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Consider the set $E=\mathbb{Q}\cap[0,1]$, and let $\{q_{j}\}_{j=1}^{\infty}$ be some enumeration of this countable set. For every $\epsilon>0$, the cubes $\{Q_{j}\}_{j=1}^{\infty}$ of length $\ell_{j}=\frac{\epsilon}{2^{j}}$ centered at each $q_{i}$ clearly cover $E$, and we have $\sum_{j=1}^{\infty}|Q_{j}|=\epsilon$. This implies that $m(E)=0$.

This example is typical. But here's my question. The closure of $E$ is $\bar{E}=[0,1]$. In particular, $\bar{E}\backslash E=\mathbb{I}\cap[0,1]$. If we examine our cover, we see that to each $q_{j}$ we have placed a neighborhood of radius $\frac{\epsilon}{2^{j+1}}>0$. Because $A$ is dense in $[0,1]$, for every $x\in[0,1]$, $B(x;\delta)\cap A\neq\emptyset$ for any $\delta>0$.

In other words, the collection of balls $\{B(q_{i};\frac{\epsilon}{2^{j}})\}_{j=1}^{\infty}$ ought to cover $[0,1]$. But if it did, then it would contradict what was shown above.

How do you resolve this (for general countably dense subsets).

This question arose from a related problem on a post I made regarding a question as to the relationship of outer Jordan measure a set $E$ and its closure $\bar{E}$, and can be found here: Closure, Interior, and Boundary of Jordan Measurable Sets.. In particular, I want to know if it's possible to justify "$\epsilon$-fattening" a cover of $E$ to a cover of $\bar{E}$. In other words, can we take a finite cover $\{Q_{j}\}_{j=1}^{N}$ of $E$ by cubes of length $\ell_{j}$, and fatten each cube by no more than $|Q_{j}\leq|Q'_{j}|\leq|Q_{j}|+\frac{\epsilon}{N}$ and then claim $\{Q'_{j}\}_{j=1}^{N}$ is a cover of $\bar{E}$? Apparently this should be true for finite covers, but it's clearly not true for countable covers because of the obvious counter example presented at the beginning. But my justification for why it's true for finite covers carries over to countable covers (in that $\bar{E}\backslash E$ is a collection of limit points of $E$, so arbitrarily fattening up a cover should capture these points, hence cover $\bar{E}$).

There must be something wrong with my thinking here, and I really need to get it resolved! So much thanks in advance to anyone who can help me with this!

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I think like Sargera, and I will fail analysis if I continue to. I need to know exactly what the inconsistency in my thinking is so I can correct it. These answers amount to "your thinking isn't rigorous," which is false in my case. – user97468 Sep 28 '13 at 23:00
If, as you claim, your thinking is rigorous, then it does not include the part of Sargera's question saying that a certain collection of balls "ought to cover $[0,1]$", because "ought to" is not a rigorous notion. What replaces that part in your rigorous thinking? – Andreas Blass Sep 29 '13 at 0:18
up vote 3 down vote accepted

This is an issue of quantifiers. Taking a closer look at them shows that there is a huge difference between being a countable dense set and forming an open cover with any balls centered at these points.

Separability mean that there is a countable set $D$, such that for every point $x$ and every $\epsilon>0$, there is a $y\in D$ such that $x\in B(y,\epsilon)$. This means, there is essentially a function $(x,\epsilon)\mapsto y(x,\epsilon)$ that maps points in the space and positive radii to elements of $D$ (no choice is required, there exists a well-ordering of $D$ since $D$ is countable).

Now, having a countable family of balls with the points in $D$ as centers and some fixed radii is different. Essentially, there is a map $y\mapsto \epsilon_y$ defined on $D$. That this gives you an open covering, means that there is a function $x\mapsto y_x$ such that $x\in B(y_x,\epsilon_{y_x})$ for all $x$. So why may such a function fail to exist?

First, it is clear that separability is easier to satisfy, since for any $x$ and $\epsilon$, it might well be that $\epsilon_{y_x}<\epsilon$. Now, we see that the similarity breaks down completely when we take into account that the quantifiers in the condition of separability say "For all $x$ and $\epsilon>0\ldots$". So we have to cover the case $\epsilon=\epsilon_{y_x}/2$ and there is no good reason why there should be another point $x'$ such that $x\in B(x,\epsilon_{y_{x'}})$.

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So basically, for any $(x,\epsilon)$, I am guaranteed the existence of some $y_{x,\epsilon}$ such that $x\in B(y;\epsilon)$. However, $\epsilon_{x}$ is fixed (quantifier 1), and there are only a finite number of $y_{i}$ with radii (in this construction at least) $r_{i}>\epsilon_{x}$. Hence, once $r_{i}=\frac{\delta}{2^{i}}<\epsilon_{x}$ (quantifier 2), there is no longer any reason not to have $B(x;\epsilon)\cap B(y;r_{i})=\emptyset$. – Taylor Martin Oct 14 '12 at 21:26
@JTian Yes. $~~~~$ – Michael Greinecker Oct 14 '12 at 21:33
On the other hand, and maybe all that's needed is some time, I still can't get over the fact that $B(q_{j};\frac{\epsilon}{2^{j}})$ intersections infinitely many rationals and irrationals, no matter how far along in the sequence we are. And how I am thinking about this, is that as each rational is enumerated in the sequence, the collection of balls starts a "chain reaction", in that the first ball (let's say $q_{1}=0$) hits another rational, which hits another, and so on until all rationals have been hit, and all irrationals in between are hit. I dunno...kind of naive I guess, eh.... – Taylor Martin Oct 14 '12 at 21:34
@JTian I think the basic problem is that visual intuition is of no use here. Since one cannot differentiate between rationals and irrationals "with the eye", it is hard to make sense of the counterexample visually. – Michael Greinecker Oct 14 '12 at 21:39
Okay, so here's my modified thinking then. Tag a radii $r_{j}=\frac{\delta}{2^{j}}$ to every $q_{j}$ for some small fixed $\delta$. Fix a pair $(x,\epsilon)$. By density, there exists a $q_{j_{x}}$ such that $x\in B(q_{j_{x}}; \epsilon)$. If $\epsilon<r_{j_{x}}$, then great. But by construction, there are only finitely many such radii where $\epsilon>r_{j}$, hence finitely many rationals $q_{j}$, hence only finitely many $x\in[0,1]$ where such an $\epsilon$ will work. So there are some $x$ where $\epsilon>r_{j_{x}}$ corresponding to the $q_{j_{x}}$ which would normally work. – Taylor Martin Oct 14 '12 at 22:16

For $\varepsilon$ small enough (say, below $\frac12$) we do the following, for every $i\in\mathbb N$ let $C_i$ be the following closed set: $$[0,1]\setminus\bigcup_{j=0}^i B\left(q_j,\frac\varepsilon{2^j}\right)$$

It is clear that $C_i$ is a non-empty closed set, since $[0,1]$ is compact we have that $C_i$ is compact. Furthermore it is obvious that $C_i\subseteq C_j$ for $i\geq j$. Therefore this is a decreasing sequence of compact sets.

Cantor's theorem tells us that the intersection is non-empty, therefore the so-called open cover is not actually an open cover as you suspected.

The reason fattening should work for finite covers and fail for infinite covers is exactly the same reason that there are the same number of rationals as integers; but there are more, many more, real numbers. It is the same reason that we can find all sort of strange behaviors with infinite sets.

Infinite sets are weird! The idea is that you cannot split an interval into finitely many cubes that one of them is not big, but when you split it to infinitely many parts you can manipulate the sizes better.

This is why Hilbert's hotel can always accept more people, even when there are never any vacancies.

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Very cool and slick. I think you meant $i\geq j$. But okay, yes, I see that $\cap_{i=1}^{\infty}C_{i}\neq\emptyset$ by Cantor's theorem, which incidentally is easy to prove even for non-countable collections of sets. But having a rigorous justification to the example I posed is only part of what I am looking for. I am trying to gain an intuitive understanding of why my first argument is the correct one, and my second one is not. Moreover, I am trying to get an answer to my question about Jordan outer measure and its relation to a set $E$ and its closure $\bar{E}$ (see above). – Taylor Martin Oct 14 '12 at 20:38
@JTian: I countered your edit with an edit which is directed to the intuition, I hope it will help. – Asaf Karagila Oct 14 '12 at 20:42
I guess I am still having conceptual difficulties. Tagging a ball of positive radius to each rational still leaves an uncountable number of points in $[0,1]$ untouched. How does this not contradict density? We set up a neighborhood to each rational, took the union, and failed to cover (the majority of) $[0,1]$. Or, as infinite sets are indeed weird, is this just something that no one has good intuition for, and why we must just accept the purely mathematical arguments available? – Taylor Martin Oct 14 '12 at 20:52
One last thing, how would one rigorously justify the fattening of a finite cover for a general set $E$ to $\bar{E}$? We at least have a counter-example for the countable case. – Taylor Martin Oct 14 '12 at 20:55
@JTian: Because you fixed those radii. Once the obstacles are set, it is easy to avoid them if you know how. General density arguments states just that the obstacles exist, not where they are exactly. If you understand the analogy... – Asaf Karagila Oct 14 '12 at 20:55

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