Show that the following pairs of sets have the same cardinality.
(a) Integers divisible by 3, and the even positive integers.
(b) $Bbb {R}$, and the interval $(0,infty)$.
(c) The interval $[0, 2)$, and the set [5,6)∪[7,8).
(d) The intervals $(-infty, -1)$ and $(-1, 0)$.
I know that cardinality means that there is a bijection between the two sets, and that means there is a surjection and injection. For the first one I think you can simply make a function that is a bijection and prove that it is an injection and surjection. I am not sure how to do the last three.
Sorry for the poor formatting in advance and if you need more information about the question I am more than happy to provide it.
Thank you.
real-analysis functions elementary-set-theory
edited Nov 28 '18 at 21:28
greedoid
38.2k114797
asked Nov 27 '18 at 2:44
smith
93
add a comment |
(a) Integers divisible by 3, and the even positive integers.
(b) $Bbb {R}$, and the interval $(0,infty)$.
(c) The interval $[0, 2)$, and the set [5,6)∪[7,8).
(d) The intervals $(-infty, -1)$ and $(-1, 0)$.
I know that cardinality means that there is a bijection between the two sets, and that means there is a surjection and injection. For the first one I think you can simply make a function that is a bijection and prove that it is an injection and surjection. I am not sure how to do the last three.
Sorry for the poor formatting in advance and if you need more information about the question I am more than happy to provide it.
Thank you.
real-analysis functions elementary-set-theory
edited Nov 28 '18 at 21:28
greedoid
38.2k114797
asked Nov 27 '18 at 2:44
smith
93
ok thank you, fixed it
– smith
Nov 27 '18 at 2:47
Do you mean the set $[5,6)cup [7,8)$?
– TonyK
Nov 27 '18 at 2:50
yes, it has been fixed
– smith
Nov 27 '18 at 2:51
add a comment |
(a) Integers divisible by 3, and the even positive integers.
(b) $Bbb {R}$, and the interval $(0,infty)$.
(c) The interval $[0, 2)$, and the set [5,6)∪[7,8).
(d) The intervals $(-infty, -1)$ and $(-1, 0)$.
I know that cardinality means that there is a bijection between the two sets, and that means there is a surjection and injection. For the first one I think you can simply make a function that is a bijection and prove that it is an injection and surjection. I am not sure how to do the last three.
Sorry for the poor formatting in advance and if you need more information about the question I am more than happy to provide it.
Thank you.
real-analysis functions elementary-set-theory
edited Nov 28 '18 at 21:28
greedoid
38.2k114797
asked Nov 27 '18 at 2:44
smith
93
(a) Integers divisible by 3, and the even positive integers.
(b) $Bbb {R}$, and the interval $(0,infty)$.
(c) The interval $[0, 2)$, and the set [5,6)∪[7,8).
(d) The intervals $(-infty, -1)$ and $(-1, 0)$.
I know that cardinality means that there is a bijection between the two sets, and that means there is a surjection and injection. For the first one I think you can simply make a function that is a bijection and prove that it is an injection and surjection. I am not sure how to do the last three.
Sorry for the poor formatting in advance and if you need more information about the question I am more than happy to provide it.
Thank you.
real-analysis functions elementary-set-theory
real-analysis functions elementary-set-theory
edited Nov 28 '18 at 21:28
greedoid
38.2k114797
asked Nov 27 '18 at 2:44
smith
93
edited Nov 28 '18 at 21:28
greedoid
38.2k114797
asked Nov 27 '18 at 2:44
smith
93
edited Nov 28 '18 at 21:28
greedoid
38.2k114797
edited Nov 28 '18 at 21:28
greedoid
38.2k114797
edited Nov 28 '18 at 21:28
greedoid
38.2k114797
38.2k114797
asked Nov 27 '18 at 2:44
smith
93
asked Nov 27 '18 at 2:44
smith
93
asked Nov 27 '18 at 2:44
smith
93
93
ok thank you, fixed it
– smith
Nov 27 '18 at 2:47
Do you mean the set $[5,6)cup [7,8)$?
– TonyK
Nov 27 '18 at 2:50
yes, it has been fixed
– smith
Nov 27 '18 at 2:51
add a comment |
ok thank you, fixed it
– smith
Nov 27 '18 at 2:47
Do you mean the set $[5,6)cup [7,8)$?
– TonyK
Nov 27 '18 at 2:50
yes, it has been fixed
– smith
Nov 27 '18 at 2:51
ok thank you, fixed it
– smith
Nov 27 '18 at 2:47
ok thank you, fixed it
– smith
Nov 27 '18 at 2:47
Do you mean the set $[5,6)cup [7,8)$?
– TonyK
Nov 27 '18 at 2:50
Do you mean the set $[5,6)cup [7,8)$?
– TonyK
Nov 27 '18 at 2:50
yes, it has been fixed
– smith
Nov 27 '18 at 2:51
yes, it has been fixed
– smith
Nov 27 '18 at 2:51
add a comment |
4 Answers
4
active
oldest
votes
Hints:
(b) Map $x in mathbb{R}$ by $x to e^x$
(c) Map $x in [0, 2)$ by $x to x + 5$ if $x in [0,1)$ and $x to x + 6$ if $x in [1,2)$
(d) Map $x in (-infty,-1)$ by $x to frac{1}{x}$
Show that these are all bijections.
answered Nov 27 '18 at 3:03
AlkaKadri
1,459411
add a comment |
(b) $mathbb{R}$, and the interval (0,∞).
Consider $f : mathbb{R}rightarrow (0,infty)$ defined by $f(x)=e^x.$
(c) The interval [0,2), and the set [5,6) or [7,8).
Consider $g: [0,2)rightarrow [5,6)$ defined by $g(x)=5+frac{x}{2}$ or $g'(x)=7+frac{x}{2}$ if codomain is $[7,8)$.
(d) The intervals (−∞,−1) and (−1,0).
Consider $h: (-1,0)rightarrow (-infty,-1)$ defined by $h(x)=tan(frac{pi}{2}x)-1.$
answered Nov 27 '18 at 3:02
LeB
986217
add a comment |
a) Let $A$ be a set of integers divisible by 3 and $C$ a set of positive even integers. You can take a composition of following functions. First $f:Ato mathbb{Z}$ defined with $$nlongmapsto {nover 3}$$
and then $g:mathbb{Z}to C$ defined by:
$$
g(n)
begin{cases}
= 4n+2 & text{if } ngeq 0, \
= 4|n| & text{if } n < 0. \
end{cases}
$$
Then let $gcirc f$ will do the job.
answered Nov 27 '18 at 21:57
greedoid
38.2k114797
add a comment |
The bijections, reading left to right, can be chosen as:
(a) $|4x/3|+2[xge 0]$ (b) $x+sqrt{1+x^2}$ (c) $x+5+lfloor xrfloor$ (d) $1/x$.
answered Nov 27 '18 at 22:11
J.G.
23.2k22137
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hints:
(b) Map $x in mathbb{R}$ by $x to e^x$
(c) Map $x in [0, 2)$ by $x to x + 5$ if $x in [0,1)$ and $x to x + 6$ if $x in [1,2)$
(d) Map $x in (-infty,-1)$ by $x to frac{1}{x}$
Show that these are all bijections.
answered Nov 27 '18 at 3:03
AlkaKadri
1,459411
add a comment |
Hints:
(b) Map $x in mathbb{R}$ by $x to e^x$
(c) Map $x in [0, 2)$ by $x to x + 5$ if $x in [0,1)$ and $x to x + 6$ if $x in [1,2)$
(d) Map $x in (-infty,-1)$ by $x to frac{1}{x}$
Show that these are all bijections.
answered Nov 27 '18 at 3:03
AlkaKadri
1,459411
add a comment |
Hints:
(b) Map $x in mathbb{R}$ by $x to e^x$
(c) Map $x in [0, 2)$ by $x to x + 5$ if $x in [0,1)$ and $x to x + 6$ if $x in [1,2)$
(d) Map $x in (-infty,-1)$ by $x to frac{1}{x}$
Show that these are all bijections.
answered Nov 27 '18 at 3:03
AlkaKadri
1,459411
Hints:
(b) Map $x in mathbb{R}$ by $x to e^x$
(c) Map $x in [0, 2)$ by $x to x + 5$ if $x in [0,1)$ and $x to x + 6$ if $x in [1,2)$
(d) Map $x in (-infty,-1)$ by $x to frac{1}{x}$
Show that these are all bijections.
answered Nov 27 '18 at 3:03
AlkaKadri
1,459411
answered Nov 27 '18 at 3:03
AlkaKadri
1,459411
answered Nov 27 '18 at 3:03
AlkaKadri
1,459411
answered Nov 27 '18 at 3:03
AlkaKadri
1,459411
1,459411
add a comment |
add a comment |
(b) $mathbb{R}$, and the interval (0,∞).
Consider $f : mathbb{R}rightarrow (0,infty)$ defined by $f(x)=e^x.$
(c) The interval [0,2), and the set [5,6) or [7,8).
Consider $g: [0,2)rightarrow [5,6)$ defined by $g(x)=5+frac{x}{2}$ or $g'(x)=7+frac{x}{2}$ if codomain is $[7,8)$.
(d) The intervals (−∞,−1) and (−1,0).
Consider $h: (-1,0)rightarrow (-infty,-1)$ defined by $h(x)=tan(frac{pi}{2}x)-1.$
answered Nov 27 '18 at 3:02
LeB
986217
add a comment |
(b) $mathbb{R}$, and the interval (0,∞).
Consider $f : mathbb{R}rightarrow (0,infty)$ defined by $f(x)=e^x.$
(c) The interval [0,2), and the set [5,6) or [7,8).
Consider $g: [0,2)rightarrow [5,6)$ defined by $g(x)=5+frac{x}{2}$ or $g'(x)=7+frac{x}{2}$ if codomain is $[7,8)$.
(d) The intervals (−∞,−1) and (−1,0).
Consider $h: (-1,0)rightarrow (-infty,-1)$ defined by $h(x)=tan(frac{pi}{2}x)-1.$
answered Nov 27 '18 at 3:02
LeB
986217
add a comment |
(b) $mathbb{R}$, and the interval (0,∞).
Consider $f : mathbb{R}rightarrow (0,infty)$ defined by $f(x)=e^x.$
(c) The interval [0,2), and the set [5,6) or [7,8).
Consider $g: [0,2)rightarrow [5,6)$ defined by $g(x)=5+frac{x}{2}$ or $g'(x)=7+frac{x}{2}$ if codomain is $[7,8)$.
(d) The intervals (−∞,−1) and (−1,0).
Consider $h: (-1,0)rightarrow (-infty,-1)$ defined by $h(x)=tan(frac{pi}{2}x)-1.$
answered Nov 27 '18 at 3:02
LeB
986217
(b) $mathbb{R}$, and the interval (0,∞).
Consider $f : mathbb{R}rightarrow (0,infty)$ defined by $f(x)=e^x.$
(c) The interval [0,2), and the set [5,6) or [7,8).
Consider $g: [0,2)rightarrow [5,6)$ defined by $g(x)=5+frac{x}{2}$ or $g'(x)=7+frac{x}{2}$ if codomain is $[7,8)$.
(d) The intervals (−∞,−1) and (−1,0).
Consider $h: (-1,0)rightarrow (-infty,-1)$ defined by $h(x)=tan(frac{pi}{2}x)-1.$
answered Nov 27 '18 at 3:02
LeB
986217
answered Nov 27 '18 at 3:02
LeB
986217
answered Nov 27 '18 at 3:02
LeB
986217
answered Nov 27 '18 at 3:02
LeB
986217
986217
add a comment |
add a comment |
a) Let $A$ be a set of integers divisible by 3 and $C$ a set of positive even integers. You can take a composition of following functions. First $f:Ato mathbb{Z}$ defined with $$nlongmapsto {nover 3}$$
and then $g:mathbb{Z}to C$ defined by:
$$
g(n)
begin{cases}
= 4n+2 & text{if } ngeq 0, \
= 4|n| & text{if } n < 0. \
end{cases}
$$
Then let $gcirc f$ will do the job.
answered Nov 27 '18 at 21:57
greedoid
38.2k114797
add a comment |
a) Let $A$ be a set of integers divisible by 3 and $C$ a set of positive even integers. You can take a composition of following functions. First $f:Ato mathbb{Z}$ defined with $$nlongmapsto {nover 3}$$
and then $g:mathbb{Z}to C$ defined by:
$$
g(n)
begin{cases}
= 4n+2 & text{if } ngeq 0, \
= 4|n| & text{if } n < 0. \
end{cases}
$$
Then let $gcirc f$ will do the job.
answered Nov 27 '18 at 21:57
greedoid
38.2k114797
add a comment |
a) Let $A$ be a set of integers divisible by 3 and $C$ a set of positive even integers. You can take a composition of following functions. First $f:Ato mathbb{Z}$ defined with $$nlongmapsto {nover 3}$$
and then $g:mathbb{Z}to C$ defined by:
$$
g(n)
begin{cases}
= 4n+2 & text{if } ngeq 0, \
= 4|n| & text{if } n < 0. \
end{cases}
$$
Then let $gcirc f$ will do the job.
answered Nov 27 '18 at 21:57
greedoid
38.2k114797
a) Let $A$ be a set of integers divisible by 3 and $C$ a set of positive even integers. You can take a composition of following functions. First $f:Ato mathbb{Z}$ defined with $$nlongmapsto {nover 3}$$
and then $g:mathbb{Z}to C$ defined by:
$$
g(n)
begin{cases}
= 4n+2 & text{if } ngeq 0, \
= 4|n| & text{if } n < 0. \
end{cases}
$$
Then let $gcirc f$ will do the job.
answered Nov 27 '18 at 21:57
greedoid
38.2k114797
edited Nov 27 '18 at 22:03
answered Nov 27 '18 at 21:57
greedoid
38.2k114797
answered Nov 27 '18 at 21:57
greedoid
38.2k114797
answered Nov 27 '18 at 21:57
greedoid
38.2k114797
38.2k114797
add a comment |
add a comment |
The bijections, reading left to right, can be chosen as:
(a) $|4x/3|+2[xge 0]$ (b) $x+sqrt{1+x^2}$ (c) $x+5+lfloor xrfloor$ (d) $1/x$.
answered Nov 27 '18 at 22:11
J.G.
23.2k22137
add a comment |
The bijections, reading left to right, can be chosen as:
(a) $|4x/3|+2[xge 0]$ (b) $x+sqrt{1+x^2}$ (c) $x+5+lfloor xrfloor$ (d) $1/x$.
answered Nov 27 '18 at 22:11
J.G.
23.2k22137
add a comment |
The bijections, reading left to right, can be chosen as:
(a) $|4x/3|+2[xge 0]$ (b) $x+sqrt{1+x^2}$ (c) $x+5+lfloor xrfloor$ (d) $1/x$.
answered Nov 27 '18 at 22:11
J.G.
23.2k22137
The bijections, reading left to right, can be chosen as:
(a) $|4x/3|+2[xge 0]$ (b) $x+sqrt{1+x^2}$ (c) $x+5+lfloor xrfloor$ (d) $1/x$.
answered Nov 27 '18 at 22:11
J.G.
23.2k22137
answered Nov 27 '18 at 22:11
J.G.
23.2k22137
answered Nov 27 '18 at 22:11
J.G.
23.2k22137
answered Nov 27 '18 at 22:11
J.G.
23.2k22137
23.2k22137
add a comment |
add a comment |
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ok thank you, fixed it
– smith
Nov 27 '18 at 2:47
Do you mean the set $[5,6)cup [7,8)$?
– TonyK
Nov 27 '18 at 2:50
yes, it has been fixed
– smith
Nov 27 '18 at 2:51