Determine which of the following sets is a null set as defined in the question. 
A set $X\subseteq \Bbb R$ is said to be a null set if for every $\epsilon\gt0$ there exists a countable collection $\{(a_k,b_k)\}_{k=1}^\infty$ of open intervals such that $X\subseteq \cup_{k=1}^\infty\{(a_k,b_k)\}$ and $\sum_{k=1}^\infty(b_k - a_k)\le \epsilon$. Which of the following sets is not a null set?

*

*Every finite set

*$\Bbb Q^c$, the set of irrational numbers

*$\Bbb Q$, the set of rational numbers

*$\Bbb N$, the set of natural numbers


I haven't been able to grasp the question clearly. What I did notice is that among the sets given, only $\Bbb Q^c$ is uncountable. Is the definition of null set given in this question actually that of a countable set?
 A: You are partially correct.  Any countable set will be a null set.  If your set $X$ consists of just the points $\{x_k\}_{k=1}^\infty$ then for any $\epsilon>0$ you may let: $$a_k=x_k-{\frac\epsilon{2^{k+1}}},\qquad b_k=x_k+{\frac\epsilon{2^{k+1}}}$$
Then for all $k$, we have $x_k\in (a_k,b_k)$.  Thus $X\subseteq \bigcup_{k=1}^\infty(a_k,b_k)$.
Further we have $\sum_{k=1}^\infty (b_k-a_k)=\sum_{k=1}^\infty \frac\epsilon{2^k}=\epsilon$.
We may conclude that a countable subset of $\mathbb{R}$ does indeed satisfy the condition of being a null set.  That is enough to answer the multiple choice question.
However the definition given for null set is not the same as for countable set.  Not every null set is countable.  For example the Cantor set is a null set, but not countable.
That is the Cantor set consists of real numbers in $[0,1]$ which may written out in ternary expansion using only the digits $0$ and $2$.  Clearly this not countable (real numbers in $[0,1]$ may be written out using just two digits in binary).  However the Cantor set may be contained in a union of $2^n$ intervals each of size as close as you like to $\frac{1}{3^{n}}$ for any $n$.  For any $\epsilon>0$ you may choose $n$ sufficiently large that $\left(\frac{2}3\right)^n<\epsilon$, so the sum of the lengths of the $2^n$ intervals will be less than $\epsilon$.
