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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[ $.

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[ $.

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[ $.

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[ $.

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[ $.

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[ $.