Paradox as to Measure of Countable Dense Subsets? 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!
 A: 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. 
A: 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 $x\in B(x,\epsilon_{y_{x'}})$.
