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Let A CRbe bounded set . Prove that there exists sequence in A that converges to Sup...

Question

Let A CRbe bounded set . Prove that there exists sequence in A that converges to Sup

Let A CRbe bounded set . Prove that there exists sequence in A that converges to Sup



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Prove that a bounded non increasing sequence converges to its greatest lower bound.

Heil here we have to prove that the sequence in in the real numbers has no convergence subsequent. If I don't leave more of A N tends to infinity. Okay then we can write limit endurance, infinity model A. N is equal to infinity and which means belongs to real numbers. So there exist a number and zero. Such that for all and uh such that more a greater than more plus one for all for our and Later than end zero. Okay then we can say mode and minus uh greater than or equal to more A N minus more uh separately and which is greater than one. Okay. And for Engadget and zero. Yeah. Yeah. And by this we can say it in place that this weekend weekends of the form A. M. We're in created than or equal to one cannot have a subsequent convergent *** Virgin to a. Okay, there is no convergence, pence, crude. Thank you.

Us to prove that a sequence converges to zero if and only if the sequence of absolute values converges to zero. So let and be some sequence of real numbers and first performed direction. Suppose that limits as n approaches infinity on a NZ border zero and let Epsilon be some positive number. Then we confined natural number and such that a n in absolute value is going to be less than Epsilon for all and greater than end. Then it follows that the absolute value of the absolute value of a N minus zero is the same as the absolute value of the absolute value of a N, which is the same as simply the absolute value of a in which we have is less than Epsilon Ferral and greater than end. Therefore, we have that sequence the and converges to zero proven one direction and now prove the other direction. Suppose that the limit is an approaches infinity on the absolute value of a N equals zero and let Epsilon. This is one way to look at it. You can use another previous theorem, however, so notice that a N is going to be less than or equal to the absolute value of a N, which is going to be greater than or equal to the opposite of the absolute value of am for all natural members in. And we have that limit as any purchase. Infinity on negative AM is also equal to zero. And so, by the sandwich, the room for sequences we have the limits of the sequence A n exists and is equal to the same as the other two limits zero.


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