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Mar 29, 2010 at 21:48 comment added David Carchedi By the way, does paracompact WITHOUT Hausdorff imply the existance of partitions of unity? I feel like this might be a dumb question...
Mar 25, 2010 at 10:51 vote accept Aston Smythe
Mar 25, 2010 at 0:17 comment added David Carchedi More generally, the etale space of any sheaf over a manifold is always locally Euclidean but rarely Hausdorff.
Mar 24, 2010 at 22:54 comment added babubba Do Banach/Hilbert manifolds admit partitions of unity?
Mar 24, 2010 at 21:35 answer added Pete L. Clark timeline score: 13
Mar 24, 2010 at 18:43 answer added Chris Schommer-Pries timeline score: 10
Mar 24, 2010 at 18:34 comment added David E Speyer Hausdorff does not follow from any local condition. Consider the "line with a double point": two copies of $\mathbb{R}$ glued to each other along $\mathbb{R} \setminus \{ 0 \}$. This is locally as nice as you could want in any sense.
Mar 24, 2010 at 18:33 comment added Harald Hanche-Olsen Yes, you need to specify Hausdorff. Consider the real line with an extra zero grafted onto it. It's locally Euclidean, but not Hausdorff.
Mar 24, 2010 at 18:29 comment added Aston Smythe I mean manifolds that aren't second countable but are locally euclidean and Hausdorff .... do we need to specify Hausdorff here, surely it should from the manifold being locally euclidean.
Mar 24, 2010 at 18:12 comment added Charlie Frohman If your manifolds are second countable, Hausdorff and locally Euclidian, then they are paracompact, and hence any open cover admits a partition of unity subordinate to it, at least in the smooth category. In the analytic category, you will not have partitions of unity, but that is because there are no bump functions.
Mar 24, 2010 at 18:07 history asked Aston Smythe CC BY-SA 2.5