Talk:Boundary point: Difference between revisions
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imported>Boris Tsirelson (→every point is a boundary case: new section) |
imported>Peter Schmitt |
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"...every point is a boundary case, or there is no boundary point at all — are both possible. In the first case the set is said to be dense in the space" — Really? The whole space is a (trivial) example of a dense set, but belongs to the latter case, not the former one. The condition "every point of the space is a boundary point of the given set" is equivalent rather to "the given set and its complement are both dense". [[User:Boris Tsirelson|Boris Tsirelson]] 09:18, 4 October 2009 (UTC) | "...every point is a boundary case, or there is no boundary point at all — are both possible. In the first case the set is said to be dense in the space" — Really? The whole space is a (trivial) example of a dense set, but belongs to the latter case, not the former one. The condition "every point of the space is a boundary point of the given set" is equivalent rather to "the given set and its complement are both dense". [[User:Boris Tsirelson|Boris Tsirelson]] 09:18, 4 October 2009 (UTC) | ||
Well, I understand that implication was meant, not equivalence. But, maybe, it is better to write "In the first case the set is said to be dense in the space (as well as its complement)." [[User:Boris Tsirelson|Boris Tsirelson]] 09:23, 4 October 2009 (UTC) | |||
: Boris, thank you for pointing out my sloppy language. [[User:Peter Schmitt|Peter Schmitt]] 22:21, 4 October 2009 (UTC) | |||
Latest revision as of 16:21, 4 October 2009
every point is a boundary case
"...every point is a boundary case, or there is no boundary point at all — are both possible. In the first case the set is said to be dense in the space" — Really? The whole space is a (trivial) example of a dense set, but belongs to the latter case, not the former one. The condition "every point of the space is a boundary point of the given set" is equivalent rather to "the given set and its complement are both dense". Boris Tsirelson 09:18, 4 October 2009 (UTC)
Well, I understand that implication was meant, not equivalence. But, maybe, it is better to write "In the first case the set is said to be dense in the space (as well as its complement)." Boris Tsirelson 09:23, 4 October 2009 (UTC)
- Boris, thank you for pointing out my sloppy language. Peter Schmitt 22:21, 4 October 2009 (UTC)