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Find with proofs the infimum, supremum, maximum and minimum of the following sets or prove non-existence.

$$E = \{x \in \mathbb{R} - \{0\}: x < \frac{1}{x}\}$$

I do not know even how to think, Could anyone help me please?

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  • $\begingroup$ This looks suspiciously close to the question asked here by a different username, MathLover... $\endgroup$ Mar 30, 2017 at 6:09
  • $\begingroup$ How are the two questions similar? @BenjaminDickman $\endgroup$
    – user426277
    Mar 30, 2017 at 6:16
  • 1
    $\begingroup$ Essentially the same phrasing; exactly the same tags. $\endgroup$ Mar 30, 2017 at 6:18

2 Answers 2

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Solve that condition: $x < \frac{1}{x}$ equates to $\frac{x^2 - 1}{x} < 0$. This can only hold under two conditions:

  • $x^2 - 1 < 0$ and $x > 0$, which is equivalent to $0 < x < 1$;
  • $x^2 - 1 > 0$ and $x < 0$, which is equivalent to $x < -1$.

So $E = (-\infty,-1) \cup (0,1)$.

I'll leave it to you to find inf, sup, max, min.

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Since $x$ can be negative or positive, we know any minimum or infimum will occur for $x < 0$, and any maximum or supremum will occur for $x > 0$. And so let us tackle these two cases:

If $x < 0$, then $x < 1/x$ is equivalent to $x^2 > 1$.

Clearly there is no minimum (or infimum) since you can let $x \rightarrow - \infty$ be as negative as you want.

If $x > 0$, then $x < 1/x$ is equivalent to $x^2 < 1$.

There is no maximum here, since you can let $x \rightarrow 1$ from below with arbitrary precision, but it cannot actually equal $1$. Still, this does allow $1$ to serve as the supremum of the set in question.

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