Mixing
How to find the preimage of a set
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I'm trying to prove continuity of two different functions $fcolon Ato B$. I know that to find continuity, the inverse of all open subsets in $B$ must be open in $A$. I also know that to find the inverse of an open subset you have to find the preimage. I just don't know how to find the preimage. How do you find the preimage?
For example, if the set is the discrete topology of ${1,2,3}$, how would you find the preimage of ${1}$, ${2}$, etc. The same for the indiscrete topology. How would you find the preimage of the null set and the whole set? Thank you!
general-topology continuity
edited Nov 18 '15 at 2:01
Silvia Ghinassi
3,89671641
asked Nov 18 '15 at 1:46
user291167user291167
1
$endgroup$
add a comment |
$begingroup$
I'm trying to prove continuity of two different functions $fcolon Ato B$. I know that to find continuity, the inverse of all open subsets in $B$ must be open in $A$. I also know that to find the inverse of an open subset you have to find the preimage. I just don't know how to find the preimage. How do you find the preimage?
For example, if the set is the discrete topology of ${1,2,3}$, how would you find the preimage of ${1}$, ${2}$, etc. The same for the indiscrete topology. How would you find the preimage of the null set and the whole set? Thank you!
general-topology continuity
edited Nov 18 '15 at 2:01
Silvia Ghinassi
3,89671641
asked Nov 18 '15 at 1:46
user291167user291167
1
$endgroup$
$begingroup$
Do you understand what "preimage of a set under a function" means? You need a set and a function.
$endgroup$
– Silvia Ghinassi
Nov 18 '15 at 1:48
$begingroup$
A couple more comments: 1) the set is ${1,2,3}$ with discrete topology, not "the set is the discrete topology of" 2) preimages have nothing to do with topology, 3) we don't say "the inverse of all open sets", we say directly "the preimages of all open sets"
$endgroup$
– Silvia Ghinassi
Nov 18 '15 at 1:52
add a comment |
$begingroup$
I'm trying to prove continuity of two different functions $fcolon Ato B$. I know that to find continuity, the inverse of all open subsets in $B$ must be open in $A$. I also know that to find the inverse of an open subset you have to find the preimage. I just don't know how to find the preimage. How do you find the preimage?
For example, if the set is the discrete topology of ${1,2,3}$, how would you find the preimage of ${1}$, ${2}$, etc. The same for the indiscrete topology. How would you find the preimage of the null set and the whole set? Thank you!
general-topology continuity
edited Nov 18 '15 at 2:01
Silvia Ghinassi
3,89671641
asked Nov 18 '15 at 1:46
user291167user291167
1
$endgroup$
I'm trying to prove continuity of two different functions $fcolon Ato B$. I know that to find continuity, the inverse of all open subsets in $B$ must be open in $A$. I also know that to find the inverse of an open subset you have to find the preimage. I just don't know how to find the preimage. How do you find the preimage?
For example, if the set is the discrete topology of ${1,2,3}$, how would you find the preimage of ${1}$, ${2}$, etc. The same for the indiscrete topology. How would you find the preimage of the null set and the whole set? Thank you!
general-topology continuity
general-topology continuity
edited Nov 18 '15 at 2:01
Silvia Ghinassi
3,89671641
asked Nov 18 '15 at 1:46
user291167user291167
1
edited Nov 18 '15 at 2:01
Silvia Ghinassi
3,89671641
asked Nov 18 '15 at 1:46
user291167user291167
1
edited Nov 18 '15 at 2:01
Silvia Ghinassi
3,89671641
edited Nov 18 '15 at 2:01
Silvia Ghinassi
3,89671641
edited Nov 18 '15 at 2:01
Silvia Ghinassi
3,89671641
3,89671641
asked Nov 18 '15 at 1:46
user291167user291167
1
asked Nov 18 '15 at 1:46
user291167user291167
1
asked Nov 18 '15 at 1:46
user291167user291167
1
1
$begingroup$
Do you understand what "preimage of a set under a function" means? You need a set and a function.
$endgroup$
– Silvia Ghinassi
Nov 18 '15 at 1:48
$begingroup$
A couple more comments: 1) the set is ${1,2,3}$ with discrete topology, not "the set is the discrete topology of" 2) preimages have nothing to do with topology, 3) we don't say "the inverse of all open sets", we say directly "the preimages of all open sets"
$endgroup$
– Silvia Ghinassi
Nov 18 '15 at 1:52
add a comment |
$begingroup$
Do you understand what "preimage of a set under a function" means? You need a set and a function.
$endgroup$
– Silvia Ghinassi
Nov 18 '15 at 1:48
$begingroup$
A couple more comments: 1) the set is ${1,2,3}$ with discrete topology, not "the set is the discrete topology of" 2) preimages have nothing to do with topology, 3) we don't say "the inverse of all open sets", we say directly "the preimages of all open sets"
$endgroup$
– Silvia Ghinassi
Nov 18 '15 at 1:52
$begingroup$
Do you understand what "preimage of a set under a function" means? You need a set and a function.
$endgroup$
– Silvia Ghinassi
Nov 18 '15 at 1:48
$begingroup$
Do you understand what "preimage of a set under a function" means? You need a set and a function.
$endgroup$
– Silvia Ghinassi
Nov 18 '15 at 1:48
$begingroup$
A couple more comments: 1) the set is ${1,2,3}$ with discrete topology, not "the set is the discrete topology of" 2) preimages have nothing to do with topology, 3) we don't say "the inverse of all open sets", we say directly "the preimages of all open sets"
$endgroup$
– Silvia Ghinassi
Nov 18 '15 at 1:52
$begingroup$
A couple more comments: 1) the set is ${1,2,3}$ with discrete topology, not "the set is the discrete topology of" 2) preimages have nothing to do with topology, 3) we don't say "the inverse of all open sets", we say directly "the preimages of all open sets"
$endgroup$
– Silvia Ghinassi
Nov 18 '15 at 1:52
add a comment |
1 Answer
1
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Let $f:Ato B$ be a map and let $S$ be a subset of $B$, then the preimage of $S$ under $f$ is
$$f^{-1}(S)={a,|,f(a)in S}.$$
For example if $A={1,2,3}$, $B={a,b,c}$ and $f$ is map defined by $f(1)=f(2)=a$, $f(3)=b$, then
$$f^{-1}(emptyset)=emptyset,\
f^{-1}({a})={1,2},
f^{-1}({b})={3},
f^{-1}({c})=emptyset,\
f^{-1}({a,b})={1,2,3},
f^{-1}({a,c})={1,2},
f^{-1}({b,c})={3},\
f^{-1}({a,b,c})={1,2,3}.$$
answered Nov 18 '15 at 4:29
NexNex
3,1951622
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $f:Ato B$ be a map and let $S$ be a subset of $B$, then the preimage of $S$ under $f$ is
$$f^{-1}(S)={a,|,f(a)in S}.$$
For example if $A={1,2,3}$, $B={a,b,c}$ and $f$ is map defined by $f(1)=f(2)=a$, $f(3)=b$, then
$$f^{-1}(emptyset)=emptyset,\
f^{-1}({a})={1,2},
f^{-1}({b})={3},
f^{-1}({c})=emptyset,\
f^{-1}({a,b})={1,2,3},
f^{-1}({a,c})={1,2},
f^{-1}({b,c})={3},\
f^{-1}({a,b,c})={1,2,3}.$$
answered Nov 18 '15 at 4:29
NexNex
3,1951622
$endgroup$
add a comment |
$begingroup$
Let $f:Ato B$ be a map and let $S$ be a subset of $B$, then the preimage of $S$ under $f$ is
$$f^{-1}(S)={a,|,f(a)in S}.$$
For example if $A={1,2,3}$, $B={a,b,c}$ and $f$ is map defined by $f(1)=f(2)=a$, $f(3)=b$, then
$$f^{-1}(emptyset)=emptyset,\
f^{-1}({a})={1,2},
f^{-1}({b})={3},
f^{-1}({c})=emptyset,\
f^{-1}({a,b})={1,2,3},
f^{-1}({a,c})={1,2},
f^{-1}({b,c})={3},\
f^{-1}({a,b,c})={1,2,3}.$$
answered Nov 18 '15 at 4:29
NexNex
3,1951622
$endgroup$
add a comment |
$begingroup$
Let $f:Ato B$ be a map and let $S$ be a subset of $B$, then the preimage of $S$ under $f$ is
$$f^{-1}(S)={a,|,f(a)in S}.$$
For example if $A={1,2,3}$, $B={a,b,c}$ and $f$ is map defined by $f(1)=f(2)=a$, $f(3)=b$, then
$$f^{-1}(emptyset)=emptyset,\
f^{-1}({a})={1,2},
f^{-1}({b})={3},
f^{-1}({c})=emptyset,\
f^{-1}({a,b})={1,2,3},
f^{-1}({a,c})={1,2},
f^{-1}({b,c})={3},\
f^{-1}({a,b,c})={1,2,3}.$$
answered Nov 18 '15 at 4:29
NexNex
3,1951622
$endgroup$
Let $f:Ato B$ be a map and let $S$ be a subset of $B$, then the preimage of $S$ under $f$ is
$$f^{-1}(S)={a,|,f(a)in S}.$$
For example if $A={1,2,3}$, $B={a,b,c}$ and $f$ is map defined by $f(1)=f(2)=a$, $f(3)=b$, then
$$f^{-1}(emptyset)=emptyset,\
f^{-1}({a})={1,2},
f^{-1}({b})={3},
f^{-1}({c})=emptyset,\
f^{-1}({a,b})={1,2,3},
f^{-1}({a,c})={1,2},
f^{-1}({b,c})={3},\
f^{-1}({a,b,c})={1,2,3}.$$
answered Nov 18 '15 at 4:29
NexNex
3,1951622
answered Nov 18 '15 at 4:29
NexNex
3,1951622
answered Nov 18 '15 at 4:29
NexNex
3,1951622
answered Nov 18 '15 at 4:29
NexNex
3,1951622
3,1951622
add a comment |
add a comment |
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$begingroup$
Do you understand what "preimage of a set under a function" means? You need a set and a function.
$endgroup$
– Silvia Ghinassi
Nov 18 '15 at 1:48
$begingroup$
A couple more comments: 1) the set is ${1,2,3}$ with discrete topology, not "the set is the discrete topology of" 2) preimages have nothing to do with topology, 3) we don't say "the inverse of all open sets", we say directly "the preimages of all open sets"
$endgroup$
– Silvia Ghinassi
Nov 18 '15 at 1:52