Question

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

Answer 1

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.

EDTI: I think the question is a kind of "duplicate" . 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 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 :-) )

Answer 2

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