# Prove there are open sets in the lower limit topology, that are not open in with the absolute value metric

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.

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 $[a,a+\epsilon)\subseteq\{a\}$.

• 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? – Error 404 May 13 '15 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. – Robert Wolfe May 13 '15 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. – Error 404 May 14 '15 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.

• Please consider my argument below @Bryan's answer. What is flaw in that? – Error 404 May 13 '15 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 '15 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. – Error 404 May 13 '15 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 '15 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 '15 at 15:28