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What would be an example of an open set in the lower-limit topology that isn't open with the absolute value metric over the real numbers.

Further, how would I show that the lower limit topology is not a discrete topology?

I'm assuming that for the discrete topology, I could show that there is no singleton set in the lower-limit topology, hence it can't be discrete.

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2 Answers 2

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Sets of the form $[a,x)$ are open in the lower-limit topology by definition (at least in mine). You can show that they are not open in the metric topology as there is no neighborhood about $a$ such that $(a-\epsilon,a+\epsilon)\subseteq [a,x)$ for any $\epsilon>0$.

To show that the topology is not discrete, you can show that singletons are not open. And this is true because there is no $\epsilon>0$ such that $[a,a+\epsilon)\subseteq\{a\}$.

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By archemedian property there will exist $n_0 \in \Bbb N$ such that $[a+1/n,b)$ holds for all $ n \ge n_0 $. Now, if we take $ \cup_{n \ge n_0} [a+1/n,b)$ = $(a,b)$. Hence $(a,b)$ is open in lower limit topology on $\Bbb R$. By this we further can get that all sets in $\Bbb R$ are open as well as closed. That means lower limit topology is the discrete topology. Are these arguments right? – Vikrant Desai May 13 at 11:14
@VikrantDesai Maybe you meant to say that if $a<b$ then there is an $n_0$ such that $a+1/n<b$ for all $n\geq n_0$. Then you indeed get that $\cup_{n\geq n_0}[a+1/n,b)=(a,b)$. It is indeed true that the Euclidean topology is contained in the lower-limit topology by this reasoning. But this certainly does not imply that the lower-limit topology is the discrete topology. – Bryan May 13 at 15:51
I found that I was misusing the arbitrary union criterion and that's why landed on getting lower limit topology being equal to discrete topology. – Vikrant Desai May 14 at 9:01

Hint: What are the basic open sets of the lower limit topology? Can you find a point of such a set that is not interior to this basic set in the absolute value-induced topology?

You do indeed want to show that no singleton is open in the lower limit topology.

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Please consider my argument below @Bryan's answer. What is flaw in that? – Vikrant Desai May 13 at 11:17
Well, first of all, what do you mean when you say "$[a+1/n,b)$ holds for all $n\ge n_0$"? How can a set hold? Second of all (and most importantly), what do you mean when you say "By this we further can get that all sets in $\Bbb R$ are open as well as closed"? All you've shown is that open intervals are unions of half-open intervals. Why should it follow that all subsets of $\Bbb R$ are both open and closed? – Cameron Buie May 13 at 11:45
First, I wanted to eliminate the case where $ a+1/n $ will be greater that $ b$ so I used Archimedian property to get an $ n_0 $ such that $ [a+1/n,b) $ is an interval. Secondly, I showed that intervals of the type $(a,b)$ are open. Hence $[a,b] = (- \infty,a) \cup (b,infty) $. hence [a,b] is closed. now consider $ \cap_{n \in \Bbb N} [a-1/n,b+1/n] = (a,b) $ hence $(a,b)$ is closed also because arbitrary intersection of closed sets is closed(?). Then every singleton set will be complement of two open intervals which are closed as I got above. which implies all sets are both open and closed. – Vikrant Desai May 13 at 11:57
You are very welcome. Incidentally, the Archimedean property will be needed to prove that $(a,b)=\bigcup_{n\in\Bbb N}[a+1/n,b),$ but not the way you're using it. It's not a big deal which (if any) $[a+1/n,b)$ are empty. First, just note that if $a+1/n\le x,$ then since $0<1/n,$ we have $a<a+1/n\le x,$ and so $a<x.$ Hence, $[a+1/n,b)\subseteq (a,b)$ for all $n\in\Bbb N,$ and so $\bigcup_{n\in\Bbb N}[a+1/n,b)\subseteq(a,b).$ – Cameron Buie May 14 at 15:22
On the other hand, if $a<x,$ then by Archimedean property, there is some $n_0\in\Bbb N$ such that $a+1/n_0<x.$ (Can you see why?) Thus, if $x\in(a,b),$ we have $x\in[a+1/n_0,b)$ for some $n_0\in\Bbb N,$ and so $x\in\bigcup_{n\in\Bbb N}[a+1/n,b).$ Hence $(a,b)\subseteq\bigcup_{n\in\Bbb N}[a+1/n,b),$ and so $(a,b)=\bigcup_{n\in\Bbb N}[a+1/n,b),$ as desired. Observe in particular that this proof works even when $b\le a$! In that case, we're simply saying that the union of countably-many empty sets is still empty. – Cameron Buie May 14 at 15:28

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