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I encountered a problem a few days ago: what is an example of a homogeneous $T_0$ space which is not $T_1$? I tried to solve this using some symmetrical properties of separation axioms which I saw before, but I couldn't. I also find this on web, but I can't prove that the so called space in example 4.4 is homogeneous (the article says that it is an example of what I'm searching for.) because I have some problems taking the greater element to the smaller, applying order topology on an ordered set.

Any help would be appreciated!

Mahan Mehravard
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1 Answers1

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Let $X$ be the set $(0,\infty)$ of non-negative real numbers and give it the left-order topology. Thus it has a base of open sets given by $\{(0,x)\mid x\in(0,\infty)\}$.

Then $X$ is $T_0$, since if $x<y\in X$, then $x\in (0,z)\not \in y$ whenever $x<z<y$. On the other hand $X$ is not $T_1$, since for any $x\in X$ we have $\overline{\{x\}}][x,\infty)$.

If $\lambda>0$ is any real number, then the map $f_\lambda:X\rightarrow X$, $x\mapsto \lambda\cdot x$, is continuous. It has a continuous inverse given by $f_{1/\lambda}$. To see that $X$ is homogeneous consider $x\neq y\in X$ and take $f_\lambda$ with $\lambda=x/y$.

Edit: In fact a slightly stronger conclusion is true: the multiplication $X\times X\rightarrow X$, $(x,y)\mapsto x\cdot y$, is a continuous group operation. The inversion, however, fails to be continuous. In any case $X$ is an example of a first-countable $T_0$ paratopological group which fails to be $T_1$.

Tyrone
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  • Thanks for your illuminating answer. :) – Mahan Mehravard Nov 04 '20 at 15:22
  • @MahanMehravard I just had a chance to look at the example in your paper a bit better. It seems to be working with the same idea, this time with $X=\mathbb{R}$ and basis ${(-\infty,x)\mid x\in\mathbb{R}}$. To show homogeneity, instead of using multiplication you should consider the addition $x\mapsto x+\lambda$. – Tyrone Nov 04 '20 at 15:28
  • Yes, I came back to it and got the idea using your answer. My mistake was a silly one, I was trying to define an order topology using the example of the paper on an totally ordered set and not on the real line. Thanks again! – Mahan Mehravard Nov 04 '20 at 15:34
  • Of course I think the example in the paper is a little different. It takes such intervals closed, so their complement give us the basis for our intended topology. – Mahan Mehravard Nov 04 '20 at 15:38