Mixing
Understanding explicit bijection between two sets
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I'm having a lot difficulty understanding the concept of explicit bijection and how to show an explicit bijection between two sets. My professor rushed through the topic during class and now I'm trying some practice problems and I'm so lost :(
How do I go about showing an explicit bijection? For example, (0,1) and [4,8)? What difference does ( and [ make?
Thanks so much.
functions discrete-mathematics
edited Jan 10 at 22:01
Ethan Bolker
42.6k549113
asked Oct 12 '15 at 23:20
CandiceCandice
162
$endgroup$
add a comment |
$begingroup$
I'm having a lot difficulty understanding the concept of explicit bijection and how to show an explicit bijection between two sets. My professor rushed through the topic during class and now I'm trying some practice problems and I'm so lost :(
How do I go about showing an explicit bijection? For example, (0,1) and [4,8)? What difference does ( and [ make?
Thanks so much.
functions discrete-mathematics
edited Jan 10 at 22:01
Ethan Bolker
42.6k549113
asked Oct 12 '15 at 23:20
CandiceCandice
162
$endgroup$
$begingroup$
The ( and [ mean that you cannot use for example $f(x)=4x+4$ as that will not send any $x in (0,1)$ to $4 in [4,8)$
$endgroup$
– Henry
Oct 12 '15 at 23:23
$begingroup$
$(0, 1) = {x : 0 < x < 1}$ and $[4, 8) = {x : 4 le x < 8}$; parenthesis are with points that aren't included (note the strict inequalities) while brackets are for points that are included. See interval notation
$endgroup$
– pjs36
Oct 12 '15 at 23:27
add a comment |
$begingroup$
I'm having a lot difficulty understanding the concept of explicit bijection and how to show an explicit bijection between two sets. My professor rushed through the topic during class and now I'm trying some practice problems and I'm so lost :(
How do I go about showing an explicit bijection? For example, (0,1) and [4,8)? What difference does ( and [ make?
Thanks so much.
functions discrete-mathematics
edited Jan 10 at 22:01
Ethan Bolker
42.6k549113
asked Oct 12 '15 at 23:20
CandiceCandice
162
$endgroup$
I'm having a lot difficulty understanding the concept of explicit bijection and how to show an explicit bijection between two sets. My professor rushed through the topic during class and now I'm trying some practice problems and I'm so lost :(
How do I go about showing an explicit bijection? For example, (0,1) and [4,8)? What difference does ( and [ make?
Thanks so much.
functions discrete-mathematics
functions discrete-mathematics
edited Jan 10 at 22:01
Ethan Bolker
42.6k549113
asked Oct 12 '15 at 23:20
CandiceCandice
162
edited Jan 10 at 22:01
Ethan Bolker
42.6k549113
asked Oct 12 '15 at 23:20
CandiceCandice
162
edited Jan 10 at 22:01
Ethan Bolker
42.6k549113
edited Jan 10 at 22:01
Ethan Bolker
42.6k549113
edited Jan 10 at 22:01
Ethan Bolker
42.6k549113
42.6k549113
asked Oct 12 '15 at 23:20
CandiceCandice
162
asked Oct 12 '15 at 23:20
CandiceCandice
162
asked Oct 12 '15 at 23:20
CandiceCandice
162
162
$begingroup$
The ( and [ mean that you cannot use for example $f(x)=4x+4$ as that will not send any $x in (0,1)$ to $4 in [4,8)$
$endgroup$
– Henry
Oct 12 '15 at 23:23
$begingroup$
$(0, 1) = {x : 0 < x < 1}$ and $[4, 8) = {x : 4 le x < 8}$; parenthesis are with points that aren't included (note the strict inequalities) while brackets are for points that are included. See interval notation
$endgroup$
– pjs36
Oct 12 '15 at 23:27
add a comment |
$begingroup$
The ( and [ mean that you cannot use for example $f(x)=4x+4$ as that will not send any $x in (0,1)$ to $4 in [4,8)$
$endgroup$
– Henry
Oct 12 '15 at 23:23
$begingroup$
$(0, 1) = {x : 0 < x < 1}$ and $[4, 8) = {x : 4 le x < 8}$; parenthesis are with points that aren't included (note the strict inequalities) while brackets are for points that are included. See interval notation
$endgroup$
– pjs36
Oct 12 '15 at 23:27
$begingroup$
The ( and [ mean that you cannot use for example $f(x)=4x+4$ as that will not send any $x in (0,1)$ to $4 in [4,8)$
$endgroup$
– Henry
Oct 12 '15 at 23:23
$begingroup$
The ( and [ mean that you cannot use for example $f(x)=4x+4$ as that will not send any $x in (0,1)$ to $4 in [4,8)$
$endgroup$
– Henry
Oct 12 '15 at 23:23
$begingroup$
$(0, 1) = {x : 0 < x < 1}$ and $[4, 8) = {x : 4 le x < 8}$; parenthesis are with points that aren't included (note the strict inequalities) while brackets are for points that are included. See interval notation
$endgroup$
– pjs36
Oct 12 '15 at 23:27
$begingroup$
$(0, 1) = {x : 0 < x < 1}$ and $[4, 8) = {x : 4 le x < 8}$; parenthesis are with points that aren't included (note the strict inequalities) while brackets are for points that are included. See interval notation
$endgroup$
– pjs36
Oct 12 '15 at 23:27
add a comment |
1 Answer
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The main difference that [ makes is, it complicates the example ;)
First, an uncomplicated example: here's a bijection $(0,1) to (4,8):
$$ x mapsto 4 + 4x$$
This first maps $x in (0,1)$ to $4x in (0,4)$ – that itself is a bijection – and then (via another bijection) translates $(0,4)$ to $(4,8)$.
You can biject any two open, closed, open-closed, or closed-open intervals in a similar way, with a simple linear function.
If $f colon A to B$ and $g colon B to C$ are bijections, then $g circ f colon A to C$ is a bijection. To show that a bijection exists from $(0,1) to [4,8)$, you only need to show that there is one from $(4,8)$ to $[4,8)$.
It's harder to specify a bijection from $(4,8)$ to $[4,8)$. Certainly it can't be order-preserving. It's best in this case to settle for its mere existence: the union of an infinite set $S$ with a finite set $F$ is an infinite set of the same size, which means: there exists a bijection $S to S cup F$.
answered Oct 12 '15 at 23:46
BrianOBrianO
14.3k1822
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$begingroup$
The main difference that [ makes is, it complicates the example ;)
First, an uncomplicated example: here's a bijection $(0,1) to (4,8):
$$ x mapsto 4 + 4x$$
This first maps $x in (0,1)$ to $4x in (0,4)$ – that itself is a bijection – and then (via another bijection) translates $(0,4)$ to $(4,8)$.
You can biject any two open, closed, open-closed, or closed-open intervals in a similar way, with a simple linear function.
If $f colon A to B$ and $g colon B to C$ are bijections, then $g circ f colon A to C$ is a bijection. To show that a bijection exists from $(0,1) to [4,8)$, you only need to show that there is one from $(4,8)$ to $[4,8)$.
It's harder to specify a bijection from $(4,8)$ to $[4,8)$. Certainly it can't be order-preserving. It's best in this case to settle for its mere existence: the union of an infinite set $S$ with a finite set $F$ is an infinite set of the same size, which means: there exists a bijection $S to S cup F$.
answered Oct 12 '15 at 23:46
BrianOBrianO
14.3k1822
$endgroup$
add a comment |
$begingroup$
The main difference that [ makes is, it complicates the example ;)
First, an uncomplicated example: here's a bijection $(0,1) to (4,8):
$$ x mapsto 4 + 4x$$
This first maps $x in (0,1)$ to $4x in (0,4)$ – that itself is a bijection – and then (via another bijection) translates $(0,4)$ to $(4,8)$.
You can biject any two open, closed, open-closed, or closed-open intervals in a similar way, with a simple linear function.
If $f colon A to B$ and $g colon B to C$ are bijections, then $g circ f colon A to C$ is a bijection. To show that a bijection exists from $(0,1) to [4,8)$, you only need to show that there is one from $(4,8)$ to $[4,8)$.
It's harder to specify a bijection from $(4,8)$ to $[4,8)$. Certainly it can't be order-preserving. It's best in this case to settle for its mere existence: the union of an infinite set $S$ with a finite set $F$ is an infinite set of the same size, which means: there exists a bijection $S to S cup F$.
answered Oct 12 '15 at 23:46
BrianOBrianO
14.3k1822
$endgroup$
add a comment |
$begingroup$
The main difference that [ makes is, it complicates the example ;)
First, an uncomplicated example: here's a bijection $(0,1) to (4,8):
$$ x mapsto 4 + 4x$$
This first maps $x in (0,1)$ to $4x in (0,4)$ – that itself is a bijection – and then (via another bijection) translates $(0,4)$ to $(4,8)$.
You can biject any two open, closed, open-closed, or closed-open intervals in a similar way, with a simple linear function.
If $f colon A to B$ and $g colon B to C$ are bijections, then $g circ f colon A to C$ is a bijection. To show that a bijection exists from $(0,1) to [4,8)$, you only need to show that there is one from $(4,8)$ to $[4,8)$.
It's harder to specify a bijection from $(4,8)$ to $[4,8)$. Certainly it can't be order-preserving. It's best in this case to settle for its mere existence: the union of an infinite set $S$ with a finite set $F$ is an infinite set of the same size, which means: there exists a bijection $S to S cup F$.
answered Oct 12 '15 at 23:46
BrianOBrianO
14.3k1822
$endgroup$
The main difference that [ makes is, it complicates the example ;)
First, an uncomplicated example: here's a bijection $(0,1) to (4,8):
$$ x mapsto 4 + 4x$$
This first maps $x in (0,1)$ to $4x in (0,4)$ – that itself is a bijection – and then (via another bijection) translates $(0,4)$ to $(4,8)$.
You can biject any two open, closed, open-closed, or closed-open intervals in a similar way, with a simple linear function.
If $f colon A to B$ and $g colon B to C$ are bijections, then $g circ f colon A to C$ is a bijection. To show that a bijection exists from $(0,1) to [4,8)$, you only need to show that there is one from $(4,8)$ to $[4,8)$.
It's harder to specify a bijection from $(4,8)$ to $[4,8)$. Certainly it can't be order-preserving. It's best in this case to settle for its mere existence: the union of an infinite set $S$ with a finite set $F$ is an infinite set of the same size, which means: there exists a bijection $S to S cup F$.
answered Oct 12 '15 at 23:46
BrianOBrianO
14.3k1822
answered Oct 12 '15 at 23:46
BrianOBrianO
14.3k1822
answered Oct 12 '15 at 23:46
BrianOBrianO
14.3k1822
answered Oct 12 '15 at 23:46
BrianOBrianO
14.3k1822
14.3k1822
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add a comment |
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$begingroup$
The ( and [ mean that you cannot use for example $f(x)=4x+4$ as that will not send any $x in (0,1)$ to $4 in [4,8)$
$endgroup$
– Henry
Oct 12 '15 at 23:23
$begingroup$
$(0, 1) = {x : 0 < x < 1}$ and $[4, 8) = {x : 4 le x < 8}$; parenthesis are with points that aren't included (note the strict inequalities) while brackets are for points that are included. See interval notation
$endgroup$
– pjs36
Oct 12 '15 at 23:27