Does Uryshon's lemma require the use of the axiom of depending choice?
- From: "Rupert" <rupertmccallum@xxxxxxxxx>
- Date: 25 Mar 2007 17:42:35 -0700
It seems to be possible to prove Urysohn's lemma without any use of
choice for second countable spaces, but in general the axiom of
depending choice seems to be required. Is this right?
.
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