Possible errata in Nicolas Bourbaki's General Topology -I, Chapter 1 Exercise 2 ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:34:23Z http://mathoverflow.net/feeds/question/87026 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87026/possible-errata-in-nicolas-bourbakis-general-topology-i-chapter-1-exercise-2 Possible errata in Nicolas Bourbaki's General Topology -I, Chapter 1 Exercise 2 ? Reetesh Mukul 2012-01-30T13:03:11Z 2013-05-23T23:21:59Z <p>Here is the text of Exercise:</p> <p>2 a) Let $X$ be an <em>ordered</em> set. Show that the set of intervals</p> <p>$\left[x, \rightarrow\right[$ (resp. $\left]\leftarrow, x\right]$)</p> <p>is a base of topology on $X$; this topology is called the <em>right</em> (resp. <em>left</em>) topology of $X$. In the right topology, any intersection of open sets is an open set, and the closure of $\{x\}$ is the interval $\left]\leftarrow, x\right]$. </p> <hr> <p>The above one was from English edition. I translated French edition and found the same text.</p> <hr> <p>Should not be $X$ a <strong>totally</strong> ordered set ? And is not that the set of intervals should be $\left]x, \rightarrow\right[$ in place of $\left[x, \rightarrow\right[$ ?</p> <p>Is this an errata ?</p> http://mathoverflow.net/questions/87026/possible-errata-in-nicolas-bourbakis-general-topology-i-chapter-1-exercise-2/87031#87031 Answer by Gerald Edgar for Possible errata in Nicolas Bourbaki's General Topology -I, Chapter 1 Exercise 2 ? Gerald Edgar 2012-01-30T14:18:05Z 2012-01-30T15:15:05Z <p>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...</p> http://mathoverflow.net/questions/87026/possible-errata-in-nicolas-bourbakis-general-topology-i-chapter-1-exercise-2/131665#131665 Answer by Wlodzimierz Holsztynski for Possible errata in Nicolas Bourbaki's General Topology -I, Chapter 1 Exercise 2 ? Wlodzimierz Holsztynski 2013-05-23T23:01:53Z 2013-05-23T23:21:59Z <p>Bourbaki was right :-) &nbsp; On the other hand, let &nbsp; $(X\ \le)$ &nbsp; be a partially ordered set. In general the family</p> <p>$$\mathbf B\ \ :=\ \ \{\ ]x,\rightarrow[\ :\ x\in X\ \}$$</p> <p>is NOT a topological base for any topology in &nbsp; $X$. &nbsp; One reason is trivial: no minimal element belongs to any member of &nbsp; $B$; &nbsp; thus if there is any minimal element then &nbsp; $X$ &nbsp; would not be open.</p> <p>OK, one could define:</p> <p>$$\mathbf B\ \ :=\ \ \{X\}\ \cup\ \{\ ]x,\rightarrow[\ :\ x\in X\ \}$$</p> <p>It will not help. Indeed, here is a characterisation of a topological base:</p> <p><strong>THEOREM</strong> &nbsp; A family &nbsp; $\mathbf B$ &nbsp; of subsets of &nbsp; $X$ &nbsp; is a topological base for a topology in &nbsp; $X\quad\Leftrightarrow$ &nbsp; the following two conditions hold:</p> <ul> <li><p>&nbsp; $\bigcup \mathbf B\ =\ X$</p></li> <li><p>&nbsp; $\forall_{G\ H\in\mathbf B}\quad G\cap H\ =\ \bigcup\ \{K\in \mathbf B : K\subseteq G\cap H\}$</p></li> </ul> <p>Now consider a 5-element set</p> <p>$$X := \{b\ \ d\ \ A\ \ C\ \ E\}$$</p> <p>where there are exactly four sharp inequalities &nbsp; $b &lt; A$ &nbsp; &amp; &nbsp; $b &lt; C$ &nbsp; &amp; &nbsp; $d &lt; C$ &nbsp; &amp; &nbsp; $d &lt; E$. &nbsp; Then the intersection</p> <p>$$]b,\rightarrow[\ \ \cap\ \ ]d,\rightarrow[\quad=\quad \{ C \}$$</p> <p>is not a union of any family of open rays &nbsp; $]x,\rightarrow[$.</p>