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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 ?

edited Jan 30 2012 at 15:14

Emil Jeřábek
10.7k●1●24●41

asked Jan 30 2012 at 13:03

Reetesh Mukul
33●3

Look for Bourbaki's definition of "ordered". – Gerald Edgar Jan 30 2012 at 14:05

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I looked it up, and "order" does, indeed, mean partial order. – Gerald Edgar Jan 30 2012 at 14:13

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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 2012 at 14:27

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erratum is the singular form. :) – Jim Conant Jan 30 2012 at 15:55

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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. – quid Jan 30 2012 at 17:50

Emil Jeřábek · Accepted Answer · 2012-01-30 15:15:05Z UTC

3

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...

edited Jan 30 2012 at 15:15

Emil Jeřábek

answered Jan 30 2012 at 14:18

Gerald Edgar
15.2k●1●28●67

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. – Reetesh Mukul Jan 30 2012 at 14:45

Wlodzimierz Holsztynski · Answer 2 · 2013-05-23 23:21:59Z UTC

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[ $.

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

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