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Why would $\inf A = 0$ if $A \cap [- \infty, \epsilon] \not = \emptyset$, $A \subset [-\epsilon,\infty]$? Where $\epsilon > 0$.

My attempt:

By assumption we know that $A$ is non-empty. We have a result:

$$A \subseteq B \implies \inf B \leq \inf A$$

So $$-\epsilon \leq \inf A$$


How can I argue further?


By def. of infimum. $s$ is infimum of $A$, if for every $\epsilon >0$ there exists $a \in A$ s.t. $a < s + \epsilon$.

This could suggest:

Let $a=-\epsilon$, then assume $\inf A = s$. Then must hold $a < s + \epsilon$. However, now it's possible that $s=-\epsilon$, since $a < -\epsilon + \epsilon$ is true.

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    $\begingroup$ What is $\epsilon$? Does the first statement hold for every positive $\epsilon$? $\endgroup$ – David Sep 9 '19 at 10:45
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If your $\varepsilon$ is fixed then the statement is false. Just take $A = [\varepsilon,\infty)$.

If however, you have $A \cap (-\infty,\varepsilon] \neq \emptyset$ for every $\varepsilon$ that means that there exists $x_\varepsilon \in A$ with $x_\varepsilon\leq \varepsilon$. Therefore, $\inf A \leq \varepsilon$ for every $\varepsilon>0$. Taking the limit as $\varepsilon \to 0$ we get that $\inf A \leq 0$.

The other implication comes from $A \subset [-\varepsilon,\infty)$ for every $\varepsilon$. This means that $\inf A \geq -\varepsilon$ for every $\varepsilon>0$. Taking $\varepsilon \to 0$ you obtain $\inf A \geq 0$.

Finally this gives you $\inf A = 0$.

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  • $\begingroup$ Okay, I didn't understand the difference between "fixed" and "all". $\endgroup$ – mavavilj Sep 9 '19 at 10:54
  • $\begingroup$ However, what you mean by "This means that $\inf A \leq 0$"? Where does this come from? From definition of $\inf$? $\endgroup$ – mavavilj Sep 9 '19 at 10:55
  • $\begingroup$ If for every $\varepsilon >0$ there is a smaller element in $A$ then $\inf A \leq \varepsilon$. Taking $\varepsilon \to 0$ gives $\inf A \leq 0$. $\endgroup$ – Beni Bogosel Sep 9 '19 at 11:35

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