Maths - Lower Bounds & Upper Bounds(Real Numbers)

Question

What is the upper bound of the infinite sequence 1/x^2, where xE{R\0}

We know that x={1, 1/4, 1/9, 1/16,.....}

Since all x here are greater than 0 or less than equal to 1.

Can't we take 1 as an upper bound here(My tutorial instructor has mentioned so in an answer sheet) I'm confused. Can someone explain? Thanks!

Answer 1

x is an element of real numbers, excluding zero

(-infinity,0) U (0,infinity) x= any real number except zero

1/x^2 is always positive, lower bound = 0, there is no upper bound, as 1/x^2 approaches infinity

maybe the problem was

x is an integer, x is an element of the set of integers, excluding zero

x is an element of Z/0 = set of all integers except 0

then x is in the set {..-3,.-2,-1, 1, 2, 3, ...}

1/x^2 is then ...1/9, 1/4, 1/1, 1, 1/4, 1/9, ...

then lower bound = 0 and upper bound = 1

all integers are real numbers

but infinity is not an integer or a real number and is excluded

infinity and -infinity are surreal numbers