Recall that the closure of a set $S$ is the union of $S$ and its all limit points. For a given topological space $X$, $p$ is a point of $X$, is it possible that the closure of a single point set $\{p\}$ ,denoted by $\overline{\{p\}}$, has many points? And any continuous map $f$ from $\overline{\{p\}}$ to a topological space $Y$ is determined by $f(p)$, is this claim correct? Any help will be greatly appreciated, thanks in advance!

  • $\begingroup$ Hint: en.wikipedia.org/wiki/Trivial_topology $\endgroup$ – Miguel May 17 '17 at 14:54
  • $\begingroup$ No, the claim is not correct. Again, the indiscrete space is a reason why. $\endgroup$ – William Elliot May 17 '17 at 20:14
  • $\begingroup$ @WilliamElliot With trivial topology, this is to say any point in the closure of {p} minus {p} is also a limit point, thank you! $\endgroup$ – Jiabin Du May 18 '17 at 0:16
  • $\begingroup$ @MiguelAtencia thank you! $\endgroup$ – Jiabin Du May 18 '17 at 0:17
  • $\begingroup$ @Jabin Provided the space S is multipoint , (cl p) - {p} = S. $\endgroup$ – William Elliot May 18 '17 at 20:17

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