Let $X=\{a,b\}$ with the indiscrete topology $T=\{X,\emptyset\}$. Consider the subset $A=\{a\}$. What is the limit point of $A$? I understand that limit point $x$ of $A$ means every neighborhood of $x$ intersects $A$ in some point other than $x$ itself. However, for a set $A=\{a\}$, how could I find the limit point of $A$? Thank you in advance!

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Well, you could start by listing the open sets in the topology, then looking at the ones that intersect $A$ ... there are not many open sets you need to look at. – Old John Sep 19 '13 at 21:39
I think you got your definition of limit point wrong. A point $x \in X$ is a limit point of $A \subseteq X$ when all neighbourhoods of $x$ (not $A$) have non-empty intersection with $A \setminus \{x\}$. – kahen Sep 19 '13 at 21:42
I think that's just a typo – Stefan Hamcke Sep 19 '13 at 21:42
So the open sets in the topology is {a,b}? – James Sep 19 '13 at 21:43
No. The open sets are those in $T$. That's the definition of a topology. – kahen Sep 19 '13 at 22:02

An open set of the topology $T$ is simply an element of $T$. For a point $x$ to be a limit point of $A$, every neighborhood of $x$ must intersect $A$ in a point other than itself. Since your space is finite we can check each point individually.
• The point $a$ is not a limit point of $A$ because all its neighborhoods only intersect $A$ at $a$.
• The point $b$ is a limit point of $A$ because it has only one neighborhood, and that is the whole space $X$ which intersects $A$ on a point different than $b$ (namely $a$).
There are no more points so $b$ is the only limit point of $A$.