Let $(X,\tau)$ be a cofinite topology, where $X$ is infinite. Then, for every $A\subseteq X$, we define a Kuratowski closure operator with: $$cl(A)=\left\{ \begin{array}{cc} A; & A \; \text{finite}\\ X; & A\; \text{infinite} \end{array} \right.$$

I proved all of the Kuratowski closure axioms, but I don't know if my proof is correct. Any comments?

$\boxed{(1) \; cl(\emptyset)=\emptyset}$

$$A \; \text{finite}\Rightarrow cl(A)=\emptyset\Rightarrow cl(\emptyset)=\emptyset $$

$\boxed{(2) \; A\subseteq B\Rightarrow cl(A)\subseteq cl(B)}$

$(i)\; A,B$ finite $\Rightarrow cl(A)=A,\; cl(B)=B$ $$A\subseteq B\Rightarrow cl(A)\subseteq cl(B)$$

$(ii)\; A,B$ infinite $\Rightarrow cl(A)=cl(B)=X$

$$A\subseteq B\Rightarrow A\cup X\subseteq B\cup X\Rightarrow X\subseteq X\Rightarrow cl(A)\subseteq cl(B)$$

$(iii)$ $A$ finite, $B$ infinite $\Rightarrow cl(A)=A,\; cl(B)=X$

$$ A\subseteq B \wedge A\subseteq X \Rightarrow A\subseteq B\cup X\Rightarrow A\subseteq X\Rightarrow cl(A)\subseteq cl(B)$$

$\boxed{(3)\; A\subseteq cl(A)}$

$(i)$ $A$ finite $\Rightarrow cl(A)=A \Rightarrow A\subseteq cl(A)$

$(ii)$ $A$ infinite $\Rightarrow cl(A)=X$ $$A\subseteq X\Rightarrow A\subseteq cl(A)$$ $\boxed{(4)\; cl(A\cup B)=cl(A)\cup cl(B)}$

$(i)$ $A$ finite, $B$ infinite $\Rightarrow A\cup B$ infinite $$cl(A\cup B)=X=A\cup X=cl(A)\cup cl(B)$$ $(ii)$ $A,B$ finite $\Rightarrow A\cup $ finite $$cl(A\cup B)=A\cup B=cl(A) \cup cl(B)$$ $(iii)$ $A,B$ inifinite $\Rightarrow A\cup B$ inifite $$cl(A\cup B)=X=X\cup X=cl(A)\cup cl(B)$$

$\boxed{(5) \; cl(cl(A))=cl(A)}$

$(i)$ $A$ finite $\Rightarrow cl(A)=A$ $$cl(cl(A))=cl(A)$$ $(ii)$ $A$ infinite $\Rightarrow cl(A)=X$ $$cl(cl(A))=cl(X)\overset{X \; \text{infinte}}{=} X=cl(A)$$

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    $\begingroup$ It looks correct. You could probably prove it with a few less cases but that doesn't matter. $\endgroup$ – Nex Nov 21 '15 at 9:30

Your proof is correct. It could be shortened a bit, with fewer cases to distinguish, and using that some properties are redundant. For example, the inclusion $cl(A\cup B)\supseteq cl(A)\cup cl(B)$ is equivalent to monotonicity. Here's how I would write it:


This holds as $\emptyset$ is finite.

$\boxed{A\subseteq cl(A):}$

If $A$ is finite, then $A=cl(A)$. If $A$ is infinite, then $A\subseteq X=cl(A)$

$\boxed{cl(A\cup B)=cl(A)\cup cl(B):}$

If $A\cup B$ is finite, then $A$ and $B$ are finite, thus $cl(A)=A$ and $cl(B)=B$. Now $$cl(A\cup B)=A\cup B = cl(A)\cup cl(B)$$ If $A\cup B$ is infinite, then wLoG $A$ is infinite, thus $cl(A)=X$. Now $$cl(A\cup B) = X = cl(A) = cl(A)\cup cl(B)$$


If $A$ is finite, then $cl(A)=A$, thus $$cl(cl(A)=cl(A)$$ If $A$ is infinite, then $cl(A)=X$, thus $$cl(cl(A))=cl(X)=X=cl(A)$$


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