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This is the question:

Which of these sets are open sets on the lower limit topology on $\mathbb{R}$, whose basis elements are $[a,b),a<b$? $$[4,5)\qquad\left\lbrace3\right\rbrace\qquad [1,2]\qquad(7,8)$$

The first one clearly is open, as it belongs to the basis. The secon one can't be, as all basis elements have more than one element, and so no union of basis elements can have only one element. The fourth can be built by:


The third one, I can't prove it's not (it's clearly not open). I'm trying something like this:

If it was open, it would be the result of union of sets like $[1,a)$. I can't prove though, that the union of those sets can't have a closed limit.

EDIT: Could it be: The union of those sets is defined by setting a set of values for $a$. $a\in\mathbb{R}$, so there must be a maximum $a$: $a_{max}$, and so the final result of the union would be $[1,a_{max})$. That works if the set of values for $a$ is finite. If it's not, then if it's bounded the same thing works, if its unbounded the set would be $[1,\infty)$

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The lower limit topology on $\mathbb R$ is strictly finer than the Euclidean topology on $\mathbb R$, as you figured out by your construction of $(7,8)$. If $[1,2]$ were open, then its complement would be closed. Can you reach a contradiction? – Ian Coley Mar 22 '13 at 1:07
It seems like $[1,2]$ should be closed since it's complement $(-\infty,1) \cup (2,\infty)$ is open. I'm not sure thought. – Wintermute Mar 22 '13 at 1:08
@FrankMcGovern You opened my eyes :) Thanks! – MyUserIsThis Mar 22 '13 at 1:31
@MyUserIsThis No problem! Glad to help. – Ian Coley Mar 22 '13 at 1:32
up vote 1 down vote accepted

If the set $[1,2]$ was open, then $2$ would have an open neighborhood $[a,b)$ such that $2\in[a,b)\subset[1,2]$. Now take any element $p$ such that $2<p<b$. It belongs to $[a,b)$, hence to $[1,2]$, and is greater than $2$...

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Thanks for the different approach. Really simple proof. Thanks again! I'm accepting this answer (all 3 were just perfect, there's no really reasonable reason for that) – MyUserIsThis Mar 22 '13 at 1:38

Note that the union of intervals of the form $[a,b)$ is never $[x,y]$. To see this, first note that it is sufficient to prove this fact for increasing unions of $[a,b_i)$.

Suppose now that $b_i$ is a strictly increasing sequence of real numbers, and $a<b_0$, then $(a,b_i)$ are open sets in the Euclidean topology, and so $\bigcup(a,b_i)=(a,\sup b_i)$ and it is easy to see that $\bigcup[a,b_i)=\{a\}\cup\bigcup(a,b_i)$.

Similarly $\{3\}$ is not open either, because it is the interval $[3,3]$.

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Thanks, so that's really what I did in my EDIT part of the question, plus for unbounded sequences of $b_i$. – MyUserIsThis Mar 22 '13 at 1:36

If $[a, b]$ is open in the lower limit topology, it is expressible as a union of sets of the form [c, d). Clearly no $[c, d)$ where $d > b$ can be in the union or else the union would contain elements not in [a, b]. So the sets must be of the form $[c, d)$ where $d \leq b$. But then the union would not contain $b$, and so $[a, b]$ cannot be expressed as a union of basic open sets. Hence $[a, b]$ is not open.

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For {3} I would I like to add, this is not open in lower limit topology on R. Because if it is open then it is arbitary union of intervals like [a,b). And this is not possible because [a,b) is either empty or infinite, how can there union be a finite non empty set. Same Argument shows no finite set(except empty set) is open in lower limit topology on R.

I think i don't need to prove all open sets in lower limit topology can be written as arbitary union intervals like [a,b). (Because these form a basis for topology)

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