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I have an example of a set

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I decided to use $f(a) = a + (2a)^{-1} \implies f'=1 -2(2a)^{-2} = 1-1/a^2=0 \iff a=\frac{\pm1}{\sqrt{2}}$

Now I realize the critical values aren't even in the set, but taking $f(a)$ at those values anyways I get

$f(1/\sqrt{2}) =\sqrt{2}$

$f(-1/\sqrt{2}) =0$

Now evaluating through the end points, we get

$f(0.1) = 51/10$

$f(5) = 51/10$

So although the critical values don't belong to $\mathbb{Q}$, we do get that

sup(B) = 51/10

inf(B) = 0

max(B)=min(B) = empty

1 Answers 1

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For all real $x$ in our interval, let $f(x)=x+(2x)^{-1}$.

You can see from the derivative that as $x$ travels from $0.1$ to $5$, $f(x)$ decreases, then increases.

The supremum (and maximum) of the set $B$ is therefore, by your calculations, $51/10$.

The negative $x$ at which the derivative is $0$ is irrelevant. The positive $x$ at which $f'(x)=0$ is very relevant.

In our interval, the function, if interpreted as being defined for all reals in the interval, attains a minimum at $x=1/\sqrt{2}$. But this is regrettably not rational.

Since there are rationals arbitrarily close to $1/\sqrt{2}$, the set $B$ has no minimum. But the infimum of $B$ does exist. It is obtained by evaluating $f(x)=x+(2x)^{-1}$ at $x=1/\sqrt{2}$.

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    Yes. Imagine for example looking at the decimal expansion of $1/\sqrt{2}$. Truncate after $20$ places to get $r$. Then $f(r)$ is very close to $f(1/\sqrt{2})$. Continue, truncating after $100$ places. The elements of $B$ are always $\gt f(1/\sqrt{2})$, but we can find elements of $B$ as close to $f(1/\sqrt{2}$ as we wish. So the infimum of $B$ is $f(1/\sqrt{2})$.2012-09-16