Timeline for relation between inclusion and embedding [closed]
Current License: CC BY-SA 2.5
17 events
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| Apr 1, 2011 at 15:48 | comment | added | Syang Chen | I guess you are working with function spaces and find someone deduce bouondedness of operators in this way. Roughly, for function spaces defined by integral, this is always true (consider pointwise convergent subsequences and apply closed graph theorem). | |
| Mar 31, 2011 at 23:35 | history | edited | Shaoming Guo | CC BY-SA 2.5 |
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| Mar 31, 2011 at 22:05 | comment | added | Yemon Choi | The new version of this question is still ill-defined. What does one mean by ``if $a\in X$ then $a\in Y$? This does not make sense without first embedding $X$ and $Y$ into some common set. | |
| Mar 31, 2011 at 21:38 | history | edited | Shaoming Guo | CC BY-SA 2.5 |
added 141 characters in body; edited tags
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| Oct 21, 2010 at 21:11 | comment | added | Mariano Suárez-Álvarez | I wish the question were enlarged with definitions of the terms! | |
| Oct 21, 2010 at 21:06 | history | edited | Willie Wong |
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| Oct 21, 2010 at 20:05 | comment | added | Nate Eldredge | @Andrew Stacey and others: It is common in the literature of this field to use the term "embedding" for "continuous injection", instead of the perhaps more logical meaning of "homeomorphism onto its image". Indeed, one often talks about one Banach space being "densely embedded" into another, which would be absurd under the latter sense. I am not sure where this usage originated; it is perhaps unfortunate but it is standard now. | |
| Oct 21, 2010 at 20:02 | comment | added | Nate Eldredge | @Shaoming: perhaps what you want to ask is: if a map $i : X \to Y$ (thought of as "inclusion") is injective, must it be continuous? The answer is no; indeed, with the axiom of choice, one can show there exists a bijective linear map of $X$ to itself which is not continuous. | |
| Oct 21, 2010 at 18:20 | history | closed |
Bill Johnson Andrew Stacey Martin Brandenburg Andrey Rekalo Yemon Choi |
not a real question | |
| Oct 21, 2010 at 18:17 | answer | added | Dick Palais | timeline score: 2 | |
| Oct 21, 2010 at 17:38 | comment | added | Andrew Stacey | I would have gone for "inclusion" meaning "continuous injection" and "embedding" meaning "isometric injection". "Embedding" for me carries the overtone that all the structure on the sub-object is inherited from the ambient one. I agree with Bill that it is not well-posed. | |
| Oct 21, 2010 at 16:42 | comment | added | Shaoming Guo | sorry, the definition is just what Wilie Wong said. | |
| Oct 21, 2010 at 16:36 | comment | added | Bill Johnson | This question is not well posed, so I vote to close. | |
| Oct 21, 2010 at 16:15 | comment | added | Willie Wong | (But looking at Wikipedia, I may be the only one...) | |
| Oct 21, 2010 at 16:12 | comment | added | Willie Wong | I don't know if this is common place, but I generally consider 'inclusion' as in set relations, and 'embedding' being a continuous inclusion (that $\|u\|_Y \leq C\|u\|_X$ for some fixed constant $C$). | |
| Oct 21, 2010 at 15:54 | comment | added | Mariano Suárez-Álvarez | Is there a technical definition of 'embedding' in the context of Banach spaces? | |
| Oct 21, 2010 at 15:36 | history | asked | Shaoming Guo | CC BY-SA 2.5 |