User david r. maciver - MathOverflow

Inequalities for uniformly convex normed spaces

2012-03-03T10:55:34Z

When reading "Chebyshev centers and uniform convexity" by Dan Amir I encountered the following result which is apparently "known and easy to prove". I'm sure it is, but I can't find a proof and am failing to prove it myself.

The result (slightly simplified) is

If $X$ is a uniformly convex space (i.e. if $||x|| = ||y|| = 1$ with $||x - y|| \geq \epsilon$ then there exists $\delta(\epsilon) > 0$ such that $||\frac{x + y}{2}|| \leq 1 - \delta(\epsilon)$) then for any $x, y$ with $||x|| \leq 1$ and $||y|| \leq 1$, and $||x - y|| \geq \epsilon$, $||\frac{x + y}{2}|| \leq 1 - \delta(\epsilon)$.

Part of the problem is that I think this isn't true without making some additional restrictions to reduce the value of $\delta(\epsilon)$. e.g. by considering $||x|| = 1$ and $y = (1 - \epsilon) x$ you can see that this requires that $\delta(\epsilon) \leq \frac{1}{2} \epsilon$. So I think the true result is probably just that you can choose $\delta$ so that this is true.

I'm sure this should be easy and I'm just missing an obvious trick, but oh well.

Probability estimates for pairwise majority votes

2010-05-27T16:40:41Z

I have items $I_1, \ldots, I_N$ and the observations of a number of pairwise trials which pit pairs $I_i$ and $I_j$ against eachother and select a "winner". Let $W_{ij}$ be the number of times i beats j.

Note that the number of trials between i and j is very much dependent on i and j: In some cases there may be none, in some there may be very many.

I am trying to estimate a matrix $P_{ij}$ corresponding to the probability that i beats j in a trial (I consider $P_{ii} = \frac{1}{2}$ for convenience reasons). My current, somewhat unprincipled, approach is a bayesian average

\[ P_{ij} = \frac{ \frac{1}{2} S + W_{ij} }{S + W_{ij} + W_{ji}} \]

where S is some smoothing constant (I currently have S = 5). This corresponds to a bayesian approach with a prior for $P_{ij}$ of $\beta(\frac{1}{2}S, \frac{1}{2}S)$ and then taking the expected value of the posterior distribution.

My problem with this is the following:

This is effectively treating each pair i, j as independent, whereas in fact we "believe" that there is a consistency between them. In particular if i tends to beat j and j tends to beat k, this should count as evidence that i tends to beat k even in the absence of pairwise trials between i and k.

There may be circumstances where we have very many trials for i, j and j, k and conclude that both $P_{ij}$ and $P_{jk}$ are high, but we have very few trials for i, k and thus conclude that $P_{ik}$ is very close to $\frac{1}{2}$ (possibly even concluding it's less than $\frac{1}{2}$ if e.g. there was only one trial and it had a surprising result). This is non-optimal.

So I'd like some sort of reasonably principled way of introducing intermediate results as evidence that the majority prefers one to the other. There are various plausible sounding things I could try, but I'd like to do this "properly" if at all possible, and most of my ideas involve more hand waving than solid mathematics.

One example of something plausible but possibly nonsensical I'm considering trying is iterating an expand/collapse process of:

Expand: $P \to P^2$

Collapse: $P \to Q$, where $Q_{ij} = \frac{P_{ij}}{P_{ij} + P_{ji}}$

The idea being that we inflate probabilities where there are a lot of large intermediate results and then collapse down to the symmetry condition that $P_{ij} + P_{ji} = 1$ .

This seems to produce semi-tolerable results (I've not tested extensively yet), but it's not actually clear to me that this process converges or why it should work.

Suggestions?

Edit:

On having thought about this a little more carefully, I think the following may capture what I am trying to achieve:

I want to assume that there is some distribution on the permutations of 1..N, with a strong prior belief that this distribution is close to uniform, and that each pairwise trial consists of sampling from this distribution and comparing the positions of i and j.

Distribution on permutations derived from probability of pairwise orderings

2010-06-06T15:13:31Z

A followup question to Probability estimates for pairwise majority votes - I think it doesn't actually give an answer in any terribly precise sense, but it would give something I'd be happy to use in lieu of one. :-)

I'm basically looking at a certain class of distribution of permutations and trying to determine the probability of it putting two items in a given order. I suspect I'm treading on well worn ground here, but haven't been able to find anything.

P is a $N \times N$ matrix with P > 0 and $P_{ij} = 1-P_{ji}$ . Define a random variable $T$ taking values in $S_N$ (the permutations of $1, \ldots, N$) by

$P(T = \sigma) \propto \prod_{\sigma(i) < \sigma(j)} P_{ij}$

For fixed i, j I'd like to calculate $P(T(i) < T(j))$.

It's clear that this can't simply be $P_{ij}$ : If you have e.g. $P_{12} = P_{23} = P_{31} = 0.9$ then $P(T(1) < T(2)) = 0.5$.

Unfortunately it's not clear to me what a general solution should look like. I suspect there may be no nice closed form solution, so I'd be happy with a reasonably efficient way to calculate a numeric approximation.

One thing worth noting is that if we let $R_{ij} = P(T(i) < T(j))$ then for all k we have the constraint

$R_{ik} \geq R_{ij} + R_{jk} - 1$

I suspect but haven't yet been able to prove that if P satisfies this constraint then P = R. If this is the case then it seems likely that R can be calculated as a solution to these constraints (plus that $R_{ij} = 1 - R_{ji}$ ) which minimises some distance function from P.

How to assign a score to items based on a set of partial rankings

2010-05-08T15:22:48Z

I have the following setup:

There is a collection of items I and a collection of partial rankings V. That is, an element of V is a total ordering on a subset of I. There is no expectation of consistency among the elements of V: it may be that x < y for one element and y < x for another.

I would like to assign a score $s : I \to \mathbb{R}$ which in some sense captures these rankings. That is, I would like s(x) < s(y) to mean "x tends to be less than y for elements of V which have both in their domain". I'm not sure of what a good way to do this is.

Arrow's impossibility theorem puts some constraints on what can be achieved here, because given a set of votes and a scoring function like this we could use the scoring function to define a total order on the items, which is then constrained by the theorem.

I suppose I'm really looking for references rather than an answer to this question (although both would be appreciated): I'm sure there's a body of theory around this, but I have no idea what it is like or what it's called, so I'm at a bit of a loss as to where to start looking for a solution.

Conditions useful for proving paracompactness

2010-04-20T07:14:09Z

I have a family of properties which I want to show taken together imply paracompactness (I can show that they are all implied by paracompactness). I can prove a whole bunch of things which are consequences of paracompactness, and was wondering if given any of these there's some simpler property equivalent to paracompactness I could prove. Any suggestions would be welcome.

Given X with these properties I can prove:

X is normal
X is countably paracompact
X is collectionwise normal
Every open cover $\{ U_a \}$ can be shrunk to a closed cover $\{ F_a \}$ with $F_a \subseteq U_a$ . (I assume this property isn't equivalent to paracompactness? I know it's equivalent to countable paracompactness when the set of $U_a$ is countable, and I know if you add "locally finite" to the condition it becomes equivalent to paracompactness)
Every open cover of X by $\kappa$ many open sets, where $\kappa$ is regular, has an open refinement which is locally $< \kappa$.

I don't think together these are sufficient to prove paracompactness, though I don't have a counter example. I believe $\omega_1$ satisfies all the properties but the last, though I've not confirmed you can shrink open covers to closed (it looks plausible though).

Any suggestions of avenues to pursue?

Answer by David R. MacIver for the delta system lemma outside set theory

2010-04-15T09:41:24Z

There's a nice application of the lemma to point-set topology (whether or not you consider this to be set theoretic is up to you :-) )

Define the following generalisation of separability:

A topological space X is ccc if every pairwise disjoint collection of non-empty open sets is countable.

It's clear that every separable space is ccc (because you can pick an element of the dense set in each open set), but ccc has the following theorem which makes it better behaved on products (a product of $> 2^{\aleph_0}$ spaces with more than one point is not separable):

Theorem: Let $\{ F_a : a \in A \}$ be a family of topological spaces such that every finite product is ccc. Then $\prod F_a$ is ccc.

Proof:

Let T be an uncountable family of pairwise disjoint non-empty open sets. We can without loss of generality assume T consists of basis elements. Each of these basis elements is a product of sets of the form $U_a \subseteq F_a$ with at most finitely many not equal to $F_a$.

Let $G = \{ \{ a : U_a \neq F_a \} : \prod U_a \in T \}$ . This is an uncountable collection of finite sets, so contains a delta system, say with root R.

But we must have the projection of $T$ to $\prod_{a \in R} F_a$ still be disjoint: For any $a \not\in R$ we have $U_a \neq F_a$ for at most one element of T. But products of finitely many $T_a$ are ccc, thus we must have the projection of T (and thus T itself) being countable, contradicting the hypothesis.

QED

This in particular gives us the following corollary:

Theorem: An arbitrary product of separables spaces is ccc.

Proof: A finite product of spearable spaces is separable and thus ccc.

which is interesting because e.g. it tells us that there are compact hausdorff spaces which are not the continuous image of $\{0, 1\}^\kappa$ for any $\kappa$.

Answer by David R. MacIver for Shape of long sequences in C(ω_1)

2010-04-12T09:17:50Z

Edit: This proof is wrong, but preserving for posterity.

Oh! In a very wizard of oz manner, the answer was within my power all along (it's within $\epsilon$ of a proof I'd already written for something else).

Here's a cute little proof that C(K) has the property that diam(A) = 2 * r(A), and thus has the chain-radius condition and thus has no bad sequences.

Let $A \subseteq C(K)$ be non-empty. Define

$ g(x) = \sup_{f \in A} f(x)$

$ h(x) = \inf_{f \in A} f(x)$

$g$ is upper semicontinuous: $g(x) > a$ iff there exists $f \in A$ such that $f(x) > a$. Similarly $h$ is lower semicontinuous.

Further, $g(x) - h(x) \leq \textrm{diam}(A)$ .

Therefore $g(x) - \frac{1}{2}\textrm{diam}(A) \leq h(x) + \frac{1}{2}\textrm{diam}(A)$

But now we have an upper semicontinuous function which is $\leq$ a lower semicontinuous function. Thus by the katetov tong insertion theorem there is a continuous function $f$ with

$g(x) - \frac{1}{2}\textrm{diam}(A) \leq f \leq h + \frac{1}{2}\textrm{diam}(A)$

But this means that $A \subseteq B(f, \frac{1}{2}\textrm{diam}(A))$. Therefore $r(A) \leq \frac{1}{2}\textrm{diam}(A)$.

But we already know that $r(A) \geq \frac{1}{2}\textrm{diam}(A)$, so the two are equal and the result is proved.

Shape of long sequences in C(ω_1)

2010-04-11T19:21:00Z

Apologies for the vague title - I couldn't come up with a single sentence that summarised this problem well. If you can, please edit or suggest a better one!

This question is also rather specific and contains lots of annoying technical detail. I must admit to not really expecting an answer unless there's an obvious solution I'm missing (which is very possible - I feel like any solution is either going to be obvious or very deep), but some pointers in plausible sounding directions would be greatly appreciated. I suspect the answer will depend on the combinatorics of $\omega_1$, which I know relatively little about.

Let $V$ be a normed space. For $A \subseteq V$, define $r(A) = \inf \{ r : \exists V, A \subseteq B(v, r) \}$ . Define a bad sequence in $V$ to be a sequence ${ v_\alpha : \alpha < \omega_1 }$ with the properties that:

$\forall \beta, r(\{ v_\alpha : \alpha < \beta \}) \leq 1$

$\inf_\beta r(\{ v_\alpha : \alpha \geq \beta \}) > 1$

An example of a space with a bad sequence is $c_0(\omega_1)$ (the set of all bounded real-valued sequences of length $\omega_1$ such that $\{ \alpha : |x_\alpha| > 0 \}$ is countable). The sequence $2 * 1_{\{\alpha\}}$ is bad. The radius of any tail is $2$ because the center must be eventually 0. The radius of the initial segments is $\leq 1$ because the segment up to $\alpha$ is contained in the closed ball of radius 1 around $1_{[0, \alpha]}$ , which is in $c_0(\omega_1)$ because $\alpha < \omega_1$.

I have two (three depending on how you count it) major examples of spaces which have no bad sequences:

Any separable space: you can choose centers to lie in the countable dense set, so one center must work as a radius for the initial segment for unboundedly many and thus for all $\alpha$.
Any space which has what I'm imaginatively calling the chain-radius condition: The union of a chain of sets of radius $\leq r$ has radius $\leq r$. This includes:
- Any reflexive space: If $U_\alpha$ forms a chain, the sets $F_\alpha = \bigcap_{v \in U_\alpha} \overline{B}(v, r + \epsilon)$ form non-empty closed and bounded convex sets with the finite intersection property, so compactness in the weak topology implies they have non-empty intersection. Any element of the intersection contains the union of the chain in $\overline{B}(c, r + \epsilon)$
- any space with the property that $\textrm{diam}(A) = 2 r(A)$ (in particular the $l^\infty$ space on any set) because it's clear that unions of chains of diameter $\leq 2r$ have diameter $\leq 2r$.

So... that's all the backstory for this question. Given that, my actual question is very simple: Does $C(\omega_1)$ contain a bad sequence?

I feel like the answer "must" be no. In particular note that the projection of any sequence onto the first $\alpha$ entries is not bad (because it's a sequence in a separable space) and that if you drop the restriction for continuity the answer is immediately yes. So it sits right between two classes of examples where there are no bad sequences, and I feel that one really should be able to take advantage of that. But on the other hand, functions in $C(\omega_1)$ are eventually constant, so maybe you can take advantage of that to construct some sets with arbitrary bad tails.

For bonus kudos, I'd love to know for what compact Hausdorff spaces $K$, $C(K)$ contains a bad sequence.

Is there a "natural" characterization of when X × βN is normal?

2010-04-08T09:42:24Z

As per a recent question of mine, $\omega_1 \times \beta \mathbb{N}$ is not normal. I'm wondering whether there's some sort of "natural" condition that describes when a space has a normal product with $\beta \mathbb{N}$, analagous to Dowker's characterisation that $X$ is countably paracompact iff $X \times [0, 1]$ is normal.

The context here is that I'm looking for something I can weaken to something sensible, as I have a property implied by $X \times \beta \mathbb{N}$ being normal which I am "surprised" isn't an equivalence ($\omega_1$ has this property) and would like to see if I can show its equivalenct to some slightly weaker topological condition.

Is ω1 × βN normal?

2010-04-03T11:14:01Z

Once upon a time I asked whether $\omega_1 \times \beta \mathbb{N}$ is normal. I got the answer no and a fairly convincing proof of this here

However I'm currently in a situation where I have three plausible proofs of plausible results at most two of which can be true, and of the three this is the one of which I'm currently the least sure (there are a bunch of details I haven't yet fully checked as I'm in the process of dusting off some of my recently rather unused knowledge about general topology), so was hoping someone could confirm.

Answer by David R. MacIver for Polish spaces in probability

2010-04-10T19:39:18Z

One simple thing that can go wrong is purely related to the size of the space (polish spaces are all size $\leq 2^{\aleph_0}$). When spaces are large enough product measures become surprisingly badly behaved. Consider Nedoma's pathology: Let $X$ be a measure space with $|X| > 2^{\aleph_0}$. The diagonal in $X^2$ is not measurable.

We'll prove this by way of a theorem:

Let $U \subseteq X^2$ be measurable. $U$ can be written as a union of at most $2^{\aleph_0}$ spaces of the form $A \times B$.

Proof: First note that we can find some countable collection $A_i$ such that $U \subseteq \sigma(A_i \times A_j)$ (proof: The set of $V$ such that we can find such $A_i$ is a sigma algebra containing the basis sets).

For $x \in \{0, 1\}^\mathbb{N}$ define $B_x = \bigcap \{ A_i : x_i = 1 \} \cap \bigcap \{ A_i^c : x_i = 0 \}$ .

Consider all sets which can be written as a (possibly uncountable) union of $B_x \times B_y$ for some $y$. This is a sigma algebra and obviously contains all the $A_i \times A_j$, so contains $A$.

But now we're done. There are at most $2^{\aleph_0}$ of the $B_x$, and each is certainly measurable in $X$, so $U$ can be written as a union of $2^{\aleph_0}$ sets of the form $A \times B$.

QED

Corollary: The diagonal is not measurable.

Evidently the diagonal cannot be written as a union of at most $2^{\aleph_0}$ rectangles, as they would all have to be single points, and the diagonal has size $|X| > 2^{\aleph_0}$.

Radii and centers in Banach spaces

2010-04-07T09:32:44Z

Suppose I have a Banach space $V$ and a set $A \subseteq V$ such that for all $\epsilon > 0$ there exists $v$ such that $A \subseteq \overline{B}(v, r + \epsilon)$. Does there exist $c$ such that $A \subseteq \overline{B}(c, r)$?

The answer is clearly yes for finite dimensional normed spaces: Define $T_\epsilon = \bigcap_{a \in A} \overline{B}(a, r + \epsilon)$. The $T_\epsilon$ form a chain of closed sets and for $\epsilon > 0$ are non-empty, so have the finite intersection property. Thus when $V$ is finite dimensional they have non-empty intersection, and any element of the intersection works as $c$.

For more general Banach spaces I feel like you should be able to choose a cauchy sequence $x_n$ such that $x_n \in T_{\epsilon_n}$ with $\epsilon_n \to 0$, but I can't seem to make it work.

Note that an arbitrary choice of $x_n \in T_{\epsilon_n}$ can't be guaranteed to be Cauchy: If $V$ is $l^\infty$ and $A = { x : x_0 = 0, ||x|| \leq 1 }$ then diam$(T_\epsilon) \geq 2$ because you can choose $c_0$ arbitrarily in $[-1, 1]$

Note also that the assumption of $V$ a Banach space is essential: If $V$ is not Banach and $c$ is an element of the completion which is not in $V$ then $A = \overline{B}(c, 1) \cap V$ has no center.

Approximate selection theorems for factoring through perfect maps

2010-04-07T08:20:45Z

I have the following setup:

$X, Y$ are topological spaces (if it helps, they can both be $T_1$ and normal. They can even be countably paracompact. They can't be assumed paracompact). $V$ is a normed space (it can be Banach if you like). $f : X \to Y$ is a perfect surjection.

I have continuous and bounded $g : X \to V$ and given $\epsilon > 0$ would like to find continuous $h : Y \to V$ such that $d(h(x), g(f^{-1}(x))) < \epsilon$

Is there some sort of selection theorem that will let me do this? I've used the Michael selection theorem to good effect elsewhere, but it doesn't apply here due to the lack of convexity of the target sets (even if they were convex the hypotheses don't apply due to potential non-paracompactness of Y, but one might be able to work something out using countable paracompactness and compactness of the targets).

$2^{\omega_1}$ separable?

2010-04-03T10:41:06Z

I was rereading an answer to an old question of mine and it included a reference to the fact that $2^{\omega_1}$ was separable. I'm having a hard time finding a reference for this fact, and the proof is not immediately obvious to me. Can anyone provide me with a cite and/or a proof?

Answer by David R. MacIver for Paracompact but not Hausdorff

2010-04-03T12:15:06Z

It's worth noting that any $T_1$ space which admits partitions of unity for finite (two element even) covers is Hausdorff:

Proof: Let $x, y \in X$. Let $U = X \ {x}, V = X \ {y}$. Then let ${f, g}$ form a partition of unity with $f$ subordinate to $U$ and $g$ subordinate to $V$. Then $A = { t : f(t) > \frac{1}{2} }$ and $B = { t : g(t) > \frac{1}{2} }$ A and B are disjoint open sets with $y \in A$ and $x \in B$.

Edit: On closer inspection, this if of course just the standard proofs that the existence of partitions of unity for finite covers implies normality + the fact that $T_1$ normal spaces are hausdorff

Answer by David R. MacIver for $2^{\omega_1}$ separable?

2010-04-03T11:38:52Z

Should have searched a bit harder before asking this one. This is an immediate consequence of the Hewitt-Marczewski-Pondiczery theorem:

Let $m \geq \aleph_0$. If ${X_s : s \in S}$ are topological spaces with $d(X_s) \leq m$ and $|S| \leq 2^m$ then $d(\prod_s X_s) \leq m$.

Countable paracompactness, normality and locally countable open covers

2010-03-29T23:40:05Z

(repost from the topology Q&A board)

I have a (T_1), Normal, countably paracompact space X. I would like to know if every locally countable open cover of X (i.e. an open cover such that every x in X has a neighbourhood which intersects only countably many members of the cover) has a locally finite refinement.

My suspicion is that the answer is a resounding no, but every time I try to construct a counterexample it starts to seem more plausible.

If the answer does turn out to be yes I'd love to know if it generalises from aleph_0 to arbitrary cardinals.

Comment by David R. MacIver

2012-03-04T11:11:31Z

Thanks. This was a useful reference.

Comment by David R. MacIver

2012-03-04T11:06:00Z

Thanks for the proofs. I must confess I don't understand the first one at all, but that's because I hadn't even heard of the Banach Mazur compaction until now. (this isn't really an area that I know that much about, though I'd like to fix that)

Comment by David R. MacIver

2012-03-04T11:01:12Z

Beats me. I've never entirely understood the role of community wiki.

Comment by David R. MacIver

2012-03-03T20:27:48Z

Sorry, I missed the $||x - y|| \geq \epsilon$ criterion in the claim (without which it makes no sense because $\epsilon$ didn't appear in it at all). Edited to fix now. Basically the claim is that you can relax the requirement in the definition of uniform convexity that the norm is exactly 1 to that it is $\leq 1$

Comment by David R. MacIver

2010-06-06T21:56:39Z

I suspect you may be right that the problem is intractable in general. Unfortunately I don't think there's likely to be much deeper structure in the problem unfortunately. However the chances are good that the original P may be close to R, as they're derived from data which is likely to be "close" to being modelled by the described process. So if there's some iterative process which converges on the result perhaps it would converge quickly in my observed case even if it has bad general performance?

Comment by David R. MacIver

2010-05-29T14:44:40Z

The pairwise preferences genuinely are inconsistent. I'm aware of this and already deal with it elsewhere, so I'm not interested in trying to deal with it at this point, only in trying to find a better way of estimating the "true" preferences.

Comment by David R. MacIver

2010-05-28T09:56:48Z

I'm not convinced by this solution for a couple reasons: * It doesn't appear to admit a nice solution (you end up with a very overconstrained and slightly thorny equation in the matrix $M_ij = \frac{s_i}{s_i + s_j}$ ) * It doesn't handle the case where there genuinely are inconsistent results (it's certainly possible for majorities to prefer A to B to C to A) * If it were computationally feasible to compute the MLE then rank aggregation would be essentially trivial, which it isn't.

Comment by David R. MacIver

2010-05-27T21:05:33Z

Hmm. That might work. It's not obvious how to assign a useful prior distribution to $S_i$ though, or am I missing something?.

Comment by David R. MacIver

2010-05-27T17:30:42Z

I think you may have misunderstood the question, as that is effectively what I'm already doing (I'm taking the expected value of the posterior, which is cheating but useful). The question is about how to introduce the fact that the pairs are not independent and that we expect there to be a certain amount of consistency between them as evidence.

Comment by David R. MacIver

2010-05-08T15:42:56Z

Ah, thanks. I looked but failed to find that. Possibly worth closing my question as a duplicate then.

Comment by David R. MacIver

2010-05-08T15:41:29Z

Unfortunately the rankings are not likely to be of the same size - they're probably all going to be of similar size (and small compared to the size of I) though.

Comment by David R. MacIver

2010-04-22T08:28:40Z

Is it not the case that $\aleph_1 \leq 2^{\aleph_0}$ without the axiom of the choice? I think the following should work: By constructing with transfinite induction we can find $g_\alpha : [0, \alpha) \to \mathbb{Q}$ an injection (using the standard argument that countable ordinals embed in the rationals), and so $f_\alpha : \alpha \to \omega$ a bijection. We now define `$f : \omega_1 \to \{0, 1\}^\omega$ inductively as a diagonalisation of $f(\beta)$ for $\beta < \alpha$, which we can do without AC because we can explicitly count $\alpha$

Comment by David R. MacIver

2010-04-21T08:26:29Z

To elaborate, the proof of shrinkability is the one I think most likely to be extendable. The construction is clearly providing more than I'm using in concluding shrinkability (because I can demonstrate that it also rules out $\omega_1$, which is shrinkable), but I can't get a handle on how much more.

Comment by David R. MacIver

2010-04-21T07:40:38Z

Well, first off there's no way I can prove my space is not paracompact because it's not a single space, and I know all paracompact spaces qualify. :-) But additionally the problem is that (as mentioned elsewhere) it's a parameterised property. The choice of parameter for the case locally < k isn't usefully shrinkable.

Comment by David R. MacIver

2010-04-20T22:11:41Z

(that is, it works for showing that it's locally < k. And locally < $\omega$ is the same as locally finite, so it implies countable paracompactness)