10-327/Classnotes for Thursday November 4: Difference between revisions

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Question: Regarding the proof of the Lebesque Number Lemma, let T(x) be the function we were working with in the proof. I am confused with how we reached the conclusion that if d(x,y)<E, then T(y)>= T(x) - E. I know that it was said that this is just an application of the triangle inequality, but I am having a bit of trouble seeing that. Hopefully someone can make this point a bit clearer for me. Thanks! Jason.
* Question: Regarding the proof of the Lebesque Number Lemma, let T(x) be the function we were working with in the proof. I am confused with how we reached the conclusion that if d(x,y)<E, then T(y)>= T(x) - E. I know that it was said that this is just an application of the triangle inequality, but I am having a bit of trouble seeing that. Hopefully someone can make this point a bit clearer for me. Thanks! Jason.
** If you could find a ball of radius 7 around <math>x</math> which fits inside some set <math>U</math>, and you move <math>x</math> just a 1 unit away to <math>y</math>, then by the triangle inequality the ball of radius 6 around <math>y</math> is entirely contained inside the ball of radius 7 around <math>x</math> so it is entirely contained in <math>U</math>. [[User:Drorbn|Drorbn]] 18:37, 6 November 2010 (EDT)

Revision as of 17:37, 6 November 2010

See some blackboard shots at BBS/10_327-101104-142342.jpg.

Dror's notes above / Student's notes below
  • Question: Regarding the proof of the Lebesque Number Lemma, let T(x) be the function we were working with in the proof. I am confused with how we reached the conclusion that if d(x,y)<E, then T(y)>= T(x) - E. I know that it was said that this is just an application of the triangle inequality, but I am having a bit of trouble seeing that. Hopefully someone can make this point a bit clearer for me. Thanks! Jason.
    • If you could find a ball of radius 7 around which fits inside some set , and you move just a 1 unit away to , then by the triangle inequality the ball of radius 6 around is entirely contained inside the ball of radius 7 around so it is entirely contained in . Drorbn 18:37, 6 November 2010 (EDT)