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So I know that sets of the form $(0,b)$ for $b<1$ are open in the $(0,1)_l,$ where $l$ denotes the lower limit topology. But my question is, why is the set say $(0,0.5)$ open in $(0,1)_l?$ Because from how I understand open sets, given any $a\in (0,0.5)$, we can find some $B_{\epsilon}(a)\subset (0,0.5)$.

Indeed this seems to be true, but what I have trouble seeing is that for any $x\in (0,1)_l,$ say $x=0$, then $x \notin B_{\epsilon}(a)\subset (0,0.5)$.

My question is, if we choose some $x$ in the set $R$ where the set $U$ is open in, does it have to hold $\forall x \in R$, such that $x\in B_{\epsilon}(x)\subset U$ and $x \in U?$

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  • $\begingroup$ What is $B_\varepsilon(a)$? $\endgroup$ – José Carlos Santos Apr 5 '18 at 6:18
  • $\begingroup$ That is the open ball with respect to $a \in U$, with $\epsilon >0.$ $\endgroup$ – Aurora Borealis Apr 5 '18 at 6:19
  • $\begingroup$ Ball?! With respect to what distance? $\endgroup$ – José Carlos Santos Apr 5 '18 at 6:22
  • $\begingroup$ So basically it is $|x-a|<\epsilon$, where $x,a\in U$ with $a $ fixed , in the lower limit topology. I am sorry I dont think I can make it any clearer from the knowledge that I have at this point in time. $\endgroup$ – Aurora Borealis Apr 5 '18 at 6:23
  • $\begingroup$ Are you saying that you cannot tell me what the set $B_\varepsilon(a)$ is? And what does it mean to add “in the lower limit topology” to the end of the sentence? $\endgroup$ – José Carlos Santos Apr 5 '18 at 6:26
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$[x,y)$ is open in $(0,1)_l$ when $0<x<y<1.$ And $(0,0.5)=\bigcup_{(0<x<0.5)}[x,0.5).$

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    $\begingroup$ The notion of an "open ball" in $(0,1)_l,$ if defined at all, should be used with caution, because the topology cannot be generated by a metric. $\endgroup$ – DanielWainfleet Apr 5 '18 at 19:36

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