The key is that the sets Ej are all disjoint. Let's suppose we just take two such sets. E1 and E2, disjoint, then E1 is a subset of the complement of E2 and E2 is a subset of the complement of E1. If both sets had countable complement, then the sets themselves would be countable (because they are subsets of these countable complements) but then their complement would be uncountable, which is a contradiction!
@@osamaarab6811 suppose E1 and E2 are of countable complement. Then since E1 is a subset of E2^c and E2 is subset of E1^c, E1 and E2 are countable (for being subsets of countable sets). The complement of a countable set is uncountable, then E1^c and E2^c are uncountable, which is a contradiction
Maam can i get the solution for q Let X be non empty set Measure (A) = 0 ( if A is countable) Measurre (B ) = 1( if A is uncountable) Check for Outer Measure?
We can not take the case Ai's are countable except two uncountable sets A1 and A2 since the measure of the union will still 1 but the sum will be 2 so we must take one uncountable set not two .. ok why ? Is the reasons are the formula of the segma algebra & Ai's are disjoint ? I proved this using contapositive l said let Ai's are countable except two sets A1 and A2 where their completens are countable.. Since Ai's are disjoint then A1 is a subset of the completent of the union of Ai's (with A2) And if we take the complement of each side then we will replace each side I mean the union will be subset of the complement of A1 wich is a contradiction ( uncountable subset of countable is a contradiction)... the union is uncountable since it has A2 which is uncountabl and the complement of A1 is a countable from the formula of the segma algebra ..ok is it a logic reason or proof ? So there exist no case which say Ai's are countable except two ? If yes what about if we take all A are uncountable? .. we will have the same problem What about this case ?
Hi, exactly as you say in your comment and what I say in the video. There can only be at most one index j for which Aj is uncountable. So the only two possible cases are - All Aj are countable - All but one Aj are countable (and that other one is of course uncountable) So the case that you mention, with two uncountable sets is not a valid option, it never happens!
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I want to thank you about your great videos
I found many things that are not found anywhere else only here ❤
Thanks! I'm so glad you find it helpful! 🥰
I can't understand why there exist only one element with countable complement 💔
The key is that the sets Ej are all disjoint. Let's suppose we just take two such sets. E1 and E2, disjoint, then E1 is a subset of the complement of E2 and E2 is a subset of the complement of E1. If both sets had countable complement, then the sets themselves would be countable (because they are subsets of these countable complements) but then their complement would be uncountable, which is a contradiction!
@@Problemathic why their complement would be uncountable?😶
@@osamaarab6811 suppose E1 and E2 are of countable complement. Then since E1 is a subset of E2^c and E2 is subset of E1^c, E1 and E2 are countable (for being subsets of countable sets). The complement of a countable set is uncountable, then E1^c and E2^c are uncountable, which is a contradiction
@@Problemathic Yes I understand now . Thank you for your videos. I shared it with my friends❤
@@osamaarab6811 Great! Thanks 🥰
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Maam can i get the solution for q
Let X be non empty set
Measure (A) = 0 ( if A is countable)
Measurre (B ) = 1( if A is uncountable)
Check for Outer Measure?
@@BabarAli-jv4fc It's here! th-cam.com/video/ukje21qhx58/w-d-xo.html
🫣❤️👆🙏😢@@Problemathic
We can not take the case
Ai's are countable except two uncountable sets A1 and A2 since the measure of the union will still 1 but the sum will be 2
so we must take one uncountable set not two .. ok why ?
Is the reasons are the formula of the segma algebra & Ai's are disjoint ?
I proved this using contapositive l said let Ai's are countable except two sets A1 and A2 where their completens are countable..
Since Ai's are disjoint then A1 is a subset of the completent of the union of Ai's (with A2)
And if we take the complement of each side then we will replace each side I mean the union will be subset of the complement of A1 wich is a contradiction ( uncountable subset of countable is a contradiction)... the union is uncountable since it has A2 which is uncountabl and the complement of A1 is a countable from the formula of the segma algebra ..ok
is it a logic reason or proof ?
So there exist no case which say Ai's are countable except two ?
If yes what about if we take all A are uncountable? .. we will have the same problem
What about this case ?
Hi, exactly as you say in your comment and what I say in the video. There can only be at most one index j for which Aj is uncountable. So the only two possible cases are
- All Aj are countable
- All but one Aj are countable (and that other one is of course uncountable)
So the case that you mention, with two uncountable sets is not a valid option, it never happens!
@Problemathic
Thank you so much..lots of love ❤️❤️❤️