Do paracompact nonHausdorff spaces admit partions of unity? I'm just curious.

The answer is no. Take the "classical" example of the line with two origins. This space is nonHausdorff, paracompact and doesn't admit partitions of unity. EDTI: I think the question is a kind of "duplicate" . Ok, but if you have an example for a nonHausdorff manifold, which doesn't admit partitions of unity, you have an example for a nonHausdorff paracompact space with the same property. First the definition: For the nonexistence of a partition of unitiy: take the open covering $ U = (\infty,0) \cup \{ 0_a \} \cup (0,\infty)$ and $\tilde{U} = (\infty,0) \cup \{ 0_b \} \cup (0,\infty)$. Assume, there is a partition of unity subordinate to this cover. Then the value of each origin would have to be $1$ which cannot be true. (Edit: villemoes was a little faster :) ) 


It's worth noting that any $T_1$ space which admits partitions of unity for finite (two element even) covers is Hausdorff: Proof: Let $x, y \in X$. Let $U = X \ \{x\}, V = X \ \{y\}$. Then let $\{f, g\}$ form a partition of unity with $f$ subordinate to $U$ and $g$ subordinate to $V$. Then $A = \{ t : f(t) > \frac{1}{2} \}$ and $B = \{ t : g(t) > \frac{1}{2} \}$ A and B are disjoint open sets with $y \in A$ and $x \in B$. Edit: On closer inspection, this if of course just the standard proofs that the existence of partitions of unity for finite covers implies normality + the fact that $T_1$ normal spaces are hausdorff 

