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Is is possible to have a bounded feasible region that does not optimize an objective function?
NO
How do you prove that if a real sequence is bounded and monotone it converges?
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|
What is an example of a function that is of bounded variation but not Lipschitz continuous?
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).
Is the standard square root function bounded?
If the range is the real numbers, it has a lower bound (zero) but no upper bound.
What point is infinity?
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.
