What does it mean for a linear order to be dense or without endpoint?
So What does it mean for a linear order to be
- Dense
- Without EndPoint
Example (R, <), which one is this and why?
order-theory
edited Nov 26 at 13:34
Servaes
22.3k33793
asked Nov 26 at 12:08
MF DOOM
13
add a comment |
So What does it mean for a linear order to be
- Dense
- Without EndPoint
Example (R, <), which one is this and why?
order-theory
edited Nov 26 at 13:34
Servaes
22.3k33793
asked Nov 26 at 12:08
MF DOOM
13
add a comment |
So What does it mean for a linear order to be
- Dense
- Without EndPoint
Example (R, <), which one is this and why?
order-theory
edited Nov 26 at 13:34
Servaes
22.3k33793
asked Nov 26 at 12:08
MF DOOM
13
So What does it mean for a linear order to be
- Dense
- Without EndPoint
Example (R, <), which one is this and why?
order-theory
order-theory
edited Nov 26 at 13:34
Servaes
22.3k33793
asked Nov 26 at 12:08
MF DOOM
13
edited Nov 26 at 13:34
Servaes
22.3k33793
asked Nov 26 at 12:08
MF DOOM
13
edited Nov 26 at 13:34
Servaes
22.3k33793
edited Nov 26 at 13:34
Servaes
22.3k33793
edited Nov 26 at 13:34
Servaes
22.3k33793
22.3k33793
asked Nov 26 at 12:08
MF DOOM
13
asked Nov 26 at 12:08
MF DOOM
13
asked Nov 26 at 12:08
MF DOOM
13
13
add a comment |
add a comment |
1 Answer
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A linear order is dense when for all $x,y$ such that $x < y$ there is some $z$ with $x < z < y$. In $mathbb{R}$ we can take $frac{x+y}{2}$ e.g. So there are always points between any two distinct points.
Without endpoint means it has no minimum or maximum. So for all $x$ there is some $y$ with $y>x$ and some $z$ with $z < x$. In $mathbb{R}$ we can always take $x+1$ or $x-1$ for that, e.g.
answered Nov 26 at 12:12
Henno Brandsma
104k346113
add a comment |
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1 Answer
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1 Answer
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votes
A linear order is dense when for all $x,y$ such that $x < y$ there is some $z$ with $x < z < y$. In $mathbb{R}$ we can take $frac{x+y}{2}$ e.g. So there are always points between any two distinct points.
Without endpoint means it has no minimum or maximum. So for all $x$ there is some $y$ with $y>x$ and some $z$ with $z < x$. In $mathbb{R}$ we can always take $x+1$ or $x-1$ for that, e.g.
answered Nov 26 at 12:12
Henno Brandsma
104k346113
add a comment |
A linear order is dense when for all $x,y$ such that $x < y$ there is some $z$ with $x < z < y$. In $mathbb{R}$ we can take $frac{x+y}{2}$ e.g. So there are always points between any two distinct points.
Without endpoint means it has no minimum or maximum. So for all $x$ there is some $y$ with $y>x$ and some $z$ with $z < x$. In $mathbb{R}$ we can always take $x+1$ or $x-1$ for that, e.g.
answered Nov 26 at 12:12
Henno Brandsma
104k346113
add a comment |
A linear order is dense when for all $x,y$ such that $x < y$ there is some $z$ with $x < z < y$. In $mathbb{R}$ we can take $frac{x+y}{2}$ e.g. So there are always points between any two distinct points.
Without endpoint means it has no minimum or maximum. So for all $x$ there is some $y$ with $y>x$ and some $z$ with $z < x$. In $mathbb{R}$ we can always take $x+1$ or $x-1$ for that, e.g.
answered Nov 26 at 12:12
Henno Brandsma
104k346113
A linear order is dense when for all $x,y$ such that $x < y$ there is some $z$ with $x < z < y$. In $mathbb{R}$ we can take $frac{x+y}{2}$ e.g. So there are always points between any two distinct points.
Without endpoint means it has no minimum or maximum. So for all $x$ there is some $y$ with $y>x$ and some $z$ with $z < x$. In $mathbb{R}$ we can always take $x+1$ or $x-1$ for that, e.g.
answered Nov 26 at 12:12
Henno Brandsma
104k346113
answered Nov 26 at 12:12
Henno Brandsma
104k346113
answered Nov 26 at 12:12
Henno Brandsma
104k346113
answered Nov 26 at 12:12
Henno Brandsma
104k346113
104k346113
add a comment |
add a comment |
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