No, they can only be jump continuous.
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No. For example, y = 7 is monotonic. It may be a degenerate case, but that does not disallow it. It is not a bijection unless the domain and range are sets with cardinality 1. Even a function that is strictly monotonic need not be a bijection. For example, y = sqrt(x) is strictly monotonic [increasing] for all non-negative x. But it is not a bijection from the set of real numbers to the set of real numbers because it is not defined for negative x.
It is because the logarithm function is strictly monotonic.
No. For example, consider the discontinuous bijection that increases linearly from [0,0] to [1,1], decreases linearly from (1,2) to (2,1), increases linearly from [2,2] to [3,3], decreases linearly from (3,4) to (4,3), etc.
It is a square number but not a perfect square. The nearest perfect squares, on either side, are 4^2 = 16 and 5^2 = 25. Since there is no integer between 4 and 5 and the square is a strictly monotonic function, 18 cannot be a perfect square.
I Stictly told you not to do that. (strictly- strongly recommended)