# Possible errata in Nicolas Bourbaki's General Topology -I, Chapter 1 Exercise 2 ?

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 ?

• I looked it up, and "order" does, indeed, mean partial order. – Gerald Edgar Jan 30 '12 at 14:13
• 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. – Henno Brandsma Jan 30 '12 at 14:27
• erratum is the singular form. :) – Jim Conant Jan 30 '12 at 15:55
• 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. – user9072 Jan 30 '12 at 17:50
• 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' ". – Steven Landsburg Jun 25 '13 at 4:02

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   $]x,\rightarrow[$.

• 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? – Włodzimierz Holsztyński May 24 '13 at 0:47
• I am asking because my "Answer" has received a down-vote. – Włodzimierz Holsztyński May 24 '13 at 1:39

Say we have a partially ordered set. What so 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...

• 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. – nature1729 Jan 30 '12 at 14:45