NO
We prove that if an increasing sequence {an} is bounded above, then it is convergent and the limit is the sup {an }Now we use the least upper bound property of real numbers to say that sup {an } exists and we call it something, say S. We can say this because sup {an } is not empty and by our assumption is it bounded above so it has a LUB.Now for all natural numbers N we look at aN such that for all E, or epsilon greater than 0, we have aN > S-epsilon. This must be true, because if it were not the that number would be an upper bound which contradicts that S is the least upper bound.Now since {an} is increasing for all n greater than N we have |S-an|
How about f(x) = floor(x)? (On, say, [0,1].) It's monotone and therefore of bounded variation, but is not Lipschitz continuous (or even continuous).
If the range is the real numbers, it has a lower bound (zero) but no upper bound.
Infinity, by definition, is not a point. Infinite means unending, so it cannot be located at a point. The moment at which a function reaches infinity is when that function ceases to be bounded.
(0,1,0,1,...)
((-1)^n)
For the statement "convergence implies boundedness," the converse statement would be "boundedness implies convergence."So, we are asking if "boundedness implies convergence" is a true statement.Pf//By way of contradiction, "boundedness implies convergence" is false.Let the sequence (Xn) be defined asXn = 1 if n is even andXn = 0 if n is odd.So, (Xn) = {X1,X2,X3,X4,X5,X6...} = {0,1,0,1,0,1,...}Note that this is a divergent sequence.Also note that for all n, -1 < Xn < 2Therefore, the sequence (Xn) is bounded above by 2 and below by -1.As we can see, we have a bounded function that is divergent. Therefore, by way of contradiction, we have proven the converse false.Q.E.D.
Wrong answer above. A limit is not the same thing as a limit point. A limit of a sequence is a limit point but not vice versa. Every bounded sequence does have at least one limit point. This is one of the versions of the Bolzano-Weierstrass theorem for sequences. The sequence {(-1)^n} actually has two limit points, -1 and 1, but no limit.
They are divided by divergent, convergent AND transform boundaries.
NO
We prove that if an increasing sequence {an} is bounded above, then it is convergent and the limit is the sup {an }Now we use the least upper bound property of real numbers to say that sup {an } exists and we call it something, say S. We can say this because sup {an } is not empty and by our assumption is it bounded above so it has a LUB.Now for all natural numbers N we look at aN such that for all E, or epsilon greater than 0, we have aN > S-epsilon. This must be true, because if it were not the that number would be an upper bound which contradicts that S is the least upper bound.Now since {an} is increasing for all n greater than N we have |S-an|
A mountain belt bounded by cratons typically forms as a result of the collision of tectonic plates. The initial collision leads to the formation of a convergent plate boundary, where intense compression causes the crust to thicken and form mountain ranges. Over time, erosion and other geological processes continue to shape the landscape and maintain the mountain belt.
No. y = 1/x is continuous but unbounded.
What is the area bounded by the graph of the function f(x)=1-e^-x over the interval [-1, 2]?
Continent-continent convergenceThe ice age and the subsequent sliding and melting of glaciers account for the mountain belt bounded on either side by craters
How about f(x) = floor(x)? (On, say, [0,1].) It's monotone and therefore of bounded variation, but is not Lipschitz continuous (or even continuous).