>>11856223as others have said, the ability to choose such a number is a secondary property of the function being continuous.
It is a big result relating to the notion of continuous functions being bounded on any compact set. The compactness is really important here (set closed and bounded)
1/x is continuous on (0, inf) because from the definition of the continuity (lim of a function as x -> a is f(a)) and the fact that the interval is open (no matter how close to 0, we will find that that lim of 1/x as x-> a is 1/a, because 0 is not included in (0, inf)).
and for the pic related: you need to go back to the chapter on real numbers and least upper bounds if you find this confusing. f on [0,1] is x^2 and suddenly jumps to 0 at 1. so its not continuous at 1, limit at 1 is not f(1) because f(1) is 0, and the limit is 1. As for the second notion that there is no f(y) that bounds every f(x) on [0,1], it follows from the properties of real numbers that you can choose real number z as close as you like to 1, that have f(z)= z^2, always. If f(1) was 1, then it would be continuous on [0,1] because the limit would be f(1), but f is 0 on 1 so you cannot choose f(1) as stated. You cannot choose something smaller than f(1) that would bound every f(x) in the interval, because f is increasing as x is increasing and from the properties of real numbers it follows that as soon as you choose some number close to 1 but smaller, you could find another one closer to 1, so f(another number) would be bigger that the bound you have found.
