TITLE: A question on Hamming metric/distance
QUESTION [6 upvotes]: Suppose $\sf{X}=\{0,1\}$, and $\sf{X}^n$ is the set of all binary sequences of length $n$. So the first question is that what does it mean by the convex closure of a subset $\sf{A}$ of $\{0,1\}^n$, which is denoted by $\sf{\bar{A}}$?
As usual, we define the Hamming distance of two subsets $\sf{A}$ and $\sf{B}$ of $\sf{X}^n$ can be defined as $d_H(\sf{A},\sf{B}) = \min_{\bf{x}\in \sf{A},\hat{\bf{x}}\in\sf{B}}$ $d_H(\bf{x},\bf{\hat{x}})$, where $d_H(\bf{x},\bf{\hat{x}})$ is the number of differences between binary sequences $\bf{x}$ and $\bf{\hat{x}}$. 
Now define $\Gamma\sf{A} = \{\bf{x}: \bf{x} \in \sf{X}^n$, $d_H(\{\bf{x}\},\sf{A})\leq1\}$, and then define the Hamming boundary $\partial\sf{A}$ of $\sf{A}\in X^n$ as $\partial\sf{A} = \sf{A} \cap$ $\Gamma\sf{\bar{A}}$.
So the second question is that why do we use $\Gamma\sf{\bar{A}}$ instead of $\Gamma\sf{A}$ in the above definition?
Thanks very much in advance.

REPLY [2 votes]: Are you sure that $\overline{A}$ means the convex closure for the second part?
What would make lots of sense would be if $\overline{A}$ is the complement of $A$ (which is a pretty standard notation in many set theory books):
Then
$$\partial A= \{ x \in A | \exists y \notin A \, \mbox{such that} \, d_H(x,y)=1 \} \,.$$
And this would represent the boundary in some sense.