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This question arise when i read the proof of Whitney Approximation Theorem for Function in Lee's book.

As i noticed, in the usual partition of unity arguments, e.g. extending smooth functions over a closed subset of a manifold, existence of Riemannian metric, etc., we only need a partition of unity of an uncountable open cover of the manifold. However, in the proof of Whitney Approximation Theorem (Lee's ISM), he explicitly use partition of unity over a countable cover.

I've looked at the details for a while now, but i can't find any reason why we should use the countable one. Did anyone know why ?

Any help will be appreciated. Thank you.

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This appears to just be an oversight. The choice of the countable subcover $\{U_{x_i}\}$ is totally unnecessary, and you can just take a partition of unity subordinate to the open cover $\{U_0\}\cup\{U_x\}_{x\in M\setminus A}$.

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  • $\begingroup$ I think you're right. I looked over that proof, and for the life of me I can't figure out why I insisted on reducing to a countable open cover. $\endgroup$ – Jack Lee Mar 27 '18 at 5:07
  • $\begingroup$ Thank you Eric Wofsey,@Jack Lee. Let me know if there are any particular reason for that Prof Lee. $\endgroup$ – Sou Mar 27 '18 at 5:32

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