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So I just started learning topology and I came across this problem.

Consider the topological space $(X,\tau)$where the set $X=\{a,b,c,d,e\}$ and $\tau=\{X,\emptyset,\{a\},\{c,d\},\{a,c,d\},\{b,c,d,e\}\}$

Determine the limit points of

i) $\{a\}$

Is $a$ a limit point for itself since the singleton a is an open set in the topology. From the definition of limit point of a set every neighbourhood of the point must contain a point of the set that id different from a however when considering the singleton there are no different points of a so what is the correct answer im very confused.

Thanks in advance

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  • $\begingroup$ It is not a limit point. $\endgroup$ Jan 22 at 9:24
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The limit points of a set $S$ are not an intrinsic property of $S$. They depend on how the set $S$ is immersed into an ambient topological space $X$.

To answer your question, you have to check if there are points $x\in X$ such that every punctured neighborhood of $x$ (i.e., the result of removing from an open neighborhood of $x$ the point $x$) passes through $a$.

  • Now the punctured neighborhoods of ${a}$ cannot contain $a$.
  • So let's examine the punctured neighborhoods of ${b}$. These are $\{c,d,e\}$ and $X-\{b\}$. At least one does not contain $a$. So $b$ is not a limit point of $\{a\}$.
  • The punctured neighborhoods of ${c}$ are $\{d\}$ and others. But you already found one that does not contain $a$ so $c$ cannot be a limit point of $a$.

And so on...

If you want some intuition about the concept of a limit point, then you can think that an element $x\in X$ is a limit point of a subset $S\subseteq X$ if $x$ can be well approximated by $S-\{x\}$.

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  • $\begingroup$ just being curious isn't $\{a,c,d,b\}$ also a nbhd of b? $\endgroup$ Jan 22 at 11:26
  • $\begingroup$ No, the neghbs of b are all superset of {b,c,d,e}. Because they must contain an open set containing b. $\endgroup$
    – Kosh
    Jan 22 at 11:26
  • $\begingroup$ isn't $\{a,c,d\}$ open in this topology? $\endgroup$ Jan 22 at 11:30
  • $\begingroup$ Yes, it is open. You listed the open sets. But this open set does not contain b. $\endgroup$
    – Kosh
    Jan 22 at 12:04
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The set of limit points of $\{a\}$ is empty.

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