show that the complement R \ [a, b] = (, a) (b, ) = {r R r < a or r > b} is open. But I can write this as a union of open intervals: (, a) = n N(a n, a) and similarly for (b, ). In particular, we can take a = b. Then the closed interval [a, a] is the single point {a}, and so {a} is a closed subset of R. This is very typical behavior: for many nice topological spaces, it s the case that single points are closed subsets. 5
6 Let s now recall an extra bit of set theory. Lemma 17. Taking complements turns intersections into unions and unions into intersections: Let X be a set and let (U i ) i I be a collection of subsets of X. Then X \ ( U i ) = \ U i ) i I i I(X and X \ ( i I U i ) = i I(X \ U i ). These two statements are known as demorgan s laws. If they re not immediately obvious to you, I recommend you try to prove them at home as an exercise. The upshot is that we can define a topology just as well with closed sets as with open sets: Lemma 18. Let X be a set and let T be a collection of subsets of X. Let T c = {X \ U U T } be the set of complements of elements of T. Then T is a topology if and only if T c satisfies the following conditions: 1. T c. 2. X T c. 3. An arbitrary intersection of elements of T c is in T c. 4. A finite union of elements of T c is in T c. Of course, if you know what the closed sets are, you also know what the open sets are: X \ (X \ U) = U, so (T c ) c = T. The point I want to make right now is that in some cases, such as the finite complement topology, it s easier to check the topology axioms with closed sets than with open sets. Indeed, a subset Z X is closed in the finite complement topology if and only if it s finite or equal to X. But 1. the empty set is finite; 2. X is equal to X; 3. an arbitrary intersection of finite sets is finite; 4. a finite union of finite sets is finite. Although the indiscrete topology is pretty useless, the finite complement topology actually shows up in the wild (for example, in algebraic geometry). It s the coarsest topology which satisfies the reasonable condition that one-point sets are closed. 6
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