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I'm having trouble of thinking of such a set (real or complex) because:

Let X $\subset \mathbb{R}$ or X $\subset \mathbb{C} $, then by definition the closure is:

$\bar{X}$:=X $\cup \partial X$

While by definition $\partial X$ := $\bar{X}$ \ int(X),

where int(X) is the interior of X.

My initial thought was to use the Cantor set, but this has no interior.

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  • $\begingroup$ You're correct that the definition of the boundary is $\partial X := \bar X \setminus \mathop{\rm int}(X)$; but the definition of the closure is not $\bar X := X \cup \partial X$ (that would make the definitions circular). The definition of closure is the usual fundamental definition; $\bar X = X \cup \partial X$ is a consequence of the definition of $\partial X$. $\endgroup$ Feb 9, 2014 at 2:41
  • $\begingroup$ @Daniel, you're right. The original title was wrong, it is now corrected. $\endgroup$
    – BBaire
    Feb 9, 2014 at 3:47
  • $\begingroup$ Look at a dense set in $[0, 1]$. $\endgroup$ Feb 9, 2014 at 3:48
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    $\begingroup$ You're thinking way too complicated: $X = (0,1) \cup (1,2)$. $\endgroup$ Feb 9, 2014 at 9:18

1 Answer 1

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Let $X = \mathbb R \setminus \{0\}$, in $\mathbb R$. Its boundary is $\{0\}$, closure is $\mathbb R$, and so boundary of its closure is empty.

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