Why us an+1 - an strictly greater than zero? Could the n+1 term in the x sequence and y sequence not be equal to each other and thus give us a modulus of zero?
I guess in the specific example he gave, if bn is the sequence on the right then the inequality is indeed strict, since |xn-yn|/(1+|xn-yn| < 1 always, hence every term in the sequence on the left will be strictly smaller than every term on the right. He should have been more precise.
Epsilon should be kept the same for the contradiction, and just reverse "there exist" in front of N by "for all", instead of "not exist", and at the end, reverse the inequality |an - supT| should be greater or equal than epsilon
![Open Balls, Closed Balls and Spheres in C[0,1]](http://i.ytimg.com/vi/TSRu095deB4/mqdefault.jpg)
Why us an+1 - an strictly greater than zero? Could the n+1 term in the x sequence and y sequence not be equal to each other and thus give us a modulus of zero?
I don't think you should use i in the metric defitinion because it confused me between index and imaginary number
Exactly
I swear to god I was like: "ooook I guess we have an esponential complex function to start with"
At 15:56 an < bn implies that lim an
I guess in the specific example he gave, if bn is the sequence on the right then the inequality is indeed strict, since |xn-yn|/(1+|xn-yn| < 1 always, hence every term in the sequence on the left will be strictly smaller than every term on the right. He should have been more precise.
although conclusion is true, your contradiction proof was wrong
Thank you!
Epsilon should be kept the same for the contradiction, and just reverse "there exist" in front of N by "for all", instead of "not exist", and at the end, reverse the inequality |an - supT| should be greater or equal than epsilon
it is just the same.
2:30 what exactly is that formula called?
Have you found out what it is called?
You are so rigorous sometimes I just have to skip forward :-P But good videos non the less.