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Apr 22, 2016 at 12:28 comment added fosco I am a child indeed (-:
Apr 22, 2016 at 12:11 vote accept fosco
Apr 22, 2016 at 12:11 comment added fosco Ah, now I see the edit; nice. I think I'll try to do the exercise, do you have an answer for it?
Apr 22, 2016 at 9:49 history edited Paul Taylor CC BY-SA 3.0
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Apr 22, 2016 at 7:55 history edited Paul Taylor CC BY-SA 3.0
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Apr 21, 2016 at 21:53 comment added fosco (a rapid check after: I'm convinced by your argument; the same proof of yours seems valid in case $R\subset M_1$ ,using $(L,R)$ instead of $(L', R')$, simply because in that case $r,l\in E_0\cap M_1$)
Apr 21, 2016 at 21:37 comment added fosco I'm not able to see why you define $L', R'$. I also actually realized that the generality I need is $E_1\subset L\subset E_0$, I think that in this case there are simpler proofs. Anyways, your general argument is neat and elegant. Thanks! If you spend two more words on the definition of $L', R'$ I'll be happy to accept the answer.
Apr 21, 2016 at 17:56 history edited Paul Taylor CC BY-SA 3.0
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Apr 21, 2016 at 17:24 history edited Paul Taylor CC BY-SA 3.0
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Apr 21, 2016 at 16:57 history edited Paul Taylor CC BY-SA 3.0
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Apr 21, 2016 at 16:33 history answered Paul Taylor CC BY-SA 3.0