# Why would $\inf A = 0$ if $A \cap [- \infty, \epsilon] \not= \emptyset$, $A \subseteq [-\epsilon,\infty]$?

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.

• What is $\epsilon$? Does the first statement hold for every positive $\epsilon$? – David Sep 9 '19 at 10:45

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 $$\inf A = 0$$.

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