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$E_2$ is the closed semi-infinite interval $[0,\infty)$ in the set of real numbers $\mathbb{R}.$

Find a continuous function $g_2: E_2 \to \mathbb{R}$ which is bounded, but which does not attain a maximum value.

I guessed $g_2= \sum_0^n \frac{x}{2^n} $ for $x>0$ but this does not work for $0.$ And I am not sure if this is a proper function which satisfies the condition. Thank you.

Best regards,

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  • $\begingroup$ Do you mean $E^2$ (${} = E \times E$), $E_2$, or something else, when you write "$E2$"? $\endgroup$ – Eric Towers Nov 21 '17 at 3:28
  • $\begingroup$ E2 is just $E_2$ it's not E^2 $\endgroup$ – Kaan Yolsever Nov 21 '17 at 3:29
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$x\rightarrow \frac{x+n}{x+n+1}$ where $n\in \mathbb N$ .

for each choice of n you can get one function.

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How about the function defined by

$$x \mapsto \frac{x}{|x|+1}$$

which is bounded below by $0$ and above by $1$, yet never achieves a maximum?

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THere are several functions: $-e^{-x}$, $\arctan(x)$,$-1/x$, plus scaling, adding constants, and applying compositions...

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  • $\begingroup$ E2 is $E_2$ not $E^2$ $\endgroup$ – Kaan Yolsever Nov 21 '17 at 3:29
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  • $g_2(x) = 1-\frac{1}{x+1}$.
  • $g_2(x) = \tan^{-1}(x)$.
  • $g_2(x) = -\mathrm{e}^{-x}$.
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  • $\begingroup$ is $1-1/x$ continuous at 0 with domain $[0,\infty]$? $\endgroup$ – Kaan Yolsever Nov 21 '17 at 17:23
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    $\begingroup$ @KaanYolsever : Good catch. Thanks! Fixed. $\endgroup$ – Eric Towers Nov 21 '17 at 18:25

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