Do paracompact non-Hausdorff spaces admit partions of unity? I'm just curious.



Paracompact but not Hausdorff
The answer is no.
Take the "classical" example of the line with two origins. This space is non-Hausdorff, paracompact and doesn't admit partitions of unity.


EDIT: I think the question is a kind of ["duplicate"](https://mathoverflow.net/questions/19219/smooth-manifolds-that-dont-admit-a-partition-of-unity) . 
Ok, but if you have an example for a non-Hausdorff manifold, which doesn't admit partitions of unity, you have an example for a non-Hausdorff paracompact space with the same property.


First the definition:  

The line with two origins is the quotient space of two copies of the real line
$\mathbb{R} \times {a}$ and $\mathbb{R} \times {b}$.  

with equivalence relation given by
$(x,a) \sim (x,b)\text{ if }x \neq 0$.  

Since all neighbourhoods of $0\_a$ intersect all neighbourhoods of $0\_b$, it is non-Hausdorff.  

However, this space is paracompact, since $\mathbb{R}$ is paracompact.


For the non-existence of a partition of unity: 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 :-) )


