Shadow Posted July 10, 2011 Share Posted July 10, 2011 I've started reading Sid Morris's Topology Without Tears and I'm stuck. Here's the exercise: Let R be the set of all real numbers. Precisely three of the following ten collections of subsets of R are topologies. Identify these and justify your answer. T1 = R, {} and every interval (a, b), with a, b any real numbers. T2 = R, {} and every interval (-r, r) with r any positive real number. T3 = R, {} and every interval (-r, r) with r any positive rational number. T4 = R, {} and every interval [-r, r] with r any positive rational number. T5 = R, {} and every interval (-r, r) with r any positive irrational number. T6 = R, {} and every interval [-r, r] with r any positive irrational number. T7 = R, {} and every interval [-r, r) with r any positive real number. T8 = R, {} and every interval (-r, r] with r any positive real number. T9 = R, {}, every interval (-r, r) and every interval [-r, r] with r any positive real number. T10 = R, {}, every interval [-n, n] and every interval (-r, r) with r any positive real number. Now, T1 is easy; not a topology. T2 through T10 though, that's a different story. Take T2 for example. I make the (possibly incorrect) assumption that if the union of any two sets from T2 is in T2 then every finite or infinite union of any number of sets from T2 is also in T2. The same with intersections. I consider three different scenarios with two intervals (-r1, r1) and (-r2, r2); a) r1 < r2, b) r1 = r2, c) r2 > r1: a) The intersection of (-r1, r1) and (-r2, r2) with r1 < r2 is always the interval (-r1, r1), which is in T2. The union of (-r1, r1) and (-r2, r2) with r1 < r2 will always be the interval (-r2, r2), which is in T2. b) Both the intersection and union of (-r1, r1) and (-r2, r2) with r1 = r2 is always the interval (-r1, r1) = (-r2, r2), which is in T2. c) The intersection of (-r1, r1) and (-r2, r2) with r1 > r2 is always the interval (-r2, r2), which is in T2. The union of (-r1, r1) and (-r2, r2) with r1 > r2 will always be the interval (-r1, r1), which is in T2. Therefor, I conclude that T2 is a topology on R. Unfortunately, the exact same reasoning can be applied to all the collections T2 through T10, which leads to the result that all collections T2-10 are topologies, but only 3 of them are. What am I doing wrong? Link to comment Share on other sites More sharing options...

DrRocket Posted July 11, 2011 Share Posted July 11, 2011 (edited) I've started reading Sid Morris's Topology Without Tears and I'm stuck. Here's the exercise: Let R be the set of all real numbers. Precisely three of the following ten collections of subsets of R are topologies. Identify these and justify your answer. T1 = R, {} and every interval (a, b), with a, b any real numbers. T2 = R, {} and every interval (-r, r) with r any positive real number. T3 = R, {} and every interval (-r, r) with r any positive rational number. T4 = R, {} and every interval [-r, r] with r any positive rational number. T5 = R, {} and every interval (-r, r) with r any positive irrational number. T6 = R, {} and every interval [-r, r] with r any positive irrational number. T7 = R, {} and every interval [-r, r) with r any positive real number. T8 = R, {} and every interval (-r, r] with r any positive real number. T9 = R, {}, every interval (-r, r) and every interval [-r, r] with r any positive real number. T10 = R, {}, every interval [-n, n] and every interval (-r, r) with r any positive real number. Now, T1 is easy; not a topology. T2 through T10 though, that's a different story. Take T2 for example. I make the (possibly incorrect) assumption that if the union of any two sets from T2 is in T2 then every finite or infinite union of any number of sets from T2 is also in T2. The same with intersections. I consider three different scenarios with two intervals (-r1, r1) and (-r2, r2); a) r1 < r2, b) r1 = r2, c) r2 > r1: a) The intersection of (-r1, r1) and (-r2, r2) with r1 < r2 is always the interval (-r1, r1), which is in T2. The union of (-r1, r1) and (-r2, r2) with r1 < r2 will always be the interval (-r2, r2), which is in T2. b) Both the intersection and union of (-r1, r1) and (-r2, r2) with r1 = r2 is always the interval (-r1, r1) = (-r2, r2), which is in T2. c) The intersection of (-r1, r1) and (-r2, r2) with r1 > r2 is always the interval (-r2, r2), which is in T2. The union of (-r1, r1) and (-r2, r2) with r1 > r2 will always be the interval (-r1, r1), which is in T2. Therefor, I conclude that T2 is a topology on R. Unfortunately, the exact same reasoning can be applied to all the collections T2 through T10, which leads to the result that all collections T2-10 are topologies, but only 3 of them are. What am I doing wrong? A topology must be closed under finite intersections and arbitrary unions. Since the union of two intervals need not be an interval, NONE of the collections of sets that you listed are topologies. Are you asking which are bases for a topology ? Edited July 11, 2011 by DrRocket Link to comment Share on other sites More sharing options...

Shadow Posted July 11, 2011 Author Share Posted July 11, 2011 Okay, this confuses me a little; why would someone who knows enough about topology to write a book make such an obvious mistake. Also, while it's true that unions of intervals do not have to result in intervals, this is only true if the intervals have an empty intersection (except the case [math](a, b) U [b, c)[/math]), which if my logic is correct is only the case in T1. Let's be specific; could you give me an example of two intervals in T2 (ie. of the form (-r, r), r is real) who's intersection or union is not in T2? Or even better a hint so that I might find them myself? Link to comment Share on other sites More sharing options...

DrRocket Posted July 11, 2011 Share Posted July 11, 2011 Okay, this confuses me a little; why would someone who knows enough about topology to write a book make such an obvious mistake. Also, while it's true that unions of intervals do not have to result in intervals, this is only true if the intervals have an empty intersection (except the case [math](a, b) U [b, c)[/math]), which if my logic is correct is only the case in T1. Let's be specific; could you give me an example of two intervals in T2 (ie. of the form (-r, r), r is real) who's intersection or union is not in T2? Or even better a hint so that I might find them myself? You are correct, I did not read closely enough. Brain fart. Unions of disjointbintervals are not intervals, and I neglected to observe that in looking at symmetric intervals that there were no disjoint unions. Unions of symmetric open intervals about the origin are again intervals of that same form. So T2 is a topology. So are T9 and T10. The others are ruled out because unions of closed intervals can be open intervals and because irrationals can be arbitrarily approximated by rationals and vice versa. Link to comment Share on other sites More sharing options...

Shadow Posted July 11, 2011 Author Share Posted July 11, 2011 Thanks for your help DrRocket, I appreciate it. However, I'm having trouble understanding the following The others are ruled out because unions of closed intervals can be open intervals and because irrationals can be arbitrarily approximated by rationals and vice versa. Do you think you could go into more detail, ideally give a specific example? Please understand, this is relatively new to me, I've yet to become comfortable with this type of thinking. How can unions of closed intervals be open intervals? And how does approximating irrational numbers have any bearing on the problem? Link to comment Share on other sites More sharing options...

DrRocket Posted July 12, 2011 Share Posted July 12, 2011 Thanks for your help DrRocket, I appreciate it. However, I'm having trouble understanding the following Do you think you could go into more detail, ideally give a specific example? Please understand, this is relatively new to me, I've yet to become comfortable with this type of thinking. How can unions of closed intervals be open intervals? And how does approximating irrational numbers have any bearing on the problem? Take a union of closed intervals. The uperbound of the union will be the least upperbound, if it exists, of the intervals in question. That least upper bound may or may not be in the union. The open interval (-r,r) can be realized as a union of intervals [-q_n,q_n] with each q_n rational by choosing q_n converging to r, where r vcould be any real number, rational or irrational. If you like let r be pi and q_n the decimal approximation of pi to n decimal places. Link to comment Share on other sites More sharing options...

Shadow Posted July 12, 2011 Author Share Posted July 12, 2011 (edited) Take a union of closed intervals. The uperbound of the union will be the least upperbound, if it exists, of the intervals in question. That least upper bound may or may not be in the union. This makes no sense to me. If I understand the concept of upper bound correctly, the upper bound of the union will be the larger upper bound of the two, ie. if we have [a, b] and [c, d] with b < d, b > c (so overlapping intervals), the union of the two will be [a, d]; d is not the least upper bound. Furthermore, a union of two or more sets is the set containing every element which is contained in at least one of the former sets. If the intervals are closed then the bounds are included in the interval, and therefore will be included in the union. And regardless, all of the intervals considered here are symmetric, ie; one interval is always a subset of the other, ie. the union will always result in the larger of the two as demonstrated in the OP. I don't understand how a union of closed intervals can result in an open interval, let alone how a union of symmetric intervals can result in an open interval. The open interval (-r,r) can be realized as a union of intervals [-q_n,q_n] with each q_n rational by choosing q_n converging to r, where r vcould be any real number, rational or irrational. If you like let r be pi and q_n the decimal approximation of pi to n decimal places. I understand what you're saying, but I still don't see the connection to whether or not Tx is a topology. How does this prove that, let's say T3, isn't closed under arbitrary union or finite intersection? Edited July 12, 2011 by Shadow Link to comment Share on other sites More sharing options...

DrRocket Posted July 12, 2011 Share Posted July 12, 2011 This makes no sense to me. If I understand the concept of upper bound correctly, the upper bound of the union will be the larger upper bound of the two, ie. if we have [a, b] and [c, d] with b < d, b > c (so overlapping intervals), the union of the two will be [a, d]; d is not the least upper bound. Furthermore, a union of two or more sets is the set containing every element which is contained in at least one of the former sets. If the intervals are closed then the bounds are included in the interval, and therefore will be included in the union. And regardless, all of the intervals considered here are symmetric, ie; one interval is always a subset of the other, ie. the union will always result in the larger of the two as demonstrated in the OP. I don't understand how a union of closed intervals can result in an open interval, let alone how a union of symmetric intervals can result in an open interval. We have previously limited the consideration to closed intervals that are symmetric about 0, hence nested closed intervals. When you take the union of an infinite number of closed intervals, the least upper bound need not be included in the union. See the example based on decimal approximations to pi., but take the intervals to be closed. Then the union of the [-q_n,q_n] is (-pi,pi). I understand what you're saying, but I still don't see the connection to whether or not Tx is a topology. How does this prove that, let's say T3, isn't closed under arbitrary union or finite intersection? You can take open intervals(-q_n,q_n) with each q_n rational for bwhich the union is (-pi,pi) which is not in the given class of sets, so T3 is not a topology. Link to comment Share on other sites More sharing options...

Shadow Posted July 12, 2011 Author Share Posted July 12, 2011 Brilliant! Thanks Link to comment Share on other sites More sharing options...

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