TITLE: The functor $\mathrm{Haus}\to\mathrm{Set}$ sending a space to its set of open sets is not representable?
QUESTION [6 upvotes]: I know the contravariant functor $\mathrm{Top}\to\mathrm{Set}$ sending a topological space to its set of open sets is representable, with representing object being the two point space with precisely one singleton being the only nontrivial open set.
Why do things not work in the category of Hausdorff spaces? 
Towards a contradiction, I suppose the functor $F$ is representable, so $F(-)\cong\operatorname{Hom}_{\mathrm{Haus}}(-,A)$ for some representing space $A$. In particular, $F(A)\cong\operatorname{Hom}(A,A)$. Let $U\in F(A)$ be the universal element corresponding to $\mathrm{id}_A$. 
Then for any space $B$, and any $V\in F(B)$, there is a unique morphism $f\colon A\to B$ such that $F(f)(V)=U$, i.e., $f^{-1}(V)=U$. This unique morphism is $\eta_B^{-1}(\eta_A(U))$ where $\eta_X$ is the component morphism $F(X)\to\mathrm{Hom}(X,A)$.
This claim certainly seems unlikely, so that $F$ is not representable. However, I'm having trouble coming up with a concrete example to show this cannot be.

REPLY [8 votes]: Let $O : \mathbf{Haus}^\mathrm{op} \to \mathbf{Set}$ be the functor that sends a Hausdorff space $X$ to the set of open subsets of $X$. Then $O (1)$ has two elements, so any representation of $O$ must be a two-point space. But the only two-point Hausdorff space is discrete – which clearly won't work.