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| bio | website | klickverbot.at |
|---|---|---|
| location | Austria | |
| age | ||
| visits | member for | 1 year, 6 months |
| seen | Jan 19 at 10:07 | |
| stats | profile views | 6 |
Aspiring student, interested in way to many things.
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Mar 24 |
awarded | Scholar |
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Mar 24 |
accepted | Proving $(f^{-1}(U))^0 = f^*(U^0)$ |
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Mar 15 |
awarded | Supporter |
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Mar 15 |
comment |
Proving $(f^{-1}(U))^0 = f^*(U^0)$ I think ${(U^0)}^0 \cong U$ only holds in the finite dimensional case. Also – sorry for that – I edited the question removing the requirement that V and W be Euclidean, since I really want to prove the general case (it just happens that the only application I have for it right now deals with Euclidean spaces). I think, though, that the proof might still work with minor modifications, let me check… |
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Mar 15 |
revised |
Proving $(f^{-1}(U))^0 = f^*(U^0)$ deleted 3 characters in body |
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Mar 15 |
revised |
Proving $(f^{-1}(U))^0 = f^*(U^0)$ added 2 characters in body |
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Mar 15 |
awarded | Editor |
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Mar 15 |
comment |
Proving $(f^{-1}(U))^0 = f^*(U^0)$ @ArturoMagidin, Benjamin Lim: Clarified the question. |
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Mar 15 |
revised |
Proving $(f^{-1}(U))^0 = f^*(U^0)$ added 166 characters in body |
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Mar 15 |
asked | Proving $(f^{-1}(U))^0 = f^*(U^0)$ |
