Point of post: In this post I will give a proof that the algebraic numbers are countable, and thus we will prove that there are non-algebraic (transcendental) numbers.
Remark: Personally I find my proof much simpler and more intuitive than the usual method by considering the “height” of numbers (viz. Rudin, pg. 43)
Problem: Let be a second countable topological space and be such that . Then,
Proof: Suppose that is countable then is uncountable. But, by definition for each there exists some neighborhood of it such that . So, clearly is an open cover for and since is second countable Lindelof’s theorem guarantees that it must have a countable subcover . Thus,
But this contradicts that is uncountable. The conclusion follows.
Corollary: If and is the set of all isolated singularities then
Corollary: Let be a separable metric space and uncountable, then there exists a sequence in such that .
Problem: Let be a metric space and perfect and open. Then, is perfect.
Proof: The fact that follows from the fact that it’s closed. Now, let and let be any neighborhood of it. Then, by definition we have that and so there exists some . Now, since and is a neighborhood of it it follows that there are infinitely many values of in it. In other words is infinite and in particular there exists some such that . Thus, . The conclusion follows.
Corollary: If is a metric space and is perfect we can choose and see that and we have a perfect subset of which is also bounded.
Problem: Let be a compact metric space. Prove that is connected if and only if it cannot be written as a union with
Proof: It is tacitly assumed that . So, in any metric space suppose that with . Remember that and thus if we have that and thus . Thus, (the reverse direction gotten using the exact same logic) from where it follows that is disconnected.
Before we continue to prove the converse we prove a nice little lemma
Lemma: Let be a compact metric space and non-empty disjoint closed subspaces of . Then, .
Proof: Since is compact and closed subspaces it follows that they themselves are compact and thus so is . So, notice that is continuous since it is the distance function (trivially continuous) restricted to . Thus, by the extreme value theorem it follows that for some and since we see that .
So, now assume that were not connected. Then, where are non-empty disjoint closed subsets of . But, by the lemma this contradicts the assumption that cannot be written as with .
Problem: Let be a separable metric space, then for any subspace we have that is separable.
Proof: We prove something stronger. But, first we prove a small lemma
Lemma: Let be a second countable topological space, then any subspace is second countable.
Proof: Let be the countable base for and let . Clearly is a countable collection of open subsets of . Now, let be arbitrary and let be any neighborhood of it. Then, by definition for some open set in . So, we may find some such that and thus and since the conclusion follows.
So, if is a separable metric space, then as proven earlier it is second countable and thus so is (by the lemma) but since every second countable space is trivially separable the conclusion follows.
Problem: Let be a compact metric space and an isometry. Then, is surjective.
Proof: Suppose and consider . Since is a compact metric space this must have a convergent subsequence . In particular, that sequence must be Cauchy. So, noting that are disjoint non-empty compact subspaces of a compact space it follows by an earlier lemma that 4. So, since is Cauchy there exists such that . So,
which is a contradiction.
Problem: Let be a compact metric space and be a function such that
for all distinct . Prove that has precisely one fixed point.
Proof: Clearly if there is a fixed point it’s unique because if then . Now, to see that it must have a fixed point we note that
is continuous since evidently the diagram
commutes. So, since is continuous and it follows that obtains a minimum at some . Now, assume that then contradicting that achieved it’s minimum at . The conclusion follows.
Problem: Prove that ever closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is a set which is countable.
Proof: It is easy to see that the above proof (proof 27.) holds in any space which is second countable, in particular for separable metric spaces. Thus, if is closed we surely have that . Thus, but since and we have then that which is the union of a perfect and countable set respectively.
Problem: Prove that every open set may be written as the disjoint union of countably many open intervals.
Proof: We need to prove a quick lemma
Lemma: Let be a topological space and let be a class of connected subspace of such that then is a connected subspace of .
Proof: Suppose that is a separation of . We may assume WLOG that for some . So, now we see that otherwise would be non-empty disjoint subsets of whose union is contradicting that is connected. Thus, it easily follows that for any we have that so that and thus by a similar reasoning we see that . Thus, since was arbitrary it follows that contradicting that .
So, now for each define
And let and finally we prove that
is a countable class of disjoint open intervals whose union is . The fact that each is an open interval follows since it is the union of open connected spaces with non-empty intersection (each element of contains ) it is also an open connected subspace of (note that each element of is open in but since is open it is also open in . But, it was proven in the book the only connected subspace of are intervals and thus is an interval for each .
Now, to see that they are disjoint we show that if then from where the conclusion will follow. So, to see this we first note that if is non-empty then is an open connected subspace of containing both and and thus and but this implies that and respectively from where the conclusion follows.
Now, to see that is countable we notice that for each we have that and so if we let be a fixed but arbitrary then
is an injection since the elements of are pairwise disjoint. The fact that is countable follows immediately.
Thus, is a countable collection of open intervals and
Thus, the conclusion follows.
Problem: Prove that if where each is a closed subset of then at least one has non-empty interior.
Lemma: Let be a complete metric space and a descending sequence of non-empty closed subsets of such that . Then, contains one point.
Proof: Clearly does not contain more than one point since . So, now for each choose some and let . Now, if were somehow finite a similar argument to that used in the first case of problem 26. is applicable, so assume that is infinite. But, since it is evident that is a Cauchy sequence and thus by assumption it converges to some point . We claim that . To see this we note similarly to problem 26. that since is infinite it is easy to see that is a limit point of and so if for some then is a neighborhood of containing only finitely many points of which clearly contradicts that it is a limit point. The conclusion follows.
So, now suppose each had empty interior (i.e. nowhere dense) . Then, since is open and is nowhere dense there exists an open ball of radius less than one such that . Let be the concentric closed ball of whose radius is half that of . Since is nowhere dense contains an open ball of radius less than one-half which is disjoint from . Let be the concentric closed ball of whose radius is one-half that of . Since is nowhere dense we have that contains an open ball of radius less than one-fourth which is disjoint from . Let be the concentric closed ball of whose radius is half that of . Continuing in this we get a descending sequence of non-empty closed subsets for which . Thus, by our lemma we have that there is some . This point is clearly not in any of the ‘s from where the conclusion follows.
Problem: Let and be separated subsets of , suppose that and define
Let . a) Prove that are separated subsets of . b) Prove that there exists such that . c) Prove that every convex subset of is connected.
a) Clearly we have that and so, if we assume that then we must have that . So, we may choose such that and so it easily follows that and thus so and thud which is a contradiction.
b) We must merely note that if then it would be a path from to which is impossible since they are separated.
c) We can prove this more generally. Let be a normed vector space and let be convex, then is path connected. This clearly follows since the straight line is a path from to .
Problem: A metric space is called separable if it contains a countable dense subset. Show that is separable.
Proof: We prove the much, much deeper following theorem.
Lemma: Let be a countable class of non-empty separable topological spaces with corresponding countable dense subsets . Then, is separable with the product topology.
Proof: For each select an arbitrary but fixed and define where
Clearly we have that and thus countable (the finite product of countable sets is countable). So, let
Then , being the countable union of countable sets, is countable. We now claim that it is dense in . To see this let be arbitrary and any basic neighborhood of it. Then, there are only finitely many indices such that . So, let . Now, since for each we have that is a neighborhood of in there is some . So, doing this for we arrive at points . So, then
Which proves that is dense. The conclusion follows.
The above is kind of heavy duty (though it works since the product topology and usual topology coincide). For people who are more interested in analysis, the real way to do this is to note that is countable and note that for any you can choose corresponding such that and do the simple calculation.
Problem: A collection of open subsets of ( a metric space) is said to be a base if for every and every neighborhood of it there is some such that . Prove that every separable metric space has a countable base (in other words that it’s second countable).
Proof: Let be the countable dense subset and let . Clearly is countable . To see this we merely note that is clearly a surjection. We shall now prove that (as it’s letter suggests) is a base. So, let be arbitrary and be given. Since is dense in there exists some . So, let and choose such that . then, and if we have that and thus . The conclusion follows.
Problem: Let be a metric space in which every infinite subset of it has a limit point. Prove that is separable.
Proof: Consider the open cover and suppose that it did not have a finite subcover. Then, we may form a sequence such that , for otherwise if there existed some such that for no we have that then for every we have that
or in other words that contradicting the assumption that has no finite subcover. But, let and consider clearly contains at most one point of . To see this suppose that it contained more than one point, say then we may assume WLOG that but this contradicts the construction of the sequence since . It follows that has a finite subcover. So, let for each we may cover with finitely many open balls of the form . So, let be the finite set of points such that and let . Clearly is countable and given any and any neighborhood of it we may (by the Archimedean principle) find some such that . But, since every point of is less than in distance from some point of it follows there is some . The conclusion follows.
Remark: All we really showed above (the hard part anyways) is that every limit point compact metric space is totally bounded. As we prove in the next theorem limit point compactness is implied by compactness in metric spaces (they are in fact equivalent) and thus every compact metric space is totally bounded. It actually is true that any space such that every there is a countable -net is separable, this is done precisely as in the last part of the above.
Problem: Prove that every compact metric space has a countable base.
Proof: It suffices to show that every infinite subset of has a limit point, from where it will follow from 24. that it is separable, and thus second countable. So, let have no limit point. Then, for each there exists some open ball such that . So, we may cover and so by assumption this open cover must admit a finite subcover . So,
The conclusion follows.
Problem: Let be a metric space such that every infinite subset has a limit point. Show that is compact.
Lemma: Let be a limit point compact metric space and a sequence of descending () of non-empty closed subsets of . Then,
Proof: For each choose some . We break this into two cases, namely whether the set of all these ‘s is finite or infinite.
Case 1: Suppose that is finite but . We first note that for each there is some such that . To see this assume not and let then otherwise and thus which was assumed to be impossible. So, for all and thus which contradicts our assumption that it was empty. So, for each there exists some for which the above idea is true. Let then which contradicts the construction of the set. It follows that the intersection is non-empty.
Case 2: Suppose that is infinite. By assumption it then has a limit point . So, we show that by showing that it is contained in a fixed but arbitrary . So, suppose that then which is open. But, since it follows that is finite contradicting that is a limit point of that set. It follows that .
So, now let be an open cover for . Since is second countable we know that admits a countable subcover . Now, suppose that for every then where is a sequence of decreasing, non-empty closed sets. But,
Which contradicts the lemma. The conclusion follows.
Problem: Let be a metric space and call a condensation point of a set if every neighborhood of contains uncountably many points of . Now, suppose that is uncountable and let be the set of all condensation points of . a) Prove that is perfect and show that is countable.
a) Let but suppose that then there exists some neighborhood of it such that for every there exists a neighborhood of it such that . So, we now need a quick lemma
Lemma: Let be a second countable topological space and any subset of . Then, if is a collection of open sets in such that then there exists some countable subcollection such that
Proof: For each there is some such that but if is the countable base there is some such that . Doing this for each clearly gives a collection such that . But, by design it is clear that is countable and so for each we may choose some (since this is how they were begotten) some such that . So, doing this for each we arrive at a countable collection such that
from where the conclusion follows.
So, note that since is second countable and we clearly have that is a collection of open sets whose union contains and so by the above lemma it follows that it must admit a countable subcover so then
which contradicts that . So, this proves that . Conversely, let then every neighborhood of contains a point of , but since is also a neighborhood of that point it follows that from where it follows by the arbitrariness of that .
b) Let . We claim that . To see this let then there is some such that and thus . Conversely, let then there exists some neighborhood of such that but there exists some such that . Thus, . So, we see that
The last part gotten from the fact that is countable and each also countable. The conclusion follows.
Problem: Form a compact subset of the real numbers (with the usual topology) which has countably many limit points.
Proof: Let which is tediously, but easily verified. Sorry folks, I’m not interested enough to Tex it all out. Comment me asking for a full solution if you so wish.
Problem: Given an open cover (with the usual topology) which has no finite subcover.
Proof: How about . So, we must first show that this in fact covers . So, let be arbitrary and let . Then, we have that and so by the Archimedean principle there exists some such that and thus . Now, to see that no finite subcover works, let be any finite subclass of of . Then, let . Now, it clearly follows that and so we mus merely show that there is some . But, we merely need realize that and so the conclusion follows.
Problem: Show that the intersection of arbitrarily many closed subsets of a metric space with the finite intersection property may be empty if they are just closed, or just bounded.
Proof: Just take for example. Then, the class is a class of closed subsets of with the FIP but their intersection is empty. To see this we merely note that for each we have that
Also, if we just have bounded then the claim is also wildly false in . Take . This is a class of bounded subsets of with the FIP, but clearly since choosing we have that for some (by the Archimedean principle) and thus .
Problem: Regard as a metric space with . Let . Prove that a) is closed in , b) it is bounded in , but it is not compact in .
a) We first prove a nice little lemma (which is almost immediate from theorem 2.30)
Lemma: Let be a metric space and a subspace (in the metric space sense, don’t get too ahead of yourself with topological terms…we haven’t yet proven they are equivalent), then is closed in it if and only if for some set which is closed in .
Proof: First suppose that is open, then is open and thus by theorem 2.30 we have that for some set closed in . A little set theoretic self-convincing then shows that .
Conversely, suppose that for some set closed in . Then, (once again with a little set theoretic magic…be sure to be mindful of what set theoretic difference means here!) and thus by theorem 2.30 again we see that this means that is open and thus closed.
From this it immediately follows that is closed in since evidently and is closed in .
b) It is evidently bounded.
c) We prove that is not compact in by a roundabout but instructive way.
Lemma: Let be a topological space and a subspace. Then, is a compact subspace of if and only if it’s a compact subspace of .
Proof: First suppose that is a compact subspace of and let be an open cover as a subspace of where is open in . Then, since is a subspace of it evidently follows that is open in for every . Thus, it is evident that is an open cover of as a subspace of . So, by assumption it admits a finite subcover and so it readily follows that is a finite subcover of
Conversely, suppose that is compact as a subspace of and is an open cover for as a subspace of . Then, since is a subspace of we have that for some open in . Thus, it is evident that is an open cover for as a subspace of and so by assumption it admits a finite subcover from where a quick check shows then that is the desired finite subcover of .
From the above lemma we see that will be compact as a subspace of if and only if it is compact as a subspace of . But, it is not since evidently is a limit point of but not a point of , and thus is not closed and thus not compact.
Problem: Let ; a) Is countable? b) Is dense in ? c) Is dense in ? d) Is perfect in ?
a) No, is not countable. It is evident that . The only one that may be slightly hazy is that but it is pretty routine to check that is a bijection. (this is just the correspondence between subsets and indicator functions).
b) Of course it isn’t dense. It is clear that .
c) It suffices from prior comment to show that it’s compact as a subspace, and by virtue of the Heine-Borel theorem we must merely show that is closed and bounded. Boundedness is clear though since . So, suppose that but . Then, we have that where for some . But, note then that for any we have that and thus which of course is a contradiction. Thus, the conclusion follows.
d) It suffices to prove (since we proved in c) that is closed) that . To do this let and let be given. Then, by the Archimedean principle there exists some such that and so if then we have that
Where obviously we choose to be the opposite of what is. Then, and . It follows that .
Problem: Is there a non-empty perfect subset of (with the usual topology) which contains no rational values?
Proof: The answer is yes, but to say it right takes some measure theory. We know that (the usual measure) and so there exists countably many open intervals such that and . So, we have that is closed, non-empty (since it is non-zero measure). So, once again since it has positive measure it must be uncountable and by problem 28 it follows that the set of all condensation points of is perfect, and since each condensation point is a limit point it must be a point of and thus not a rational number.
Problem: a) If are disjoint closed sets in a metric space then they are separated. b) If are disjoint and open they are separated. c) Fix and define and , and prove that are separated. d) Prove that every connected metric space with more than one point is uncountable.
a) Clearly are separated since .
b) Suppose that then it must follow that there is some . But, since every neighborhood of intersects and thus intersects and thus which is a contradiction. An identical argument works for the other cases.
c) Merely note that is continuous (since it’s the normal metric restricted to ). And, that which are disjoint and open. The conclusion follows from b)
d) Suppose that were countable and has more than one point, then so is and then so is . So, choose to be distinct and let . But, since is uncountable and countable there exists some . So, let and . Then, are non-empty ( and disjoint and since their union is equal to . It follows that is not connected.
Problem: Are closures and interiors of connected sets always connected?
Proof: Closures are, we reprove this fact for general topological spaces.
Lemma: Let be a topological space, a connected subspace and then is a connected subspace.
Proof: Evidently . So, let (where is the two-point discrete space) be continuous. Then since is connected must have that is constant, but since is dense in , continuous and Hausdorff it follows that is constant. The conclusion follows.
The interior of connected subspaces is not always connected though. Consider the “dumbbell” shape . This is clearly connected (it is the union of three connected sets with non-empty intersection) but it’s interior is which is as classically disconnected as one can get.
Remark: It is interesting to note that the result is true for , since the only connected subsets are interiors which (as can easily be checked by case) the interiors of which are intervals.
Just for fun I’ve decided to do the entire section on topology in Rudin’s Principles of Mathematical Analysis. The problems themselves aren’t particularly difficult, but some are interesting and it will give me a chance to do some refreshing. Also, a friend of mine is starting Rudin and it will be nice to have done them in case he has any questions. I will probably prove more general results from topology if I feel like it. So…
Problem: Prove that the empty set is a subset of every set.
Proof: Suppose that we are working in some universal set and there existed some such that . Then, by definition there exists some such that . But, this is clearly preposterous since the statement is never true.
Problem: A complex number is said to be algebraic if there are integers , not all zero, such that
Let denote the set of all algebraic numbers. Prove that .
Proof: Let . By the fundamental theorem of algebra we have that . So, clearly being the countable union of countable sets is countable. Thus, we claim that
from where it will follows that being a subset of a countable union of countable sets is countable. So, let then there exists such that . Thus, clearly from where the conclusion follows.
Problem: Prove that there exists real numbers which aren’t algebraic.
Proof: Clearly since is countable so is . And thus, if were countable then so would which is of course absurd. It follows that there are uncountably many non-algebraic (transcendental) numbers.
Problem: Is the set of all irrational numbers countable?
Proof: Of course not. For using a similar argument to the above, if were countable then would be as well.
Problem: Construct a bounded set of real numbers with exactly three limit points.
Proof: Why not do it for an arbitrary natural number. Let . Evidently has precisely one limit point. Now, let then it is equally clear has exactly one limit point (namely ). So, the set has precisely limit points as one can easily check.
Problem: Let be the set of all limit points of a set (where is a metric space). Prove that a) is closed , b) , c) Do and always have the same limit points?
a) This is true in any topological space. But, first we need a quick lemma
Lemma: Let be a space, and . Then, if then for very neighborhood of we have that is infinite.
Proof: Suppose not, and there existed a neighborhood of such that . Since is for each there exists some neighborhood of such that . Thus, is a neighborhood of which does not intersect , which of course contradicts that .
So, using this let be a space, then for any it is true that is closed. To see this let and let be any neighborhood of it. By assumption there exists some and thus since is a neighborhood of also there exists infinitely many points of in . In particular there exists some such that . Since was arbitrary it follows that . The conclusion follows.
b) Let then every neighborhood contains a point of different from itself, and thus a point of different from itself. Conversely, let and let be any neighborhood of it. We must have that there is some . Now, if we’re done, and if not by the previous lemma since is a limit point of it must be that contains infinitely many points of anyways, and thus a point of different from . The conclusion follows.
c) Of course not. with the usual topology has exactly one limit point, namely . But, a singleton in Euclidean space does not have an limit points.
Problem Let be a metric space and . a) Show that . b) show that example that and the inclusion can be strict.
a) This is true in any topological space . Suppose then there exists a neighborhood of such that and thus and thus and so . Conversely, let . Then, since there exists some neighborhood of it such that . Thus, clearly is a neighborhood of which does not intersect and thus . The conclusion follows.
b) Let then there exists a neighborhood of such that and thus . It follows that and thus . But, as was noted the inclusion can be strict. Give the usual topology. Then, for each we have the singleton is closed and thus . So,
Problem: a) Is every point of every open set a limit point of ? Answer the same for closed sets in ?
a) Of course, I mean this is true in any topological space. If is open then each point of is an interior point and thus trivially a limit point.
b) No, take a singleton.
Problem: Let denote the set of all interior points for in a metric space . Prove that:
a) is always open
b) is open if and only if .
c) If and is open then .
e) Do and always have the same interiors?
f) Do and always have the same interiors?
a) Let then there exists a neighborhood of such that , but for each we also have that is a neighborhood of it which is contained in . Thus, .
b) If the above proves that is open. Conversely, since a set in a metric space is open if and only if each point is an interior point the converse is trivial.
c) Let then is automatically a neighborhood of which is contained in so that .
d) Let then for every neighborhood of we have that and so and thus . Conversely, let then for every neighborhood of we have that and thus and so or .
e) Of course not. It is trivial to check that if one gives the usual topology then (since it’s complement is dense in ) and .
f) The above works as a counterexample to this as well.
Problem: Let be an infinite set and define by
a) Prove that this is a metric. b) Which subsets of the resulting metric space are open? c) Which are closed? d) Which are compact?
a) Clearly and . Also, since equality and inequality are transitive symmetry immediately follows. The triangle inequality is just a set of three cases. If then . If then . If then .
b) For every we have that is open since . Thus, all subsets of being the union of singletons are open.
c) From b) it readily follows that every subset of is also open.
d) will be compact if and only if is finite. To see this suppose that is infinite, then is evidently an open cover of it which admits no finite subcover. The converse is trivial.
Problem: Let the following be maps from to , and determine if they determine a metric. a) , b) , c) , d) , e)
a) This is not a metric.
b) This is a metric as one (I’m to lazy to tex it) can check.
c) Nope, .
d) Negatory, .
e) Yes. In fact, (once again I omit the proof) if is a metric so is
Problem: Let with the usual topology be defined by . Prove directly (i.e. no Heine-Borel) that is compact.
Proof: This is true in any topological space . But, for the sake of convenience let us do it merely in a Hausdorff space . So, let is convergent sequence in with . Then, is a compact subspace of .
Proof: Let be an open cover for . Choose such that . Then, since all but finitely may values of lie in . So, for each of the other finitely many points of choose any element of which contains them. Clearly this will be a finite subcover.