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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.

EDIT: 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 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 :-) )

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  • $\begingroup$ 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? $\endgroup$
    – David Carchedi
    Commented Apr 1, 2010 at 8:23
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    $\begingroup$ 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. $\endgroup$
    – villemoes
    Commented Apr 1, 2010 at 10:52
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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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Paracompact non-Hausdorff spaces can admit partitions of unity, but you need a stronger property than paracompactness. A cover admits a subordinate partition of unity iff it has a locally finite refinement by cozero sets. A cozero set of a space $X$ is a set of a the form $f^{-1}(0,1]$, for a continuous function $f: X \to [0,1]$.

For a space, every open cover having a partition of unity is equivalent to being fully normal. Fully normal spaces are necessarily paracompact. Pseudometrizable spaces that are not metrizable are examples of non-Hausdorff fully normal spaces. Fully normal spaces are necessarily normal, so any compact non-normal space is an example of a paracompact space that is not fully normal.

You can find citations for these results at the nlab's page for numerable open cover. The results are ultimately due to Kiiti Morita, of Morita equivalence fame.

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