surely this was an accidental newline ?

rachel lambda samuelsson

]]>added pointer to:

- Nicolas Bourbaki,
*Topological Groups*, Chapter 3 in:*General topology*, Elements of Mathematics III, Springer (1971, 1990, 1995) [doi:10.1007/978-3-642-61701-0]

added DOI to

- Alexander Arhangel’skii, Mikhail Tkachenko,
*Topological Groups and Related Structures*, Atlantis Press 2008 (doi:10.2991/978-94-91216-35-0)

added the remark that some authors understand topological groups to be Hausdorff (e.g. Bredon 72)

]]>added pointer to:

- Glen Bredon, Chapter 0 of:
*Introduction to compact transformation groups*, Academic Press 1972 (ISBN 9780080873596, pdf)

I very much agree with Richard here.

]]>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.

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

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

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

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

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

]]>Hi Richard,

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

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

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

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

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

The closedness of $\mathcal{W}$ is used only in deducing that $A'$ is open (proving that $A'$ is open is quite a fun little argument). The local finiteness of $\mathcal{W}$ 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 $\mathcal{A}$ 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 $x \in X$, the open set $U_{x}$ which the local finiteness of $\mathcal{A}$ provides us with.

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

]]>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.

]]>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.

]]>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 $\mathcal{W}$ 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.

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

]]>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 $\mathcal{W}$ intersects only finitely many elements of $\mathcal{V}$, $\mathcal{V}$ is locally finite.

But $\mathcal{W}$ is a *closed* cover. This only implies that $\mathcal{V}$ is point-finite, i.e. that every point is contained in a finite number of the elements of $\mathcal{V}$, 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 $\mathcal{V}$, for that to be true we would need that $\mathcal{W}$ 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?

]]>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.

]]>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.

]]>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?

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

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

]]>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 $Top$. 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.

]]>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?

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