When I got back to learning pure math about three weeks ago after an almost 3 year hiatus, I initially struggled and misbelieved that pure math is much higher than theoretical physics, which would imply that even though I could without difficulty learn special relativity deeply enough to re-derive it myself, it does not mean I would have a high chance of being able to do the same with the likes of Galois theory and measure theory, which were presumably of higher cognitive threshold. In undergrad and the few years after, I honestly felt Galois theory was harder than quantum mechanics. While quantum mechanics might be in some sense “counterintuitive”, it is not technically complex, especially non-relativistic quantum mechanics, unless you make it mathematically rigorous in which case it becomes more mathematical physics than physics. However, I became much more confident of my mathematical ability after I successfully wrote these notes on Galois theory without looking up anything other than the statement of Eisenstein’s criterion (which I suspected correctly was needed to prove irreducibility of the th cyclotomic polynomial for prime ), without much difficulty. I remember doing a fair bit of technical proof to show that the multiplicative group modulo any prime number is cyclic and while that was not immediate, I eventually figured out all the details. It took me less than two days to figure out all that was needed for those notes in addition to doing the writeup. For measure theory, I initially failed to prove a simple technical lemma in Rudin’s book, after which I became pessimistic and suspected that measure theory may simply beyond my level of intelligence, especially when I saw that there was a fair bit of detail and formalism required to construct the Lebesgue measure, which I was somewhat reluctant to read through. However, after noticing the existence of pseudometric with respect to the measure on symmetric difference of sets, the motivation and inspiration behind the Lebesgue measure became much more apparent and intuitive. I became more confident that I could now reconstruct the Lebesgue measure myself, and I succeeded in doing so, writing up notes more detailed than what was presented in that section in Rudin. In particular, I remember that Rudin sort of hand waved the proof of closure under set complement of the Lebesgue -ring, and I filled in the details by proving a lemma regarding the countable intersection of -finite sets.
Though my yet another attempt at learning pure math was mostly successful and in fact far more so than all previous attempts, I did experience a few failures. For instance, after thinking for a few hours, I was still unable to prove Taylor’s theorem for remainder estimate using Cauchy’s mean value theorem, for which I looked up the solution. I also failed to prove the Heine-Borel theorem, and suspecting that I lacked the requisite intuition and knowledge familiarity to do so without too much difficulty, I also looked up its solution. The few proofs in measure theory which I failed at also involved compactness, which I never developed any serious intuition for. Given all this, it is not surprising that I failed to reprove Tychonoff’s theorem after reading it. It is my hope that in the process of writing this, I can actually figure out what I am missing as far as those finite subcovers are concerned.
I thought of re-proving the Heine-Borel theorem myself, which relies on the Bolzano-Weierstrass theorem, which I did succeed in re-proving myself, recalling the use of nested intervals for it. However, before that, we will cover more point-set topology preliminaries not already covered in .
Definition 1 A Hausdorff space is a topological space such that any two distinct points can be separated by neighborhoods.
Definition 2 A metric defined on a space is a function that satisfies the following properties.
- for all .
- iff .
- for all .
Definition 3 A topological space is metrizable iff there exists a metric on it.
Definition 4 A base of a topological space is a collection of open sets such that every open set in the space can be expressed as a union of sets from the base.
Definition 5 A topological space is second countable iff there exists a countable base.
Definition 6 A metrizable topological space (or one of its subspaces) is bounded iff there exists such that for all in the space .
Definition 7 For any topological space on , is in the interior of any subset iff there exists a neighborhood of contained in , in which case must be a neigborhood of . We denote the interior of as .
Definition 8 For any topological space on , the boundary of any subset is the set of points in the closure of but not in the interior of . Formally, the boundary is .
Definition 9 A sub-neighborhood of neighborhood of in a topological space is a neigborhood of that is a proper subset of such that is a neighborhood of all points of .
Proposition 1 For a Hausdorff topological space on , either every singleton set is an open set, or for every and every neigborhood of , , there exists a sub-neighborhood.
Proof: Suppose by contradiction the topology on is not the finest possible and that for some does not have a sub-neighborhood. In this case, every neighborhood of contains , because if not taking the intersection of it with would yield a sub-neighborhood of . Thus, because is not a singleton set, we can not separate any two pairs of distinct points in it by neighborhoods, which would imply that is not Hausdorff.
Proposition 2 if open iff .
Proof: Assuming that is open, by definition of open, is a neighborhood of every one of its points. One of the neighborhood axioms (as listed in ) tells us that for every point , contains a neighborhood of it such that is also a neighborhood is of every point of . Thus, by definition of interior if , . That by definition of interior proves the other direction.
Proposition 3 for any set , and is necessarily open.
Proof: In Proposition 2, let , then it follows directly that is open. Because is open, it is a neigborhood of all of its points. Thus every point in is also in by definition of the interior operator. The other direction of set inclusion directly follows trivially from the definition of interior operator. Thus, .
Proposition 4 If is a neighborhood of , then is a neighborhood of .
Proof: is a neighborhood of . Thus by definition of interior point, . By Proposition 3, is open. By definition of open set, is a neighborhood of every point in , which it also contains. This completes our proof.
Corollary 1 If is a neighborhood of , then there is an open sub-neighborhood.
Proof: Follows directly from Propositions 3 and 4.
Corollary 2 Every point has an open neighborhood.
Proof: Follows directly from Proposition 4.
With Corollary 2, from now on, whenever we say neighborhood of a point, one can tacitly assume that is open unless specified otherwise.
Proposition 5 For any Hausdorff space on , for any point , there exists a chain of neighborhoods of , , monotonically decreasing under the sub-neighborhood partial order, such that .
Proof: By Corollary 2, we Suppose by contradiction that for every such chain, is a proper superset of . We can construct the collection of such infinite intersections such that WLOG, no two elements in it exist such that one properly contains the other. We call this collection of sets . Obviously, no neighborhood of can be a proper subset of any . But for any , wherein is a sub-neighborhood of for all , there is a point in it not equal to , and thus because the space is Hausdorff, there exists an open neighborhood of not containing that point, which we call , with . By one of the neighborhood axioms, is a neighborhood of for all , which is also open. then is such a chain of neighborhoods and one that is a proper subset of , a contradiction, which completes our proof.
Theorem 1 (Cantor’s intersection theorem) For any topological space , a decreasing nested sequence of non-empty compact closed subsets of has a non-empty intersections.
Proof: Left to the reader. This is non-trivial, and I myself have not seen the proof either.
Proposition 6 For any chain of neighborhoods of , , monotonically decreasing under the sub-neighborhood partial order, such that , then an sequence need not converge to .
Proof: Let . Let all intervals and be open sets. Let such a chain of neighborhoods of , in which case . Let . This does not converge to since is an open yet no element in the sequence is equal to .
I got this counterexample from @Sizhe Chen after failing to prove a false proposition. However, I suspect it would hold for connected neighborhoods.
Proposition 7 The complement of an open set is always closed.
Proof: Because any open set is equal to its interior, it does not contain any of its boundary elements. We note that can be partitioned into . The elements of any sequence in is either in or in . Because is a neighborhood of all of its points, such a sequence cannot convergence to a value in , which completes our proof.
Proposition 8 For any and any , any neighborhood of contains a point in and a point in . Moreover, .
Proof: With , we know by definition of boundary that not subset of can be a neighborhood of . Thus, any neighborhood of cannot be disjoint with . Now suppose by contradiction that there is a neighborhood of disjoint with . In this case, is a subset of . Because is open, no sequence of points outside it can converge to any point in it. Thus, is disjoint with . By definition of boundary, there exists a sequence in that converges to . But no sequence in has any points in which contains . Thus, no sequence in to , which means that , which contradicts our hypothesis.
For the second part, with disjoint with , we have that for all subsets . Apply this to and we get . This completes our proof.
Proposition 9 Given any subset , can be partitioned into .
Proof: are disjoint by definition. Since by Proposition 9, are disjoint too. Obviously the two interiors are also disjoint. That the union of the three sets equals is also easy to verify.
Proposition 10 For any topological space on , for any set , is open.
Proof: The cases of are of course trivial. In general we note how
Then by Proposition 3, is open.
Proposition 11 For any set , and is closed with no interior points.
Proof: By Proposition 8, for any point , any neighborhood of it contains a point both in and in , which by Proposition 9, are both disjoint with . Thus, is not a neighborhood of any of its points. This shows that and that has no interior points. Because are both open sets, no sequence in can converge to any one of their points. This shows that is closed.