A Fallacious Argument on Open Cover of Rationals in $[0,1]$ Let $A=\mathbb{Q}\cap [0,1]$, the set of rational numbers in the interval $[0,1]$, and suppose that $\left(c_{k}\right)_{k=1}^{\infty}$ denotes an ordered enumeration of the elements of $A$. Then, given $\epsilon>0$, consider an open interval cover $\left(I_{k}\right)_{k=1}^{\infty}$ of $A$ defined by $I_{k} = \left(c_{k}-\frac{\epsilon}{2^{k}},c_{k}+\frac{\epsilon}{2^{k}}\right)$ for all $k$. * * * Question: Is it correct to argue that at most countably many irrationals could have been left out from being covered by the union $\bigcup\limits_{k=1}^{\infty} I_{k}$? If so, can someone please provide a proof of it? Thanks in advance.

There could be uncountably many irrationals not in $\bigcup_{k-1}^{\infty} I_k$ (depending on $\varepsilon$ of course). To see this, let $ \lambda$ denotes the Lebesgue measure on $[0,1]$. Then $\lambda I_k = \frac{\varepsilon}{2^{k-1}}$. Then it would follow that $ \lambda(\bigcup_{k\in\mathbb{N}} I_k) \leq \sum_{k \in \mathbb{N}} \lambda(I_k) = \varepsilon \sum_{k \in \mathbb{N}} \frac{1}{2^{k-1}} = 2\varepsilon$. Since $\varepsilon$ is arbitrary, this is a contradiction.