# Paracompact but not Hausdorff

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

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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 :-) )

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It was in fact the question to which you linked which made me wonder this. This is not a duplicate question, since Aston insisted on a Hausdorff manifold. How can I see that the example you give is paracompact, and how can I see there is no partition of unity? – David Carchedi Apr 1 '10 at 8:23
Take any open cover of the "real line" for which there are exactly one open set U_1 containing one copy of 0 (call this 0_1), and one open set U_2 containing the other copy of 0 (0_2). A partition of unity subordinate to this open covering must contain a function f_1 supported in U_1, such that f_1(0_1) = 1, since 0_1 is only contained in this open set. Similarly for 0_2. It follows by continuity that for some small epsilon, we have f_1(epsilon) + f_2(epsilon) > 1, so it is not a partition of unity. – villemoes Apr 1 '10 at 10:52

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

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