add a non-LOTS subspace of R

[StevenClontz]

Continuing from #644, here I add the space $(0,1)\cup{2}\subseteq\mathbb R$ which is not a LOTS.

I think the proof that this space is not a LOTS is beyond the scope of a contribution to $\pi$-Base without a reference, so I'll ask the question on Math.SE. https://math.stackexchange.com/questions/4915404/

[prabau]

fyi, to properly display braces in the github comment, one has to use double backslashes. So $\\{2\\}$.

[StevenClontz]

fyi, to properly display braces in the github comment, one has to use double backslashes. So $\\{2\\}$.

Yes, but I hate that and hope that one day GitHub just fixes their renderer so my laziness will one day pay off. 😤

[prabau]

Hmm, that mathse question is too general. It does not even mention that space in question, so it may be premature to use it as justification.

[StevenClontz]

I'm adding an appropriate answer now

[prabau]

This exact example is explained in https://en.wikipedia.org/wiki/Order_topology (section "Example of a subspace of a linearly ordered space whose topology is not an order topology"). Not sure if the justification is complete or could be made shorter (last time I looked at it was a while ago). But it may be a good exposition to compare with, in addition to the comments by @Jianing-Song.

[Jianing-Song]

This exact example is explained in https://en.wikipedia.org/wiki/Order_topology (section "Example of a subspace of a linearly ordered space whose topology is not an order topology"). Not sure if the justification is complete or could be made shorter (last time I looked at it was a while ago). But it may be a good exposition to compare with, in addition to the comments by @Jianing-Song.

At least the proof is correct: It essentially uses the fact that dense and bounded complete imply connected. I've largely simplified the proof, please check if it looks OK! :)

[prabau]

Sorry, I have to admit I am still a little confused in the details about both Steven's proof in mathse and Jianing's revised wikipedia proof. I have added some comment for the mathse proof, but it's maybe better if I add some here.

In the proof of the lemma, for checking the "least upper bound property", one considers sets $S$ that have an upper bound. But it's important to assume $S$ is nonempty (because every element is an upper bound for the empty set, but ...) So needs some tweaks.

And for the paragraph before the lemma, I'll try to paraphrase to make sure I understand. Let $L=(0,1)$ to make thing easier to talk about. $L$ is connected. So in the ordered space $(Y,\prec)$ with $Y=L\cup\{2\}$, $2$ cannot have elements of $L$ both to its left and to its right (otherwise $L$ would be disconnected). So $2$ is either the minimum element or the maximum element of $Y$. Wlog, assume it is the minimum. The point $2$ is topologically isolated in $Y$, so it has an immediate successor $a$ in $(Y,\prec)$. That element $a$ is then the minimum of $L$. And by the lemma $L\setminus\{a\}$ is also connected. On the other hand, every point of $(0,1)$ is a cut point of $(0,1)$. Contradiction.

[prabau]

As for the wikipedia proof, it's definitely better than before. But there are still extraneous and confusingly verbose parts, I think (not in what you added).

[Jianing-Song]

As for the wikipedia proof, it's definitely better than before. But there are still extraneous and confusingly verbose parts, I think (not in what you added).

I have again simplified the proof to some extent. Please don't hesitate to modifiy what's written there :)

[StevenClontz]

And for the paragraph before the lemma, I'll try to paraphrase to make sure I understand...

That paraphrase looks good to me.