No. This is heresy.
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All protosandwiches have the locally sandwich property, this is first year stuff
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what? no, this is incorrect
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Replying to @0xdeadbabe @St_Rev and
if you remove every subset of the protosandwich that is locally sandwich, you are left with a subset that is not locally sandwich
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You're left with the empty set, locally sandwich neighborhoods form a covering set.
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the hot dog, for example, has an subset which not locally sandwich
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Sure, but it's still covered by locally sandwich neighborhoods. The local sandwich property doesn't say that every subset is a sandwich. It says that every point has a neighborhood which is a sandwich.
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It's equivalent to saying that there's a continuous homomorphism h: [0,1] -> s and neighborhoods B, M, B' of s such that: a) B and B' are bread and M is filling b) h(0) lies in B, h(1/2) lies in M, h(1) lies in B', and c) B and B' are disjoint
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I think I follow. You're lucky I'm kinda exhausted and can't think clearly right now.
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. Banned in Sweden. SubGenius, Zhuangist, white-hat troll. Defrocked mathematician. Brain problems.