Mixing
Density of linear order where $|A| = 1$
$begingroup$
Let $le$ mean a linear order in a non-empty, finite set $A$. I need to prove that this order is dense only when $|A| = 1.$
I understand that when I assume that $|A| = 1$, then I have:
$$forall x,yin ABig((x lt y) rightarrow exists z (x<z<y)Big)$$
and that's true.
But I don't quite understand the case when $|A| ge 2$.
Because then I have:
$$forall x,yin Abig(exists zneq x,(ymid x<z<y)big). $$
And that's for some reason contradiction.
And I really want to understand and learn this.
discrete-mathematics
edited Jan 20 at 18:10
Andrés E. Caicedo
65.6k8159250
asked Jan 20 at 16:47
KarolKarol
205
$endgroup$
add a comment |
$begingroup$
Let $le$ mean a linear order in a non-empty, finite set $A$. I need to prove that this order is dense only when $|A| = 1.$
I understand that when I assume that $|A| = 1$, then I have:
$$forall x,yin ABig((x lt y) rightarrow exists z (x<z<y)Big)$$
and that's true.
But I don't quite understand the case when $|A| ge 2$.
Because then I have:
$$forall x,yin Abig(exists zneq x,(ymid x<z<y)big). $$
And that's for some reason contradiction.
And I really want to understand and learn this.
discrete-mathematics
edited Jan 20 at 18:10
Andrés E. Caicedo
65.6k8159250
asked Jan 20 at 16:47
KarolKarol
205
$endgroup$
add a comment |
$begingroup$
Let $le$ mean a linear order in a non-empty, finite set $A$. I need to prove that this order is dense only when $|A| = 1.$
I understand that when I assume that $|A| = 1$, then I have:
$$forall x,yin ABig((x lt y) rightarrow exists z (x<z<y)Big)$$
and that's true.
But I don't quite understand the case when $|A| ge 2$.
Because then I have:
$$forall x,yin Abig(exists zneq x,(ymid x<z<y)big). $$
And that's for some reason contradiction.
And I really want to understand and learn this.
discrete-mathematics
edited Jan 20 at 18:10
Andrés E. Caicedo
65.6k8159250
asked Jan 20 at 16:47
KarolKarol
205
$endgroup$
Let $le$ mean a linear order in a non-empty, finite set $A$. I need to prove that this order is dense only when $|A| = 1.$
I understand that when I assume that $|A| = 1$, then I have:
$$forall x,yin ABig((x lt y) rightarrow exists z (x<z<y)Big)$$
and that's true.
But I don't quite understand the case when $|A| ge 2$.
Because then I have:
$$forall x,yin Abig(exists zneq x,(ymid x<z<y)big). $$
And that's for some reason contradiction.
And I really want to understand and learn this.
discrete-mathematics
discrete-mathematics
edited Jan 20 at 18:10
Andrés E. Caicedo
65.6k8159250
asked Jan 20 at 16:47
KarolKarol
205
edited Jan 20 at 18:10
Andrés E. Caicedo
65.6k8159250
asked Jan 20 at 16:47
KarolKarol
205
edited Jan 20 at 18:10
Andrés E. Caicedo
65.6k8159250
edited Jan 20 at 18:10
Andrés E. Caicedo
65.6k8159250
edited Jan 20 at 18:10
Andrés E. Caicedo
65.6k8159250
65.6k8159250
asked Jan 20 at 16:47
KarolKarol
205
asked Jan 20 at 16:47
KarolKarol
205
asked Jan 20 at 16:47
KarolKarol
205
205
add a comment |
add a comment |
1 Answer
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$begingroup$
List in order, the elements a,b,.. n.
If there is more than one element, pick a,b, the first two.
Since there is no x such that a < x < b, the order is not dense.
answered Jan 20 at 21:08
William ElliotWilliam Elliot
8,4422720
$endgroup$
add a comment |
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$begingroup$
List in order, the elements a,b,.. n.
If there is more than one element, pick a,b, the first two.
Since there is no x such that a < x < b, the order is not dense.
answered Jan 20 at 21:08
William ElliotWilliam Elliot
8,4422720
$endgroup$
add a comment |
$begingroup$
List in order, the elements a,b,.. n.
If there is more than one element, pick a,b, the first two.
Since there is no x such that a < x < b, the order is not dense.
answered Jan 20 at 21:08
William ElliotWilliam Elliot
8,4422720
$endgroup$
add a comment |
$begingroup$
List in order, the elements a,b,.. n.
If there is more than one element, pick a,b, the first two.
Since there is no x such that a < x < b, the order is not dense.
answered Jan 20 at 21:08
William ElliotWilliam Elliot
8,4422720
$endgroup$
List in order, the elements a,b,.. n.
If there is more than one element, pick a,b, the first two.
Since there is no x such that a < x < b, the order is not dense.
answered Jan 20 at 21:08
William ElliotWilliam Elliot
8,4422720
answered Jan 20 at 21:08
William ElliotWilliam Elliot
8,4422720
answered Jan 20 at 21:08
William ElliotWilliam Elliot
8,4422720
answered Jan 20 at 21:08
William ElliotWilliam Elliot
8,4422720
8,4422720
add a comment |
add a comment |
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