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

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

2012-01-30T13:03:11Z

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 ?

Answer by Gerald Edgar for Possible errata in Nicolas Bourbaki's General Topology -I, Chapter 1 Exercise 2 ?

2012-01-30T14:18:05Z

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...

Answer by Wlodzimierz Holsztynski for Possible errata in Nicolas Bourbaki's General Topology -I, Chapter 1 Exercise 2 ?

2013-05-23T23:01:53Z

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 characterisation 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 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[ $.