Hartshorne (II Prop 2.6) defines a functor $t$ from the category of topological spaces to itself as follows: If $X$ is a topological space, define $t(X)=\{Z\subseteq X:Z\text{ is irreducible and closed}\}$. Assign $t(X)$ the topology with closed sets given by $t(Y)$ where $Y$ is a closed subset of $X$. If $f\colon X_1\to X_2$ is continuous, define $t(f)\colon t(X_1)\to t(X_2)$ by $t(f)(Z)=\overline{f(Z)}$. (See the question Hartshorne proposition II(2.6) for more details). Finally, he defines a function $\alpha\colon X\to t(X)$ (which one can easily show is continuous) by $\alpha(P)=\overline{\{P\}}$.

My question is concerning Abramo's answer to the above-linked question, namely that $\alpha(U)=t(X)-t(Y)$ for any open subset $U=X-Y$ of $X$. I seem to have produced a counterexample to this, and I was hoping that someone might point out where I messed up:

Let $X$ be the nonnegative integers with closed subsets given by $[n]=\{0,1,\ldots,n-1\}$ for $n\geq0$ and of course $X$ itself. Notice that each of these closed sets is irreducible. In particular, $X$ is irreducible. Now, the closure of any $n\in X$ is the set $[n+1]$, so applying $\alpha$ to $X$, we get $$ \alpha(X)=\{[1],[2],\ldots\}.$$ On the other hand, $$ t(X)-t(\emptyset) = t(X)=\{[1],[2],\ldots\}\cup \{X\}.$$ So, $t(X)-t(\emptyset)\neq \alpha(X)$. In fact, this shows that $\alpha(X)$ isn't even open, since $t(X)-\alpha(X)=\{X\}$ isn't closed (if $t(Y)=\{X\}$ for some closed $Y\subseteq X$, then $X\subseteq Y\subseteq X$, i.e. $Y=X$, a contradiction).

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    $\begingroup$ This is the soberification functor for topological spaces. $\endgroup$ – Zhen Lin Dec 16 '13 at 4:32
  • $\begingroup$ @ZhenLin What's the soberification functor? $\endgroup$ – Avi Steiner Dec 16 '13 at 4:55
  • $\begingroup$ @Avi: You can find this functor in texts about "pointless topology", for example Sketches of an elepehant, but also by googling "soberification". $\endgroup$ – Martin Brandenburg Dec 16 '13 at 10:30

I think I've figured it out:

The map $\alpha$ that Abramo defines, which for clarification purposes I'll call $\beta$, is not the map taking an open subset $U$ of $X$ to the closure of its points (i.e. the map $U\mapsto \left\{\overline{\{P\}}:P\in U\right\}$). What it is is the inverse to the pre-image map $\alpha^{-1}$ of the original map $\alpha$. To see this, let $U=X-Y\subseteq X$ be open. Then $$ \alpha^{-1}(\beta(U))=\alpha^{-1}(t(X)-t(Y))=X-\alpha^{-1}(t(Y)). $$ But $\alpha^{-1}(t(Y))$ is just $Y$: Since $Y$ is closed, and since $\overline{P}$ is necessarily irreducible, $P\in Y$ iff $\alpha(P)\in t(Y)$. So, $$ \alpha^{-1}(\beta(U)) = X-Y=U.$$ Conversely, let $V=t(X)-t(Y)$ be an open subset of $t(X)$, where $Y$ is a closed subset of $X$. Then $$ \beta(\alpha^{-1}(V))=\beta[\alpha^{-1}(t(X)-t(Y))]=\beta(X-Y)= t(X)-t(Y)=V.$$


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