Difference between revisions of "Talk:Polar body"
From Encyclopedia of Mathematics
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Another doubt. Indeed, $K^\circ$ is closed and bounded; but "compact"? Do you implicitly assume finite dimension? In fact, if $V$ is complete, then $K^\circ$ is ''weakly'' compact. [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 10:01, 24 October 2017 (CEST) | Another doubt. Indeed, $K^\circ$ is closed and bounded; but "compact"? Do you implicitly assume finite dimension? In fact, if $V$ is complete, then $K^\circ$ is ''weakly'' compact. [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 10:01, 24 October 2017 (CEST) | ||
| + | :Yes, I do. [[User:Richard Pinch|Richard Pinch]] ([[User talk:Richard Pinch|talk]]) 19:13, 26 October 2017 (CEST) | ||
Revision as of 17:13, 26 October 2017
"If $K$ is a convex set containing the zero element in its interior then $K^\circ$ ... is again a convex neighbourhood of the origin." — Really? Even if $K$ is the whole space $V$? Boris Tsirelson (talk) 22:59, 23 October 2017 (CEST)
- ...bounded...bounded... Thank you. Richard Pinch (talk) 08:29, 24 October 2017 (CEST)
Another doubt. Indeed, $K^\circ$ is closed and bounded; but "compact"? Do you implicitly assume finite dimension? In fact, if $V$ is complete, then $K^\circ$ is weakly compact. Boris Tsirelson (talk) 10:01, 24 October 2017 (CEST)
- Yes, I do. Richard Pinch (talk) 19:13, 26 October 2017 (CEST)
How to Cite This Entry:
Polar body. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Polar_body&oldid=42192
Polar body. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Polar_body&oldid=42192