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I want to prove that $(-\infty, a)$ is an open set. So if I label $U = (-\infty, a)$, I must prove that for all $x \in U$, there exists $\varepsilon > 0$ such that $(x-\varepsilon,x+\varepsilon$) is a subset of $U$.

My challenge is working out what $\varepsilon$ is. I believe I should take $\varepsilon = \min\{|a-x|,|x-a|\}$ - would this be a correct selection? I know there are a lot of different values I can pick, but would this be acceptable?

Ekene E.
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Dam
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    Hint: $(a, b)$ is open for any $a, b \in \mathbb R$ with $a < b$. Is this enough for you to finish the solution? – Ekene E. Jan 14 '24 at 13:41
  • @EkeneE. Sorry I am confused. I know what I wrote in the question isn't the full justification, I wanted to know if my value of $\epsilon$ is correct or wrong? I want to use this method rather than another method. – Dam Jan 14 '24 at 13:43
  • @stange No it doesn't. I know how to do the example for ($a$,$b$) already. It doesn't help me for the example I asked in my question. – Dam Jan 14 '24 at 13:53
  • If you know how to deal with finite intervals, then infinite intervals shouldn't be a problem. – stange Jan 14 '24 at 13:54

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Notice that if $x\in (-\infty, a)$, then $x < a$. Hence, $ε = |x - a| = |a - x| = a - x$. It's redundant to take $ε = \min \{|x - a|, |a - x|\}$, but beside that it's the correct answer.

Paulo Estêvão
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  • Thank you. I understand, I appreciate your response. – Dam Jan 14 '24 at 13:58
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    @Dam Also beware that every positive $\epsilon$ that satisfies $\epsilon\leq a-x$ can be used. So the observation that $(0,a-x]\neq\varnothing$ is enough already. – drhab Jan 14 '24 at 21:24