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say $X=\{a,b\}$ be a set. The following is a topology on $X$.

$\tau=\{\{ a\}, \{a,b\}, \varnothing\}$

Then $b$ is a limit point of $a$, as all open sets $(\{a,b\})$ intersect $\{a\}$ at points other than $b$. Then how is it that all finite sets are closed? What I am doing wrong here!

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    $\begingroup$ Note that $\{\phi\}$ is not a subset of $X$, whereas $\varnothing$ is a subset of $X$. $\endgroup$ – Asaf Karagila Apr 23 '15 at 11:44
  • $\begingroup$ @AsafKaragila: Unless one of $a,b$ is $\emptyset$. ;) $\endgroup$ – tomasz Apr 23 '15 at 22:06
  • $\begingroup$ @tomasz: Sure, but it wouldn't be a topology, and in case $\phi$ is a Greek letter, not the symbol for the empty set! :-) $\endgroup$ – Asaf Karagila Apr 23 '15 at 22:13
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All finite sets in a $T_1$ space are closed. Spaces like $\Bbb R^n$ are Hausdorff and therefore $T_1$. But the Sierpinski space (which you suggest) is not a $T_1$ space, and so not every finite set there is closed.

Note that if all finite sets were closed, universally, then every topology on a finite set would be the discrete topology. But then the trivial topology, $\{X,\varnothing\}$ wouldn't be a topology on any non-singleton finite set.

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Since $\{b\}$ is not in the topology, by definition it is not open, and hence its complement, $\{a\}$, is not closed.

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Are you asking "how are all finite sets closed" in this particular case? Or are you providing a counter example as to why all finite sets are not closed. Under the right conditions. I know it is sufficient that you require each pair of point in a space $X$ have open neighborhoods that don't contain the other (i.e., $X$ is $T_1$ as was mentioned by Asaf Karagila). But for your example this is not the case (as $b$ has no open set that does not contain $a$), and so we can not assume that all finite sets will be open.

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