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et f : R → R be a function, and consider the two setsX = {f (x)2 : x ∈ R}, Y = {f (x2) : x ∈ R}.Show that if X has a supremum, then Y has both a supremum and an infimum.Give a counterexample to show that the converse is false (that is, Y can havea supremum and infimum, even if X has no supremum)

Question

et f : R → R be a function, and consider the two setsX = {f (x)2 : x ∈ R}, Y = {f (x2) : x ∈ R}.Show that if X has a supremum, then Y has both a supremum and an infimum.Give a counterexample to show that the converse is false (that is, Y can havea supremum and infimum, even if X has no supremum)

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Solution

First, let's prove that if X has a supremum, then Y has both a supremum and an infimum.

Assume that X has a supremum, say sup(X). This means that for all x in R, f(x)^2 ≤ sup(X). Taking the square root of both sides (since the square root function is increasing), we get |f(x)| ≤ sqrt(sup(X)).

Now consider the set Y. For all x in R, -sqrt(sup(X)) ≤ f(x^2) ≤ sqrt(sup(X)). This shows that Y is bounded above by sqrt(sup(X)) and below by -sqrt(sup(X)). Hence, Y has both a supremum and an infimum.

Now, let's give a counterexample to show that the converse is false. Consider the function f(x) = x for all x in R. Then X = {x^2 : x in R} has no supremum in R, but Y = {x^2 : x in R} has both a supremum and an infimum in R (namely, 0 and +∞). This shows that Y can have a supremum and infimum, even if X has no supremum.

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