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    • CommentRowNumber1.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 21st 2010

    The entry on topological group could stand more work, but I added some stuff on the uniform structure, in particular the proposition that for group homomorphisms f:GHf: G \to H, continuity at a single point guarantees uniform continuity over all of GG. The proof is follow-your-nose, of course.

    What we really need is an entry Haar measure. I’ll get started on that soon.

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeJun 13th 2013

    added note at topological group – Protomodularity

    Also created TopGrp and linked to it from protomodular category (which is mainly empty!)

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeMay 22nd 2017
    • (edited May 22nd 2017)

    added to topological group statement and proof that locally compact topological groups are paracompact (here)

    • CommentRowNumber4.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 22nd 2017

    In the first statement you have (connected) locally compact, and then in the second you have just locally connected. Do you want locally connected locally compact?

    I made an edit to indicate how local connectedness does get used at the end, to ensure that the connected component of the identity is an open subgroup (whereupon the group is a coproduct of the cosets of that connected component). I’m not sure how you’d argue that without local connectedness though.

    • CommentRowNumber5.
    • CommentAuthorUrs
    • CommentTimeMay 22nd 2017

    Thanks, Todd, for the fixes.

    I was about to do some of them myself, but my internet connection keeps going down, and then you had already locked the entry. (At least that worked tis time! :-)

    • CommentRowNumber6.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 22nd 2017

    I was being dumb before. Local connectedness is not needed.

    I’ve therefore done some editing. I hope the (somewhat significant) edit doesn’t come across as rude, but it seems to me that the argument becomes less wordy if you make the two statements of the Theorem the punch lines at the very end of the proof. See what you think.

    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTimeMay 22nd 2017

    Ah, right, thanks. Looks good, thanks.

    • CommentRowNumber8.
    • CommentAuthorUrs
    • CommentTimeMay 22nd 2017

    As you will have deduced from my activity, I was trying to collect useful classes of “concrete” examples of paracompact spaces, e.g. CW-complexes, locally compact topological groups. What else to include?

    • CommentRowNumber9.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 22nd 2017
    • (edited May 22nd 2017)

    Metrizable spaces, of course (and I see that’s in the list of examples at paracompact space). It looks like this covers the major examples that most people talk about.

    Point-set topologists will probably remind us that regular Lindelöf spaces are paracompact, but I couldn’t even tell you the definition of Lindelöf space without looking it up. So I have no feeling for those.

    What I might like to know is what sorts of categorical operations paracompact spaces are closed under. We’ve got coproducts in TopTop. Equalizers of maps between paracompact Hausdorff spaces: yes. Countable products? (Uncountable products: no.) What sorts of colimits? I haven’t seen this sort of thing discussed.

    Reflection? Coreflection? etc.

    • CommentRowNumber10.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 22nd 2017

    Well, apparently even finite products in TopTop is a ’no’. The Sorgenfrey plane which is a product of two Sorgenfrey lines is not paracompact, so I am told.

    • CommentRowNumber11.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 22nd 2017

    One thing led to another, and I created Sorgenfrey line (to which Sorgenfrey plane redirects).

    • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeMay 22nd 2017

    Metrizable spaces, of course

    Right, so that’s why I said “classes of concrete examples”, thinking of something that one is going to test in practice. From my experience we regularly check for instance if something admits the structure of a CW-complex or if something is a locally compact topological group. But metrizability?

    • CommentRowNumber13.
    • CommentAuthorDavidRoberts
    • CommentTimeMay 22nd 2017
    • (edited May 22nd 2017)

    Smooth mapping spaces are infinite-dimensional and paracompact, which was to me slightly counter-intuitive.

    Even just Fréchet spaces in general, which are examples of metrizable spaces that don’t always come with a canonical metric - one can of course build one from the family of seminorms, but it can be slightly arbitrary.

    • CommentRowNumber14.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 22nd 2017

    But metrizability?

    Why on earth not? This is one of the best studied of all classes of spaces, and people have expended a lot of effort exploring necessary and sufficient conditions (i.e., tests) for metrizability.

    David actually raises a good point. For example, it’s not always completely obvious when say a LF-space is metrizable.

    Put differently, it seems to me believable that people “check” that something is metrizable about as frequently as they “check” that something is a locally compact topological group.

    • CommentRowNumber15.
    • CommentAuthorUrs
    • CommentTimeMay 23rd 2017
    • (edited May 23rd 2017)

    Thanks. Please bear with my ignorance.

    Now I remember why I interrupted writing out a proof that metric spaces are paracompact: the proof I wanted to do used Michael’s theorem, and I am still stuck on the proof of that. On p. 3 of Michael’s article (pdf) he says

    and since each element of the locally finite covering 𝒲\mathcal{W} intersects only finitely many elements of 𝒱\mathcal{V}, 𝒱\mathcal{V} is locally finite.

    But 𝒲\mathcal{W} is a closed cover. This only implies that 𝒱\mathcal{V} is point-finite, i.e. that every point is contained in a finite number of the elements of 𝒱\mathcal{V}, but it does not in itself imply that there is an open neighbourhood of every point which intersects only a finite number of the elements of 𝒱\mathcal{V}, for that to be true we would need that 𝒲\mathcal{W} has a refinement by an open cover.

    I suppose I must be making a fool of myself here. So be it. What am I missing?

    • CommentRowNumber16.
    • CommentAuthorDavidRoberts
    • CommentTimeMay 23rd 2017

    Is 𝒲\mathcal{W} given by closure of an open cover? Or more generally do the interiors of the elements still cover?

    • CommentRowNumber17.
    • CommentAuthorUrs
    • CommentTimeMay 23rd 2017
    • (edited May 23rd 2017)

    do the interiors of the elements still cover?

    That had been my first thought, too.

    But check that single paragraph labeled “(c) \to (a)” on p. 3 of that short pdf.

    The closed cover 𝒲\mathcal{W} is not constructed in any way that would make it inherit tacit extra properties. Instead it is taken to exist by the assumption (labeled “(c)”) that every open cover has a closed locally finite refinement.

    Next I thought that Michael is using terminology in a (nowadays, maybe) non-standard way, that maybe “closed cover” means “cover by closed neighbourhoods” for him. But on p. 2 he states his definitions in the paragraph starting with “Let us quickly recall the definitions” and from that a closed cover is nothing but a cover by closed sets.

    Or else I am missing something.

    • CommentRowNumber18.
    • CommentAuthorUrs
    • CommentTimeMay 23rd 2017

    On the other hand, the implication “(b) \to (c)” does produce a closed cover whose interiors still cover. So maybe if we rephrase item (c) appropriately, then the issue goes away.

    • CommentRowNumber19.
    • CommentAuthorUrs
    • CommentTimeMay 23rd 2017
    • (edited May 23rd 2017)

    the implication “(b) \to (c)” does produce a closed cover whose interiors still cover.

    Sorry, no, it does not. It produces a cover whose patches are closures of patches of another cover, but that other cover is by any subsets, not necessarily open.

    • CommentRowNumber20.
    • CommentAuthorDavidRoberts
    • CommentTimeMay 23rd 2017

    Hmm, ok. I haven’t got anything else right now, sorry.

  1. Hi Urs, I took a look at the proof of c) => a). I believe that it works; I will try to explain it.

    The key idea is the consideration of the covering \mathcal{B}. Note that the defining property of this covering is that every set in it intersects only finitely many sets in 𝒜\mathcal{A}. Of course we need to convince ourselves that such a covering exists, but let’s assume that for the moment. Now, 𝒲\mathcal{W} is by definition a refinement of \mathcal{B}, so it inherits the same property: every set in 𝒲\mathcal{W} intersects only finitely many sets in 𝒜\mathcal{A}.

    It now follows that a set in 𝒲\mathcal{W} also intersects only finitely many sets of the form AA' for AA in 𝒜\mathcal{A}, because even though AA' is larger than AA, it has, by construction, the property (as is explicitly pointed out) that a set WW in 𝒲\mathcal{W} intersects AA' if and only if it intersects AA.

    It follows immediately that a set WW in 𝒲\mathcal{W} also intersects only finitely many sets of the form AR AA' \cap R_{A} for AA in 𝒜\mathcal{A}, because AR AA' \cap R_{A} is smaller than AA', so WW can intersect at most as many sets of this form as of the form AA', and, as we have observed, there are at most finitely many such sets.

    The closedness of 𝒲\mathcal{W} is used only in deducing that AA' is open (proving that AA' is open is quite a fun little argument). The local finiteness of 𝒲\mathcal{W} is used only at the very end of the proof.

    It remains to justify that a covering with the defining property of \mathcal{B} can be found. To obtain the covering 𝒜\mathcal{A} from \mathcal{R}, the assumption of c) is being used (so this assumption is used twice in the proof). To get \mathcal{B}, we can just pick, for every xXx \in X, the open set U xU_{x} which the local finiteness of 𝒜\mathcal{A} provides us with.

    • CommentRowNumber22.
    • CommentAuthorUrs
    • CommentTimeMay 24th 2017

    Hi Richard,

    the issue that I am referring to is not the local finiteness of 𝒲\mathcal{W}, that is clear, but the local finiteness of 𝒱\mathcal{V}, that last step in the proof for (c) \to (a)

    We want to check that an open cover 𝒱\mathcal{V} is locally finite. This means equivalently that there is another open cover all whose elements intersect only finitely many of the elements in 𝒱\mathcal{V}. Michael’s proof suggests to use 𝒲\mathcal{W} as this other cover. But 𝒲\mathcal{W} is not an open cover. So why does it follows that 𝒱\mathcal{V} is locally finite?

    • CommentRowNumber23.
    • CommentAuthorRichard Williamson
    • CommentTimeMay 24th 2017
    • (edited May 24th 2017)

    Hi Urs, what I explained was a proof of the following claim.

    each element of … 𝒲\mathcal{W} intersects only finitely many elements of 𝒰\mathcal{U}

    From this and the fact that 𝒲\mathcal{W} is locally finite, it is immediate that 𝒰\mathcal{U} is locally finite. Indeed, take an xx in XX. Because 𝒲\mathcal{W} is locally finite, there is an open neighbourhood U xU_{x} of xx which intersects only finitely many sets in 𝒲\mathcal{W}. Now, by the claim that I quoted and explained a proof of, each of these sets intersects only finitely many sets in 𝒰\mathcal{U}. So U xU_{x} intersects only a finite union of finitely many sets in 𝒰\mathcal{U}, hence it intersects only finitely many sets in 𝒰\mathcal{U}.

    So 𝒰\mathcal{U} is locally finite, and it is certainly open and a refinement of \mathcal{R}, so we are done.

    • CommentRowNumber24.
    • CommentAuthorUrs
    • CommentTimeMay 24th 2017

    Oh, I see. All right, so I was indeed making a fool of myself. Thanks.

  2. On the contrary, I wish people would ask these kinds of question more often. I enjoyed thinking about it, and hopefully you were able to make some progress a little quicker than might otherwise have been the case. Everyone gets stuck in this way often; having the honesty to acknowledge this in particular cases is very fruitful, I feel.

    • CommentRowNumber26.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 24th 2017

    I very much agree with Richard here.

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