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] $. 




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




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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 ?



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


