In this post, it is claimed that the Hausdorff distance $d_H$ between $I=[0,1]$ and $B_n = \{0, \frac{1}{n}, \cdots, \frac{n-1}{n}, 1\}$ equals zero.
Is it not equal to $\frac{1}{2n}$ ?
For example, if $n=1$, $B_1=\{0,1\}$, and since $(B_1)_{\frac{1}{2}}=I$, I believe the Hausdorff distance equals $$ d_H(I,B_1)=\inf\{\varepsilon\;|\; I \subseteq (B_1)_\varepsilon,B_1\subseteq I_{\varepsilon} \} = \frac{1}{2} $$
With the same logic, we can show that $$ d_H(I,B_n)=\frac{1}{2n} \rightarrow0 $$
Is this correct?
Note: the notation $A_{\varepsilon}$ denotes the $\varepsilon$-thickening of the considered set $A$, that is, the set of points at a distance to any point of $A$ smaller than $\varepsilon\ge0$.