In this link, there is the example of regular space, that is not completely regular. This space is also completely Hausdorff (https://math.stackexchange.com/questions/3516497/is-mysiors-example-completely-hausdorff/3516547#3516547). But in the article, the second page is remark, where is written, that if we add one point b to the space, with its local neighborhoods, that the space is also regular and not completely regular, but also not completely Hausdorff, because there is not the continuous function f, for which f(a)=f(b). I am interested in, how to prove that (that this space is not completely Hausdorff).
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$\begingroup$ The statement is that for every continuous real $f$ on $X$ we have $f(a)=f(b)$. Hence the conclusion that $X$ is not functionally Hausdorff. $\endgroup$– Henno BrandsmaCommented Jan 26, 2020 at 18:15
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$\begingroup$ Its clear, but I cant prove that f(a)=f(b) for every realvalued function. I tried to prove as there is proved that A and a cant be separae by continuous function, that I cant do $\endgroup$– VDGGCommented Jan 26, 2020 at 19:39
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$\begingroup$ The extended Mysior space is not completely Hausdorff because for any sequence of open sets $(U_n)_{n\in\omega}$ with $+\infty\in U_n\subset\overline U_n\subset U_{n+1}$, the union $\bigcup_{n\in\omega}U_n$ contains the point $-\infty$ in its closure. $\endgroup$– Taras BanakhCommented Jan 27, 2020 at 7:33
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$\begingroup$ Writing $+\infty$ and $-\infty$ I had in mind the points denoted by $a$ and $b$ in the Mysior's paper. But it it is better to imagine those points $a,b$ as $+\infty$ and $-\infty$. $\endgroup$– Taras BanakhCommented Jan 27, 2020 at 8:09
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$\begingroup$ Thank a lot. But I cant understant how that implies that f(a)=f(b) for every realvalued function $\endgroup$– VDGGCommented Jan 27, 2020 at 8:13
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1 Answer
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Modify the proof in the paper: if $f(b)<p$ then, by continuity there is an $n$ such that $f(x,0)<p$ for all $x<-n$. The same argument as in the paper now works to establish that $f(a)\le p$. Likewise if $f(b)>p$ then $f(a)\ge p$.
