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Here is the text of Exercise:

2 a) Let $X$ be an ordered set. Show that the set of intervals

$\left[x, \rightarrow\right[$ (resp. $\left]\leftarrow, x\right]$)

is a base of topology on $X$; this topology is called the right (resp. left) topology of $X$. In the right topology, any intersection of open sets is an open set, and the closure of $\{x\}$ is the interval $\left]\leftarrow, x\right] $.

The above one was from English edition. I translated French edition and found the same text.

Should not be $X$ a totally ordered set ? And is not that the set of intervals should be $\left]x, \rightarrow\right[$ in place of $\left[x, \rightarrow\right[$ ?

Is this an errata ?

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    $\begingroup$ I looked it up, and "order" does, indeed, mean partial order. $\endgroup$
    – Gerald Edgar
    Commented Jan 30, 2012 at 14:13
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    $\begingroup$ errata = a list of errors in a published work. So it's not an "errata", but possibly (at most) an error. But I don't think it is, it's correct as stated. $\endgroup$
    – Henno Brandsma
    Commented Jan 30, 2012 at 14:27
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    $\begingroup$ erratum is the singular form. :) $\endgroup$
    – Jim Conant
    Commented Jan 30, 2012 at 15:55
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    $\begingroup$ Still I would also suggest to avoid using erratum in this form, as it is my impression its standard meaning in academic writing is a bit different from the pure translation 'error'; often referring to some (informally) published note pointing out and possibly fixing an error. $\endgroup$
    – user9072
    Commented Jan 30, 2012 at 17:50
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    $\begingroup$ Somewhere there is a book in which page 342 is headlined "Errata", and the text reads, in its entirety: "On page 342, change 'Errata' to 'Erratum' ". $\endgroup$
    – Steven Landsburg
    Commented Jun 25, 2013 at 4:02

2 Answers 2

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Bourbaki was right :-)   On the other hand, let   $(X\ \le)$   be a partially ordered set. In general the family

$$\mathbf B\ \ :=\ \ \{\ ]x,\rightarrow[\ :\ x\in X\ \}$$

is NOT a topological base for any topology in   $X$.   One reason is trivial: no minimal element belongs to any member of   $B$;   thus if there is any minimal element then   $X$   would not be open.

OK, one could define:

$$\mathbf B\ \ :=\ \ \{X\}\ \cup\ \{\ ]x,\rightarrow[\ :\ x\in X\ \}$$

It will not help. Indeed, here is a characterization of a topological base:

THEOREM   A family   $\mathbf B$   of subsets of   $X$   is a topological base for a topology in   $X\quad\Leftrightarrow$   the following two conditions hold:

  •   $\bigcup \mathbf B\ =\ X$

  •   $\forall_{G\ H\in\mathbf B}\quad G\cap H\ =\ \bigcup\ \{K\in \mathbf B : K\subseteq G\cap H\} $

Now consider a 5-element set

$$X := \{b\ \ d\ \ A\ \ C\ \ E\}$$

where by definition there are exactly four sharp inequalities   $b < A$   &   $b < C$   &   $d < C$   &   $d < E$.   Then the intersection

$$ ]b,\rightarrow[\ \ \cap\ \ ]d,\rightarrow[\quad=\quad \{ C \} $$

is not a union of any family of open rays   $ ]x,\rightarrow[ $.

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  • $\begingroup$ May I know, please, what's wrong with my answer? @Reetesh Mukul wondered if Bourbaki should consider open intervals instead of closed intervals as a base of the left (or right) topology. I tried to explain that one would not get a topological base for general partially ordered space. Is it a crime on MO? $\endgroup$
    – Włodzimierz Holsztyński
    Commented May 24, 2013 at 0:47
  • $\begingroup$ I am asking because my "Answer" has received a down-vote. $\endgroup$
    – Włodzimierz Holsztyński
    Commented May 24, 2013 at 1:39
  • $\begingroup$ Not much mystery, I think: probably the downvoter would have upvoted instead "Bourbaki was wrong" :) $\endgroup$
    – YCor
    Commented Jan 27, 2021 at 18:48
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Say we have a partially ordered set. What do you doubt? (1) The set of intervals $\left[x,\rightarrow\right[$ is a base for a topology. (2) Any intersection of open sets is open. (3) The closure of $\{x\}$ is $\left]\leftarrow,x\right]$. They all look OK to me...

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  • $\begingroup$ I solved this Exercise using the idea of Poset only, but then I got confused after reading the definition of right and left topology in Wiki: en.wikipedia.org/wiki/…. In General Topology -I, 1.4, while giving definition of Closed Set, Bourbaki says: On the rational line, every interval of the form [a,→[ is a closed set, for its complement ]←,a[ is open. But I agree that closed/open sets are subject to Topology under consideration. $\endgroup$
    – nature1729
    Commented Jan 30, 2012 at 14:45

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