0
$\begingroup$

Is there any filter in the $(\mathbb{R},\max, \min)$ lattice that is not a principal filter (a principal filter generated by an element $a$ is the set $\{ x \in \mathbb{R}\mid a \leq x \}$ )

I'm guessing it exists as the assignment states "Find a non-principal filter of $(\mathbb{R},\max, \min)$", but I cannot figure it out. The only one I can think of is the whole set $\mathbb{R}$, but I don't know if that's right.

Here's my thoughts about it:

Any filter $F$ is non-empty. If $F$ had a minimum element $r_{\rm min}$ then $F$ would be the principal filter generated by $r_{\rm min}$. Therefore, $F$ must "grow indefinitely backwards". But then, $F$ equals $\mathbb{R}$, because if $x$ belongs to $F$ then any real number $y$ larger than $x$ must belong to $F$.

I doubt this reasoning is correct, but even if I am right: is there any other filter F different from $\mathbb{R}$ that is not principal?

  • 0
    @Leandro: Yes, \{x:x>\pi\} works fine. So does \{x:x>0\}, which is what Arturo was thinking of when he wrote ‘try not [to] be so negative ...’.2011-12-12

1 Answers 1

1

Community wiki answer so the question can be marked as answered:

The set $\{x\in\mathbb R\mid x\gt a\}$ is a non-principal filter for any $a\in\mathbb R$.