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I have searched a lot of related with my question but I did not find anything useful and have decided to ask it. So please do not duplicate my question!

Question: How it follows that there is this form of the interval, namely $(c,d]$?

My approach: Since $c\in B_0$ and $B_0$ is open in $[a,b]$ then there is basis element $V$ of subspace topology of $[a,b]$ such that $c\in V\subset B_0$.

Since order topology on $L$ is generated by intervals $(s,t)$ or $(s,t_0]$ (if $L$ has maximal element $t_0$) or $[s_0,t)$ (if $L$ has minimal element $s_0$). Then typical basis element of subspace topology of $[a,b]$ will be intersection of $[a,b]$ with one above types, right?

In further reasoning we will consider that $a<c<b$.

Case 1. If $V=(s,t)\cap [a,b]$ and $c\in V$ then the following two cases are possible:

1.1 If $a<e<c$ then by definition of linear continuum we can find $d$ such that $e<d<c$ then $(d,c]\subset B_0$.

1.2. If $e\leq a<c$ then we can take $d$ such that $a<d<c$ and $(d,c]\subset B_0$.

Case 2. If $V=(s,t_0]\cap [a,b]$ and taking into account that $t_0$ is maximal element of $L$ then two cases are possible:

2.1 If $a<s<c$ then take $d$ such that $s<d<c$ and $(d,c]\subset B_0$

2.2. If $s\leq a$ then then just take $d$ such that $a<d<c$ and $(d,c]\subset B_0$.

The third case is almost the same.

Remark: The case when $c=b$ is almost the same and even easier I guess.

I would like to know is my reasoning correct? Would be very grateful for any help! enter image description here

RFZ
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  • I think everything you have said is true except for maybe a few typos with which letters you are using. However, the argument can be simplified. The author notes that the subspace topology on $[a,b]$ is the same as the order topology (this is because $[a,b]$ is convex). Since $c$ is in $B_0$ and $B_0$ is open, by definition of order topology, there is some interval or open ray of $[a,b]$ inside $B_0$ which contains $c$. But then $(d,e]$ will be a subset of this interval or ray where $d$ is the left endpoint of the interval or ray. – Anonymous Sep 28 '19 at 03:20

1 Answers1

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The reasoning is simpler than you make it out to be, following the quoted Munkres proof at the end: If $c \in B_0$ we have two cases: $c=b$ or $a < c < b$. I think that is perfectly clear.

So $c$ could be of two types: the max of $L$ (if it exists at all), and then it has (basic, also subbasic) neighbourhoods (in $L$) of the form $(t,b]$ for all $t < b$, or not $\max(L)$ and then it has basic (not subbasic) neighbourhoods of the form $(t,u)$ with $t < c < u$.

In $[a,b]$ it thus has basic neighbourhoods of the form $(t,c] \cap [a,b]$ (in the former case) or $(t,u) \cap [a,b]$ in the second case. In both cases, $(t,c]$ is a subset of such a neighbourhood, and that is the underlined statement (which uses that $B_0$ is open in $[a,b]$ and so contains a basic neighbourhood of $c \in B_0$ which stays inside $B_0$) and so $(t,c] \subseteq B_0$ for some $t < c$ for either case. So it suffices to take $d=t$.

That's all the case distinguishing that's needed.

Henno Brandsma
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  • Thanks a lot for your answer! But let me ask you the following question: Why $(t,c]\subset (t,u)\cap [a,b]$? What if $t<a$ then in this case it is not a subset, right? Just take an element $x\in (t,c]$ such that $t<x<a$. – RFZ Sep 28 '19 at 14:03
  • @ZFR $(t,u)$ is chosen so that $(t,u) \cap [a,b]$ is a neighbourhood of $c$ which is $\sup A_0$ so certainly in $[a,b]$, so $t < a$ is not something that can happen. – Henno Brandsma Sep 28 '19 at 21:13
  • Dear Henno! To be honest I still cannot understand why $t<a$ cannot be the case? How it follows from $c=\sup A_0$? – RFZ Sep 28 '19 at 21:27
  • @ZFR $t$ is not arbitrary! It is chosen so that $(t,u)$ is contained in $B_0$ (which is a subset of $[a,b]$). Read carefully. Also $a \in A_0$ and $c = \sup A_0 \ge a$ and $t$ lies in between somewhere. – Henno Brandsma Sep 28 '19 at 21:30
  • I know that $t$ is NOT arbitrary! It is chosen so that $(t,u)\cap [a,b]\subset B_0\subset [a,b]$. I am reading carefully but you are writing is not so obvious. – RFZ Sep 28 '19 at 21:35
  • I cannot see in your arguments why $t<a$ gives contradiction. Could you show it please? – RFZ Sep 28 '19 at 21:42
  • @ZFR e.g. Look at the figure 24.1 on the next page. $t=d$ (remember?). $d < a$ does not happen. – Henno Brandsma Sep 28 '19 at 21:59
  • Sorry but you did not explain why it cannot happen? You just claiming that it is not possible. And the picture just showing it without any explaining – RFZ Sep 28 '19 at 22:02
  • Let us continue this discussion in chat. – Henno Brandsma Sep 28 '19 at 22:23