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The set $A=\{1, \frac{1}{2}, \frac{1}{4}, \dots \}$ is obviously not closed in $\mathbb{R}$ with the Euclidean metric, as the sequence $1, \frac{1}{2}, \frac{1}{4} \dots$ converges to $0 \notin A$.

But if we look at the complement, $A'$ and let $p\in A'$ then there exist some $a_i$ and $a_{i+1}$ such that $a_{i+1}<p<a_i$. Letting $\epsilon=\frac{a_i-a_{i+1}}{2}$ we see that the open ball $B_\epsilon(p) \cap A = \emptyset$ so $B\epsilon(p) \subset A'$ thus proving that $A'$ is open and $A$ closed.

What am I doing wrong?

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    $\begingroup$ Let $p=0$...... $\endgroup$ – ThePortakal Apr 28 '18 at 13:42
  • $\begingroup$ Nobody has pointed this out yet: just as 0 is the exceptional "missing" point that prevents $A$ from being closed, in the exact same way the same point 0 is the exceptional "extra" point that spoils the openness of $A'$. The two concepts are exactly opposite. $\endgroup$ – MJD Apr 28 '18 at 14:36
  • $\begingroup$ Nonempty countable sets are never open, as any neighborhood of a point in the set is uncountable. $\endgroup$ – Jonathan Hebert Apr 28 '18 at 16:56
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Your complement $A'$ includes $0$ but you cannot find an open ball around it. Your mistake is tacitly assuming $p \in A'$ is in $(0,1)$ as well.

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Since $A$ is not closed, its complement cannot be open, $0$ is in the complement of $A$ and every ball which contains $0$ meets $A$.

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