TITLE: $\bigcup_{i=0}^\infty I_i $ not superset of $[0,1]$?
QUESTION [9 upvotes]: Index all rational numbers in $[0,1]$ from 1 to infinity. For each rational number $q$ in $[0,1]$, form a set $I_i=[q-\frac\epsilon{2^i},q+\frac\epsilon{2^i}]$, where $\epsilon$ is a small number, maybe $\epsilon=0.01$. Now consider the set $$A=\bigcup_{i=0}^\infty I_i $$ Set $A$ seems to be a superset of $[0,1]$ because for any number $x\in[0,1]$, there is a rational number that is very close to it. So $x\in I_i$ for some $i$. But the measure of $A$ is $\epsilon$ and the measure of $[0,1]$ is $1$. Does this mean there must be numbers in $[0,1]$ but not $A$? If so, what are those numbers or how to construct them? Also why is the argument for $x\in I_i$ for some $i$ incorrect?

REPLY [2 votes]: The answers given have pointed out the flaw in the OP's argument. Let me say a bit, though, about the intuitive content of the mistake.
The statement "if $x$ is arbitrarily close to elements of $A$ (or more precisely, if $\forall\epsilon>0\exists a\in A(d(x, a)<\epsilon)$), then $x\in A$" is just saying "$A$ is closed."
Of course not every set is closed. For example, consider $A=\mathbb{R}-\{0\}$. Of course 0 is arbitrarily close to elements of $A$, but not in $A$.
The intuitive response to this is often, "Sure, but we can make $A$ closed by just adding one point; that won't effect the size of $A$." This is true. However, this is often then extended to,

We can always make $A$ closed just by adding "a bunch of single points," so that shouldn't change the size of $A$.

This is the part that is wrong, and often surprising for those seeing analysis for the first time! The closure of a set can be vastly larger than the original set, in essentially any sense of "largeness" whatsoever. 

A dual mistake is assuming that all open dense sets are big; in fact, you can make countably many open dense sets whose intersection has measure 0!