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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?

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Hint. Does every subset of $\mathbb{R}$ that is bounded below have a minimum? –  Arturo Magidin Dec 12 '11 at 6:40
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Hint 2. Your error is thinking that just because for every $x\in F$ there is a $y\in F$ with $y\lt x$, this means that $F$ must contain real numbers that are arbitrarily large and negative. It's possible for a subset of $\mathbb{R}$ to have no minimum, and yet be bounded below. If you can't think of one, try not be so negative... –  Arturo Magidin Dec 12 '11 at 7:03
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@ArturoMagidin thanks for the hints! If I get you right I could say, for example, F = {x | x > PI} Thus F would be bounded by PI though it would be no principal filter as I cannot "reach" any certain real number g such that F=[g) –  leandrodemarco Dec 12 '11 at 7:31
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@Arturo: Surely thinking negatively would be ideal. :-) (Leandro: Ignore this comment.) –  Brian M. Scott Dec 12 '11 at 7:32
    
@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 ...’. –  Brian M. Scott Dec 12 '11 at 7:35

1 Answer 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 $a\in\mathbb R$.

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