Well, I did it. I listened to all thirteen renditions of Chestnuts. No regrets, but possibly there are one or two that I never need to hear again.
You wouldn't think that a chorus could really do justice to this song, and you'd probably be right. But it's not that simple: the Mitch Miller version and the Mormon Tabernacle Choir version are really very different. I prefer the latter, though I'll use it sparingly.
It occurs to me that it is not possible to record a version of this song without making implicit reference to the Nat King Cole version. (I am thinking of Borges' "Pierre Menard, Author of the Quixote".) For example, if you don't sing "Merry Christ-ma-has" at the end, that's a statement in itself.
A Salt Hygrometer
19 hours ago
40 comments:
What do you think Borges would have called that story? "Santa Claus, Author of The Christmas Song"? "The Three Chipmunks, Authors of Chestnuts Roasting on an Open Fire"? Those wouldn't be quixotic enough. You owe us a title.
Forgot to check the follow-up box.
I don't know what Borges would call it, but I would call it "There are No Unramified Covers".
(That was me. Keyboarding malfunction.)
That's very cute.
Are there more theorems on such quixotic albums? Is there a general category of interest, say "chestnut spaces"? These might be spaces whose coverings are necessarily ramified. That is, they have implications in every local neighborhood, from sea to shining sea.
In topology "unramified cover" usually means "covering space", and if one said "X has no unramified covers" one would undoubtedly mean "X has no (connected) covering space except the trivial one".
The good old fact about this -- you could even call it a chestnut -- is that the nonexistence of such covers is equivalent to the rather different-looking condition "X has no essential loops" (a.k.a. X is simply-connected).
I suppose one might describe a song as having or not having essential loops in it, but I cannot see that this has much to do with questions about the ramifications of recording a cover version of the song. So I don't have a theory to ptopose, just examples.
My idea was about the emotional ramifications of songs for different people in different neighborhoods - not the ramifications of recording them.
After making the comment, I read up a bit on covering spaces, having forgotten most of what I once knew, more even. According to WiPe, semilocal simply-connectedness is a necessary condition for the existence of a universal cover. That is of course weaker than simply-connectedness.
I see now that when I wrote "ramification" I was thinking of the fibers of a covering. The ramifications were the (inverse) images that people called up in their minds when they heard the songs.
So when I wondered about "necessarily ramified" covers of X, this turns out to mean: are there are non-trivial connected covers C of X, for each of which the fiber cardinality is > 1 ? This was surely a silly thing to wonder about. A non-trivial connected cover must have fiber cardinality > 1, since cardinality = 1 would imply that the cover is trivial. At least if X is simply connected. Is there a non-simply connected X that has a connected cover C with fiber cardinality = 1, such that C and X are not homeomorphic?
By the way, snooping around in other topology articles in WiPe, I encountered the Hawaiian earring and the wedge sum of countably many S1. It took me much too long to understand why they are not homeomorphic. Sigh ...
My idea was about the emotional ramifications of songs for different people in different neighborhoods - not the ramifications of recording them.
Maybe your idea is not so far from mine.
I was punning on "cover". In music, X is a cover of Y if Y is a well-known song by one artist and X is a performance (recorded or not) of the same song by another artist. In some cases Y is so much with us that we cannot hear X except in relation to Y. So a cover has ramifications. It is true that there can be different ramifications for different listeners.
In mathematics, let us confine ourselves to those topological spaces that have universal covering spaces. This includes those which are "semi-locally simply-connected", and in particular those which are locally simply-connected. Many of the spaces that one thinks about -- manifolds and cell complexes, for example -- are locally even nicer than this: locally contractible. Sticking with this case, we have the very satisfying theorem that subgroups of the fundamental group of Y correspond with connected covering spaces of Y. The covering space corresponding to the trivial subgroup is called the universal cover(ing space) of Y, and it has as many points in each of its fibers as the group has elements. At the other extreme, the one that corresponds to the whole group is Y itself (covering itself with fibers having one point apiece, as you say).
The fundamental group is a global, not local, feature of the space. The question of what covers the space has is also a global question.
(I'll get to the mathematical mening of ramification later.)
Too bad our chestnuts are extinct. Did Elvis ever do the song? He had such a nice gospel voice for Christmas music.
Did Elvis ever do the song?
Not as far as I can tell.
I can tell you this song doesn't work at all when your daughter has a nut allergy and you love roasted chestnuts.
Is there a non-simply connected X that has a connected cover C with fiber cardinality = 1, such that C and X are not homeomorphic?
Simple example: consider the countably infinite wedge sum W = V (S1(1)_i, p), and the Hawaiian earring E = U (S1(r_i)_i, p). Here S1(x) means an S1 with radius 1/x and p is a distinguished point of that S1.
Take the 1-1 surjection f: W -> E which maps each S1(1)_i loop to the S1(r_i)_i loop, preserving p. f is a continuous cover of E by W, with fibre cardinality 1. But E and W are not homeomorphic. In our case, the inverse of f is not continuous.
1-1 surjection
Forgot about the word bijection.
Is there a non-simply connected X that has a connected cover C with fiber cardinality = 1, such that C and X are not homeomorphic?
"Non-simply connected" is irrelevant.
To restate the question without the irrelevant feature: can a continuous bijection be a cover without being a homeomorphism?
No. In order for a map f:X-->Y of topological spaces to be called a cover, it is necessary (though not sufficient) that every point x in X should have a neighborhood N in X such that f maps N by a homeomorphism to its image f(N). And for this it is necessary that for every open set U of X the set f(U) is open in Y. And if f is a continuous bijection then this latter requirement (f takes open sets to open sets) makes f a homeomorphism.
So the example that you gave, of a continuous bijection which is not a homeomorphism, cannot be a cover.
By the way, here is another space that looks at first glance to be homeomorphic to those two (the Hawaiian earring and the infinite wedge of circles): The union of a n infinite sequence of circles with radii increasing to infinity. To be definite, say that the nth circle is centered at the point with coordinates (n,0) in the (x,y) plane: the union of these over all integers n greater than 0.
This is not homeomorphic to the Hawaiian earring because it is not compact. It is not homeomorphic to the wedge -- briefly, because it has a compact subset not contained in the union of finitely many of the circles.
Yeah, I keep forgetting about the "local" part of the definition of cover.
Also, as I already said, it took me quite a while to realize why my bijection between wedge and earring was not continuous in both directions - in the original formulation, why they were not homeomorphic, before I tried to analyze some specific bijection.
Thinking about it now, I see I had apparently, vaguely thought the inverse of f must be continuous if f was, because the topologies of wedge and earring were both induced from that of the plane. But that's silly because many standard examples of some pathology, or special feature, are given as subsets of the plane. One wants to regard them as independent spaces with induced topologies. However, the arguments demonstrating the pathology of these independent spaces are carried out by visualizing not only the inducing topology applied to the spaces as subsets of the plane, but also the induced topologies that makes them independent spaces. One switches back and forth between inducing and induced topologies.
I didn't say this quite right.
it is necessary (though not sufficient) that every point x in X should have a neighborhood N in X such that f maps N by a homeomorphism to its image f(N).
I should have said something like "to its image f(N), a neighborhood of f(x)".
The essence of the definition of cover(ing space) f:X-->Y is the localness in Y. For every point y in Y, if you restrict your attention to a small enough part of Y around y and to the part of X which is mapping into that part of Y, then X looks like just some number of disjoint things each getting mapped homeomorphically to Y. This number can be one (in which case the original global f is a bijection and in fact (since continuity of the inverse is a local condition) a homeomorphism, or zero, in which case X is empty.
As to how to recognize that a given continuous bijection f:X-->Y is in fact a homeomorphism, the easy case is when X and Y are compact Hausdorff spaces; in fact you only need to know in advance that X is compact and Y is Hausdorff. Then (C closed in X) implies (C compact in X) implies (f(C) compact in Y) implies (f(C) closed in Y).
The following sentence referring to your "other space", which I'll call the Hawaiian expanding-earring EE, caused me altogether many hours of not-understanding grief:
A) It [EE] is not homeomorphic to the wedge -- briefly, because it has a compact subset not contained in the union of finitely many of the circles.
But during these hours I revived basic knowledge that I had forgotten. So I am much further along than merely having identified such a compact subset of EE as you refer to in A). This is how I see things now (please tell me if I've made another stupid mistake in my thinking !):
1) First of all, the countable wedge W is not homeomorphic to a bouquet B of "equally sized", countably many copies S^1_i of S^1 embedded in R^2. I had been thinking of W as if it were (homeomorphic to) B.
The topology of W is a quotient of the product topology on P = Product(S^1_i). Since P is compact, so is W. So since B is also compact, and there is an obvious bijection, one might imagine that W and B were homeomorphic.
But the topology of W is strictly coarser than the induced topology on B (when you imagine W and B as point-sets with the obvious bijection), so W and B are not likely to be homeomorphic. To demonstrate that they are not, I note that the wedge point is in every open set of the *infinite* wedge W, so W is not Hausdorff. But B is, so W and B are not homeomorphic. A peculiar fact I noticed here is that {wedge_point} is not closed in the wedge topology, but of course {bouquet_point} is in the induced topology.
2) Since EE is Hausdorff, it can't be homeomorphic to W, which is not Hausdorff.
To be absolutely precise, the last sentence in 1) above should start:
"A peculiar fact I noticed here is that {wedge_point} is not closed in the infinite wedge topology, ..."
Even more precisely:
"A peculiar fact I noticed here is that {wedge_point} is not closed in the specific infinite wedge topology W, ..."
Otherwise I might be taken to be making a claim about {wedge_point} in wedge topologies in general.
Dammit. The wedge sum of compact spaces is not necessarily compact. I was getting confused by the notation in various things in the internet about wedge sums and smash products that I had skimmed over. The large rectangular U used for disjoint union, in the definition of wedge sum, looked too much like an upper-case Greek Pi for product, so Tychonoff's theorem was hovering in my head.
But compactness doesn't make any difference, since W, the wedge sum of countably many S^1, is still not Hausdorff, and so not homeomorphic to the R^2 bouquet B.
I am really out of practice with this stuff. A major difficulty I have is that I am not that interested in it any more, so my attention keeps flagging.
Well, I give up. I just can't piece together all the bits I find in the internet, and my topology books are still in storage.
Still under the spell of the U notation looking like Greek P, I have been thinking of the wedge topology as it were the product topology. I haven't been able to find a definition anywhere of the wedge topology, except as induced by the quotient. If that means what I think it does for W, then W will be Hausdorff after all. So my argument falls to the ground.
As you see, my attention keeps flagging, then it comes back to bite me on the shins.
Stuart, just let me know whether you're interested or not.
"Wedge" and "bouquet" are synonyms, as I use them. They both mean that quotient space of the disjoint union. The wedge of compact spaces will rarely be compact if there were infinitely many spaces involved. For example, if each of the spaces is Hausdorff and has more than one space in it then the wedge will not be compact. The wedge of Hausdorff spaces is always Hausdorff.
When you have a collection of spaces X(i) all occurring as subspaces of a given space X (for example, a collection of subsets of the plane each homeomorphic to the circle), and having no points in common excpet one point common to all, then their union, a subspace of X, is easily confused with their wedge. There is an obvious bijection from the wedge to the union, and it is certainly continuous, but it is not necessarily a homeomorphism. In the case of the union of circles with radii tending to zero, it cannot be a homeomorphism; one reason is that the union is compact, unlike the wedge. In the case of the union of circles with radii tending to infinity, it cannot be a homeomorphism, for other reasons. We could also consider other examples, like another sequence of circles in the plane. But I guess these can all be distinguished from the wedge in a uniform way. To be continued ...
I didn't mean to imply that the subject is not interesting. It's merely that my interest flags when trying to piece things together out of the internet. I do appreciate your explanations.
So, now I've got the wedge topology straightened out. It does make a difference what imagination space one has, as I'll call it. When the imagination space is the plane, the induced topology on a point-set with a bijection to a wedge sum may be different from the topology defined abstractly on the wedge sum. In any case, the imagination space of an abstract topology is empty, if I may say so. It's still there as a sort of white background - at least in my mental visualization - but is not contributing to the topology. The white background is just a counterfoil for the point-set aspect.
Anyway, it's clear that the infinite wedge W of circles is not compact. So it's not homeomorphic to the bounded R^2 embedded version of it with the induced topology, because this is compact. Also, W is not homeomorphic to the earring E because their fundamental groups are different. But I still don't quite see why W is not homeomorphic to the EE, the expanding earring. You wrote:
A) It [EE] is not homeomorphic to the wedge -- briefly, because it has a compact subset not contained in the union of finitely many of the circles.
There's some argument hiding in this sentence that I'm just not getting. Does the "it" in "because it has a compact subset ..." refer to EE or to W ?? When you say "finitely many of the circles", so you mean "interiors of the circles" ? When one is considering compactness one considers open covers. The individual circles in W and EE are closed.
I see only that there are neighborhoods of the wedge point that look different from neighborhoods of (0,0) in E. But I don't see how to take advantage of this information to show that EE and W are not homeomorphic.
Assuming that by "it" in "because it has a compact subset ..." you mean EE, I have considered sequences resulting from intersections of the circles with y = 1 and x = y. I don't think these sequences are compact subsets of EE, though.
Would - at the very least - the wedge sum of circles with radii tending to infinity be homeomorphic to EE ?
Perhaps not. It seems to me that the wedge topology is somehow finer than that of EE - when you make the obvious point-set-wise comparison between them. Base neighborhoods of the wedge point are unions of mutually independent little open intervals in each circle that pass through the point. Whereas base neigborhoods of (0,0) in EE are induced by open disks, so the little intersections with each circle are not completely independent of each other.
I don't know how to make this vague observation more precise, or exploit it in some way.
I was stingy with explanations. Here's some more.
W is the wedge.
E is the Hawaiian earring (the union of a certain sequence of shrinking circles in the plane).
EE is the union of a certain expanding sequence of circles in the plane.
The "bounded R^2 version" that you mention: what is that? It sounds like you have in mind the union of an infinite sequence of circles in R^2 again, but this time with radii bounded below by some positive number and also the whole union being bounded. That will not be compact, because it cannot possibly be a closed subset of the plane. It might be homeomorphic to EE.
I think you have to let go of the idea of an "imagination space" for the infinite wedge of circles. (Visualization space? Is there a Ger. loc. going on here? Fantasieraum ? Vorstellungsraum ?) The wedge sum of infinitely many circles is not in fact homeomorphic to any subspace of ordinary n-dimensional space; in fact, it is not homeomorphic to any metric space (not metrizable, as they say)! To see this, recall that in a metric space every point has a so-called countable neighborhood base, for example the balls of radius 1/n centered at that point: a countable set of neighborhoods such that every neighborhood contains at least one of them. But in the wedge of an infinite sequence of circles the "wedge point" has no countable neighborhood base. In fact, let X_1, X_2, ... be any infinite sequence of based spaces each homeomorphic to a circle, and let X be their wedge. Then X is not metrizable. Here's why. Suppose for contradiction that U_1, U_2, ... is a neighborhood base for the wedge point in X. For each i, choose an open subset V_i of X_i that contains the basepoint and that is not contained in U_i. (This is possible because in a circle (in this case X_i) any given neighborhood (in this case the intersection of U_i with X_i) of a point must properly contain some other neighborhhod.) Let V be the union of all the V_i's, a subset of X. It is a neighborhood of the wedge point, by definition of the topology. There is no i such that V is contained in any U_i, because for every i V contains V_i which is not contained in U_i.
So the non-metrizably is a reason why the infinite wedge is different from all these other similar-looking things. But I had a different argument in mind when I wrote that EE has a compact subset not contained in the union of finitely many of the circles. The point is that in the infinite wedge has this striking property: every compact subset is contained in the union of finitely many of the circles. All of these examples in R^n can have that property, becaus you can choose a point p_i in the i-th circle that is within distance 1/n of the common point and then the set consisting of these points p_1, p_2, ... and the common point will be compact, being both closed and bounded.
Here's a proof that in the wedge of infinitely many spaces X_i every compact set S is contained in the union of finitely many X_i's. All I need to assume about the spaces to make the proof work is that points are closed. Make a set T by taking, for every i such that S contains a point of X_i different from the basepoint, one such point. I will show that T is closed. For this it suffices to check that for every i the intersection of T with X_i is closed. Since the intersection has at most one point in it, this is OK. Now, the same argument applies also to every subset of T. Therefore every subset of T is closed. This makes T a discrete space. But, being closed in the compact space S, it is also compact. Compact plus discrete implies finite. It follows that S had nontrivial intersection with only finitely many of the X_i's.
(I hope there are no bad typos here, but I have a stiff neck and I need a shower and I think it's time to get back to family and Christmas.)
Would - at the very least - the wedge sum of circles with radii tending to infinity be homeomorphic to EE ?
It matters not what circles you use. When you make the wedge of a bunch of spaces, what matters is only the topologies of those individual spaces.
Perhaps not. It seems to me that the wedge topology is somehow finer than that of EE - when you make the obvious point-set-wise comparison between them. Base neighborhoods of the wedge point are unions of mutually independent little open intervals in each circle that pass through the point. Whereas base neigborhoods of (0,0) in EE are induced by open disks, so the little intersections with each circle are not completely independent of each other.
You've got the idea!
Got it ! I can move my toes again, after being paralyzed by forgetfulness.
1) Let C be a compact subset C of W, with w the wedge point. C \ {w} intersects S^1 \ {w} for only a finite number of the S^1 in W.
Proof: Take a compact subset C of W. Consider the following open cover M of C: for each point p of C that is not w, take an open neighborhood of p disjoint from {w}. This is possible because W is Hausdorff. Add to M any open neighborhood Z of the wedge point w.
M has a finite subcover F. By construction, C \ {w} intersects S^1 \ {w} for only finitely many S^1 in W.
2) EE contains a compact subset D that intersects each of the expanding circles.
Proof: Such a D consists of the intersection points of the line y = x with EE.
3) EE is not homeomorphic to W.
Proof: If there were a homeomorphism f: EE -> W, then f(D) would be compact. By 1), f(D) \ {w} would intersect S^1 \ {w} for only a finite number of the S^1 in W. f(-1)f(D) \ {(0,0)} would intersect only a finite number of the S^1 \ {(0,0) of EE. But this contradicts the construction of D.
That is a good Czerny exercise to limber up the mentality. By comparing abstractly defined topologies with induced ones on embeddings, one gets a good feel for the kind of thinking that is needed.
My goal is to understand what cohomology is good for, if only in a very general way.
To be more precise, in the proof of 1) above:
for each point p of C that is not w, take an open neighborhood of p disjoint from {w} that lies entirely within the S^1 in which p lies.
There's something strange going on with the timing of these comments. When I added my "Got it !" comment A, I hadn't seen your "I was stingy with explanations. Here's some more" comment B. I may simply not have looked far enough up the thread. In any case, I got the notification of your comment B by mail an hour or two after I had posted my comment A. I was well pleased at the idea that I had got my proofs in before you explained everything. Oh well ...
Whoops, I wrote some nonsense. Here's the revised version of the proof that the wedge point in W has no countable neighborhood base:
Suppose for contradiction that U_1, U_2, ... is a neighborhood base for the wedge point in X. For each i, choose a neighborhood of the basepoint in X_i that does not contain the intersection of X_i with U_i. (This is possible because in a circle (in this case X_i) no neighborhood (in this case the intersection of U_i with X_i) of a point is contained in every neighborhhood.) Let V be the union of all the V_i's, a subset of X. It is a neighborhood of the wedge point, by definition of the topology. There is no i such that V contains U_i, because if V did contain U_i then V_i, the intersection of V with X_i, would contain the intersection of U_i with X_i.
Sorry to be a wet blanket, but ...
1) Let C be a compact subset C of W, with w the wedge point. C \ {w} intersects S^1 \ {w} for only a finite number of the S^1 in W.
Proof: Take a compact subset C of W. Consider the following open cover M of C: for each point p of C that is not w, take an open neighborhood of p disjoint from {w}. This is possible because W is Hausdorff. Add to M any open neighborhood Z of the wedge point w.
M has a finite subcover F. By construction, C \ {w} intersects S^1 \ {w} for only finitely many S^1 in W.
No. If the last open set that you threw into the cover was in F then it may contain points of C from infinitely many of the circles. As far as I can tell, this argument nowhere uses the fact that we are dealing with W rather than, say, E or EE.
2) EE contains a compact subset D that intersects each of the expanding circles.
Proof: Such a D consists of the intersection points of the line y = x with EE.
No, that set is not bounded. I was thinking of making a sequence of points where the nth point is on the nth circle and where the sequence converges to (0,0). The compact set is then the sequence plus the limit point. For example, the nth point could have y coordinate 1/n and x coordinate n minus the square root of (n^2 - 1/n^2). (It's easier to draw a picture.)
No, that set is not bounded. [referring to the intersection points of the line y = x with EE]
Of course I meant the bounded part of it near (0,0), say the unit square [(0,0),(1,0)] X [(0,0),(0,1)].
say the unit square ...
intersected by the unit square ...
But I need to choose a point on the nth circle that is approaching (0,0) as n approaches infinity. That line is not doing it for me. I thought we were talking about the circle with center (n,0) and radius n. This intersects that line in just two points: (0,0) and (n,n).
Will nothing work right for me anymore ?? Of course I first wanted to define points directly on the circles. But it seemed easier and shorter to write "intersection with y = x".
What a bummer. Ever and again, I am epsilon away from the right solution.
Part 1 of 2 (your software said my initial comment was too long)
I think you have to let go of the idea of an "imagination space" for the infinite wedge of circles. (Visualization space? Is there a Ger. loc. going on here? Fantasieraum ? Vorstellungsraum ?) The wedge sum of infinitely many circles is not in fact homeomorphic to any subspace of ordinary n-dimensional space;
It's not possible to "let go" of the idea of an imagination space, for any reason. Imagination space is a very abstract, schematic notion. It do not mean n-dimensional space, nor anything specific. In certain kinds of philosophy/epistemology, imagination space is called "unmarked space". Let me say a few general words about this, then go back to the topology context.
Unmarked space is a notion bracketing the specific thing or things you're considering at a given time, along with everything else that you're not considering. To talk about or imagine something, one says that one "introduces a mark that divides unmarked space". The "mark" is a "distinction which makes a difference" (Bateson), i.e. makes a difference as to what you are talking about or imagining.
Let's take an example. Imagine a shoe, and nothing else. However you consider the shoe, you consider it as something which is a shoe, and not anything else. You are not paying attention to this "anything else", but to the shoe. However, "anything else" is still there for you to switch your attention to, if you like. This has nothing to do with assuming or claiming the existence of anything, but rather with the way we thing about things.
Now back to topology. In abstract topology, we have the notions point-set, union, topology and so on. We distinguish these things from anything that is not a point-set, union, topology and so on. The other side of the distinction "abstract topology" is "nothing" - something we ignore entirely. The way I formulated this was:
In any case, the imagination space of an abstract topology is empty, if I may say so. It's still there as a sort of white background - at least in my mental visualization - but is not contributing to the topology. The white background is just a counterfoil for the point-set aspect.
You might say that the imagination space is what you imagine as containing point-sets, unions etc. - and nothing else.
Part 2 of 2 (your software said my initial comment was too long)
We can introduce a further distinction into abstract topology itself: the distinction "is embedded". This allows us to distinguish between "embedding space" and "embedded space". Both kinds of spaces are topological spaces, but each represents a kind of imagination space for the other. We can look first at one side of the distinction - embedded spaces, say - then the other side of the distinction, namely embedding spaces, and then go back to the embedded-space side to define "induced topologies".
For a long while in the above comment thread, I was not in my mind clearly distinguishing "abstract imagination space", where a wedge is defined (and the surrounding imagination space is "empty"), and "Euclidean imagination space" , where embedded topologies are defined, and the opposite side of the distinction containing the embedding topologies. In thinking about topologies, I kept unwittingly recurring to the point-set view of things (abstract topology) in contexts where I should have been paying more attention to the induced topolopy (Euclidean embedding spaces).
The notions of mapping, homeomophism etc. allows us to keep these two issues more clearly apart. In a certain sense, you can imagine abstract topology as being more "fundamental" than Euclidean topology. But I think it was just this failure to distinguish the two adequately that contributed to my confusion - writing things like "bounded R^2 version" of the wedge.
I expect what I have written in this comment may sound like gibberish to you. But there is a concrete application of these ideas to mathematical thinking which accords with something someone said in a Hat thread many months ago, and with which I'm sure you'll agree. This person wrote that it's better not to talk about a group G1 in some context being "the same as" a group G2 in another context. He preferred to say "they are isomorphic".
Even if this is gibberish to you, it has made it clear to me I need to distinguish abstract topology more sharply from specific topological settings - so that I can switch between them without getting confused.
It's not gibberish to me. I thought
Isomorphism is a useful idea, sure, and in topology a traditional synonym for isomorphism is homeomorphism.
Pedagogically, it's usually a good idea to introduce the concepts of topology in the context of Euclidean space first, and only later to introduce the abstract concept of topological space. Quotient spaces are hard to do without, though, and if you're going to treat them then you pretty much have to go all the way to points and open sets.
In a month I start teaching an introductory undergraduate topology course. I suppose you could say that I need to do some work on imagining what sorts of imagination spaces or blank slates my various students might bring to the experience.
It was probably misleading for me to talk about imagination space in the singular, and of imagination spaces being empty. There's not much you can say about unmarked space. But every time you make a fundamental distinction (say abstract topology versus Euclidean things), you are creating something to concentrate on while ignoring the rest. What you are ignoring is not really "empty", you're just not paying attention to it for a while. Imagination spaces are created by practice. You practice moving back and forth across lines of distinction.
Not until I was in the middle of your proof about compact subspaces of the wedge did it become clear to me that, until that point in time, I had not really, seriously applied my attention to visualizing (finding my own way of dealing with) the topology of the wedge, independently of Euclidean-embedding ideas. The explanation is, I think, that I am out of practice. I have never had problems with that kind of thing, and have dealt with much more complicated matters. Unfortunately, I see now that I was sliding over the issue, thinking "oh, I know all about that, I don't need to bother with the details".
Moore did not discourage visualization. He did insist, though, that proofs make no explicit appeal to visualization. That was his kind of abstraction, and of encouraging learning how to deal with it. But there are many kinds of ability, and many ways of imagining things, and many kinds of proof. Learning topology is like learning how to spell, or learning musical notation - in order to do something with this learning.
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