Generell Topologi. Richard Williamson. May 6, 2013

Transcription

1 Generell Topologi Richard Williamson May 6,

2 Thursday 7th January. Basis o a topological space generating a topology with a speciied basis standard topology on R examples Deinition.. Let (, O) be a topological space. A basis or (, O) is a set O o subsets o belonging to O such that every subset U o belonging to O may be obtained as a union o subsets o belonging to O. Proposition.. Let be a set, and let O be a set o subsets o such that the ollowing conditions are satisied. () can be obtained as a union o (possibly ininitely many) subsets o belonging to O. () Let U and U be subsets o belonging to O. Then U U belongs to O. Let O denote the set o subsets U o which may be obtained as a union o (possibly ininitely many) subsets o O. Then (, O) is a topological space, with basis O. Proo. We veriy conditions ()-() o Deinition.. () We think o an an empty union o subsets o belonging to O, so that O. I you are not comortable with this, just change the deinition o O to include as well. () We have that O by deinition o O together with the act that O atsiies condition () in the statement o the proposition. () Let {U j } j J be a set o subsets o belonging to O. For every j J, by deinition o O we have that = k K j U k or a set K j, where U k O. Then ( ) j J U j = j J = k K j U k r ( j J K j ) U r. Thus j J U j belongs to O. is a union o subsets o belonging to O, and hence j J U j () Let U and U be subsets o which belong to O. By deinition o O, we have that U = j J U j where U j O or all j J, and that U = j J U j, where U j O or all j J. Then ( ) ( ) U U = U j = j J j J U j (j,j ) J J U j U j. 5

3 Since O satisies condition () o the proposition, we have that U j U j belongs to O or every (j, j ) J J. Thus U U belongs to O. By construction o O, we have that O is a basis or (, O). Terminology.. Let be a set, and let O be a set o subsets o satisying conditions () and () o Proposition.. Let O denote the set o unions o subsets o belonging to O, which by Proposition. deines a topology on. We reer to O as the topology on which is generated by O. Observation.. Let O = {(a, b) a, b R}. Then O satisies condition () o Proposition. with respect to R, since or example R = n N ( n, n). By Observation., we have that O satisies condition () o Proposition.. Deinition.5. The standard topology on R is the topology O R generated by O. Observation.6. All open intervals in R belong to O R. We have the ollowing cases. () I a, b R, then by deinition o O and O we have that (a, b) O R. () I a R, we have that (a, in) = n N (a, a + n). Since (a, a + n) belongs to O or every n N, we deduce that (a, in) O R. () I b R, we have that ( in, b) = n N (b n, b). Since (b n, b) belongs to O or every n N, we deduce that ( in, b) O R. () We noted in Observation. that R O R. Remark.7. As mentioned in Idea., we will prove later in the course that O R consists exactly o disjoint unions o (possibly ininitely many) open intervals. Observation.8. Let (, O) be a topological space, and let O be a basis or (, O). Let O be a set o subsets o. I every U such that U O can be obtained as a union o subsets o O, then O deines a basis or (, O). Examples.9. () For ɛ R such that ɛ >, and or any x R, let B ɛ (x) = {y R x ɛ < y < x + ɛ}. x ɛ x x + ɛ In other words, B ɛ (x) is the open interval (x ɛ, x + ɛ). Then O = {B ɛ (x) ɛ R and ɛ >, and x R} is a basis or (R, O R ). 6

4 Proo. By Observation.8, it suices to prove that or every a, b R we can obtain the open interval (a, b) as a union o subsets o R belonging to O. In act, (a, b) itsel already belongs to O. Indeed (a, b) = B b a ( a + b ). b a b a a a+b b In particular, we see that a topological space may admit more than one basis. () Let = {a, b, c, d, e}, and let O denote the topology on given by {, {b}, {a, b}, {b, c}, {d, e}, {a, b, c}, {b, d, e}, {a, b, d, e}, {b, c, d, e}, }. Then is a basis or (, O). O = { {b}, {a, b}, {b, c}, {d, e} } The same holds or any set O o subsets o such that O O. No other set o subsets o is a basis or (, O). For example, O = { {a, b}, {b, c}, {d, e} } } is not a basis or (, O), since {b} cannot be obtained as a union o subsets o belonging to O. Similarly O = { {b}, {a, b}, {d, e} } is not a basis or O, since {b, c} cannot be obtained as a union o subsets o belonging to O. () Let O = {(, b) b R}. Then O is not a basis or (R, O R ), since or example we cannot obtain the open interval (, ) as a union o open intervals o the orm (, b). But O satisies the conditions o Proposition., and thus generates a topology O on R. In the manner o Observation., one can prove that O = O {, R}. 7

5 . Continuous maps examples continuity o inclusion maps, compositions o continuous maps, and constant maps Notation.. Let and be sets, and let be a map. Let U be a subset o. We deine (U) to be {x (x) U}. Deinition.. Let (, O ) and (, O ) be topological spaces. A map is continuous i or every U O we have that (U) belongs to O. Remark.. A map R R is continuous with respect to the standard topology on both copies o R i and only i it is continuous in the ɛ δ sense that you know rom real analysis/calculus. See the Exercise Sheet. Examples.. () Let = {a, b}, and let O denote the topology on given by {, {b}, }, so that (, O) is the Sierpiński interval. Let = {a, b, c }, and let O denote the topology on given by {, {a }, {c }, {a, c }, {b, c }, }. Let be given by a b and b c. Then is continous. Proo. We veriy that (U) O or every U O, as ollows. () ( ) = () ( {a } ) = () ( {c } ) = {b } () ( {a, c } ) = {b} (5) ( ) =. 8

6 Let g be given by a c and b b. Then g is not continuous, since or example g ( {c } ) = {a}, which does not belong to O. () Let D I D be given by (x, y, t) ( ( t)x, ( t)y ). We will prove on the Exercise Sheet that is continuous. We may think o as a shrinking o D onto its centre, as t moves rom to. We can picture the image o D {t} under as ollows as t moves rom to. () Fix k R. Let I S be given by t φ(kt), where φ is the continuous map o Question 8 o Exercise Sheet. Let us picture or a ew values o k. 9

7 () Let k =. In words, begins at the point (, ), and travels exactly once around S. Don t be misled by the picture the path really travels around the circle, not slightly outside it. We may picture ( [, t] ) as t moves rom to as ollows. Recall rom Examples.8 (6) that a typical open subset U o S is as depicted below. ( ) Then (U) is as depicted below. In particular, (U) is open in I. Thus intuitively we can believe that is continuous! ( )

8 () Let k =. In words, begins at the point (, ), and travels exactly twice around S. Again, don t be misled by the picture the path really travels twice around the circle, thus passing through every point on the circle twice, not in a spiral outside the circle as drawn. We may picture ( [, t] ) as t moves rom to as ollows. Let U S be the open subset depicted below. ( ) Then (U) a disjoint union o open intervals as depicted below, so is open in I. ( ) ( )

9 ( Let k =. In words, begins at the point (, ), and travels exactly one and a hal times around S. We may picture ( [, t] ) as t moves rom to as ollows. Let U S be the open subset depicted below. ( ) Then (U) is a disjoint union o open subsets o I as depicted below, so is open in I. ( ) ( ) () Let I I

10 be given by t t. We will prove on the Exercise Sheet that is continuous. We may depict as ollows. Let U I be the open subset depicted below. ( ) Then (U) is as depicted below. In particular, (U) is open in I. ( ) (5) Let I S be the map given by t { φ ( t) i t, φ(t) i < t. As in (), φ is the map o Question 8 o Exercise Sheet. We may depict as ollows.

11 Then is not continuous. Indeed, consider an open subset U o S as depicted below. ( ) Then (U) is a hal open interval as depicted below. ( ] In particular, (U) is not an open subset o I. (5) Consider a map I D as depicted below. A precise deinition o this map is not important here the path should be interpreted as beginning on the top let o the disc, moving to the bottom let, jumping to the top right, and then moving to the bottom right. Let U D be an open subset o D depicted as a dashed rectangle below. Then (U) is a hal open interval in I as depicted below. In particular, (U) is not open in I.

12 ( ] Terminology.. Let be a set, and let A be a subset o. The inclusion map with respect to A and is the map A given by a a. We will oten denote it by A. Proposition.5. Let (, O ) be a topological space. Let A be equipped with the subspace topology O A with respect to (, O ). Then the inclusion map A i is continuous. Proo. Let U be a subset o belonging to O. Then i (U) = A U. By deinition o O A, we have that A U belongs to O A. Hence i (U) belongs to O A. Proposition.6. Let (, O ), (, O ), and (Z, O Z ) be topological spaces. Let and g Z be continuous maps. Then the map g Z is continuous. Proo. Let U be a subset o Z belonging to O Z. Then (g ) (U) = {x g ( (x) ) U} = {x (x) g (U)} = ( g (U) ). Since g is continuous, we have that g (U) O. Hence, since is continuous, we have that ( g (U) ) O. Thus (g ) (U) O. 5

13 Terminology.7. Let and be sets. A map is constant i (x) = (x ) or all x, x. Proposition.8. Let (, O ) and (, O ) be topological spaces. Let be a constant map. Then is continuous. Proo. Since is constant, (x) = y or some y and all x. Let U O. I y U, then (U) =, which belongs to O. I y U, then (U) =, which also belongs to O. 6