Mixing
Functions and sets. Show that if A is a subset of X, then $f^{-1}(f(A))supseteq A$. Find an example where...
$begingroup$
I have proved that if $A$ is subset of $X$, then $f^{-1}(f(A))supseteq A$ for any function $f:Xrightarrow Y$. However, I can't find an example where $f^{-1}(f(A))neq A$.
real-analysis discrete-mathematics
edited Jan 19 at 16:09
Yuval Gat
572213
asked Jan 19 at 15:54
Mathiaspilot123Mathiaspilot123
596
$endgroup$
add a comment |
$begingroup$
I have proved that if $A$ is subset of $X$, then $f^{-1}(f(A))supseteq A$ for any function $f:Xrightarrow Y$. However, I can't find an example where $f^{-1}(f(A))neq A$.
real-analysis discrete-mathematics
edited Jan 19 at 16:09
Yuval Gat
572213
asked Jan 19 at 15:54
Mathiaspilot123Mathiaspilot123
596
$endgroup$
2
$begingroup$
The trick is to choose an $f$ that is not injective (one to one)
$endgroup$
– Omnomnomnom
Jan 19 at 15:58
$begingroup$
Thank you, I kind of thought about it, but was too unsure.
$endgroup$
– Mathiaspilot123
Jan 19 at 15:59
add a comment |
$begingroup$
I have proved that if $A$ is subset of $X$, then $f^{-1}(f(A))supseteq A$ for any function $f:Xrightarrow Y$. However, I can't find an example where $f^{-1}(f(A))neq A$.
real-analysis discrete-mathematics
edited Jan 19 at 16:09
Yuval Gat
572213
asked Jan 19 at 15:54
Mathiaspilot123Mathiaspilot123
596
$endgroup$
I have proved that if $A$ is subset of $X$, then $f^{-1}(f(A))supseteq A$ for any function $f:Xrightarrow Y$. However, I can't find an example where $f^{-1}(f(A))neq A$.
real-analysis discrete-mathematics
real-analysis discrete-mathematics
edited Jan 19 at 16:09
Yuval Gat
572213
asked Jan 19 at 15:54
Mathiaspilot123Mathiaspilot123
596
edited Jan 19 at 16:09
Yuval Gat
572213
asked Jan 19 at 15:54
Mathiaspilot123Mathiaspilot123
596
edited Jan 19 at 16:09
Yuval Gat
572213
edited Jan 19 at 16:09
Yuval Gat
572213
edited Jan 19 at 16:09
Yuval Gat
572213
572213
asked Jan 19 at 15:54
Mathiaspilot123Mathiaspilot123
596
asked Jan 19 at 15:54
Mathiaspilot123Mathiaspilot123
596
asked Jan 19 at 15:54
Mathiaspilot123Mathiaspilot123
596
596
2
$begingroup$
The trick is to choose an $f$ that is not injective (one to one)
$endgroup$
– Omnomnomnom
Jan 19 at 15:58
$begingroup$
Thank you, I kind of thought about it, but was too unsure.
$endgroup$
– Mathiaspilot123
Jan 19 at 15:59
add a comment |
2
$begingroup$
The trick is to choose an $f$ that is not injective (one to one)
$endgroup$
– Omnomnomnom
Jan 19 at 15:58
$begingroup$
Thank you, I kind of thought about it, but was too unsure.
$endgroup$
– Mathiaspilot123
Jan 19 at 15:59
2
2
$begingroup$
The trick is to choose an $f$ that is not injective (one to one)
$endgroup$
– Omnomnomnom
Jan 19 at 15:58
$begingroup$
The trick is to choose an $f$ that is not injective (one to one)
$endgroup$
– Omnomnomnom
Jan 19 at 15:58
$begingroup$
Thank you, I kind of thought about it, but was too unsure.
$endgroup$
– Mathiaspilot123
Jan 19 at 15:59
$begingroup$
Thank you, I kind of thought about it, but was too unsure.
$endgroup$
– Mathiaspilot123
Jan 19 at 15:59
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You can find such an example by taking a non injective function, e.g. $f : mathbb R to mathbb R^+ : x mapsto x^2$. If $A = [0, 1]$, then $f(A) = [0, 1]$. But $f^{-1}(A) = [-1, 1]$. And $f^{-1}(f(A)) = f^{-1}([0, 1]) = [-1, 1] neq [0, 1]$
answered Jan 19 at 15:57
BermudesBermudes
18713
$endgroup$
add a comment |
$begingroup$
You can take any non-injective function.
A trivial example is $f:{1,2}to {1,2}$ defined as $f (1)=f (2)=1$. Then take $A={1} $.
answered Jan 19 at 15:58
Thomas ShelbyThomas Shelby
3,7192525
$endgroup$
add a comment |
$begingroup$
As you have proven $A subset f^{-1}(f(A))$ in order for $A ne f^{-1}(f(A))$ there must be and $xin f^{-1}(f(A))$ so that $x not in A$.
But $x in f^{-1}(f(A))$ means $f(x) in f(A)$ which means there is an $y in A$ so that $f(y) = f(x)$. But we know $yne x$ because $x not in A$. That's not unusual. That just means $f$ isn't injective.
If we take any old non-injective function $f$, say $f(x) = x^2$ and any pair of $x,y$ so that $f(x) = f(y); x ne y$, say $x = -1$ and $y=1$.... then we just take any set $A$ so that $yin A$ and $x not in A$.... say $A= {1}$.
The $A= {1}$, $f(A) = {f(w)|win A} = {f(1)} = 1$. And $f^{-1}(f(A)) = {y| f(y) in f(A)} ={y|f(y) in {1}} = {1, -1}$.
And that's a valid example..
Or $A = {-3,3, 1, 7}$. Or $B = mathbb R^+$ or $C = mathbb Rsetminus {-1}$.
Then $f(A) = {9,1,49}; f^{-1}(f(A)) = {pm 3, pm 1, pm 49}$.
$f(B) = mathbb R^+; f^{-1}(f(B)) = mathbb R$, $f(C) = mathbb R^+; f^{-1}(f(C)) = mathbb R^+$.
Etc.
answered Jan 19 at 16:28
fleabloodfleablood
71.7k22686
$endgroup$
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
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$begingroup$
You can find such an example by taking a non injective function, e.g. $f : mathbb R to mathbb R^+ : x mapsto x^2$. If $A = [0, 1]$, then $f(A) = [0, 1]$. But $f^{-1}(A) = [-1, 1]$. And $f^{-1}(f(A)) = f^{-1}([0, 1]) = [-1, 1] neq [0, 1]$
answered Jan 19 at 15:57
BermudesBermudes
18713
$endgroup$
add a comment |
$begingroup$
You can find such an example by taking a non injective function, e.g. $f : mathbb R to mathbb R^+ : x mapsto x^2$. If $A = [0, 1]$, then $f(A) = [0, 1]$. But $f^{-1}(A) = [-1, 1]$. And $f^{-1}(f(A)) = f^{-1}([0, 1]) = [-1, 1] neq [0, 1]$
answered Jan 19 at 15:57
BermudesBermudes
18713
$endgroup$
add a comment |
$begingroup$
You can find such an example by taking a non injective function, e.g. $f : mathbb R to mathbb R^+ : x mapsto x^2$. If $A = [0, 1]$, then $f(A) = [0, 1]$. But $f^{-1}(A) = [-1, 1]$. And $f^{-1}(f(A)) = f^{-1}([0, 1]) = [-1, 1] neq [0, 1]$
answered Jan 19 at 15:57
BermudesBermudes
18713
$endgroup$
You can find such an example by taking a non injective function, e.g. $f : mathbb R to mathbb R^+ : x mapsto x^2$. If $A = [0, 1]$, then $f(A) = [0, 1]$. But $f^{-1}(A) = [-1, 1]$. And $f^{-1}(f(A)) = f^{-1}([0, 1]) = [-1, 1] neq [0, 1]$
answered Jan 19 at 15:57
BermudesBermudes
18713
answered Jan 19 at 15:57
BermudesBermudes
18713
answered Jan 19 at 15:57
BermudesBermudes
18713
answered Jan 19 at 15:57
BermudesBermudes
18713
18713
add a comment |
add a comment |
$begingroup$
You can take any non-injective function.
A trivial example is $f:{1,2}to {1,2}$ defined as $f (1)=f (2)=1$. Then take $A={1} $.
answered Jan 19 at 15:58
Thomas ShelbyThomas Shelby
3,7192525
$endgroup$
add a comment |
$begingroup$
You can take any non-injective function.
A trivial example is $f:{1,2}to {1,2}$ defined as $f (1)=f (2)=1$. Then take $A={1} $.
answered Jan 19 at 15:58
Thomas ShelbyThomas Shelby
3,7192525
$endgroup$
add a comment |
$begingroup$
You can take any non-injective function.
A trivial example is $f:{1,2}to {1,2}$ defined as $f (1)=f (2)=1$. Then take $A={1} $.
answered Jan 19 at 15:58
Thomas ShelbyThomas Shelby
3,7192525
$endgroup$
You can take any non-injective function.
A trivial example is $f:{1,2}to {1,2}$ defined as $f (1)=f (2)=1$. Then take $A={1} $.
answered Jan 19 at 15:58
Thomas ShelbyThomas Shelby
3,7192525
answered Jan 19 at 15:58
Thomas ShelbyThomas Shelby
3,7192525
answered Jan 19 at 15:58
Thomas ShelbyThomas Shelby
3,7192525
answered Jan 19 at 15:58
Thomas ShelbyThomas Shelby
3,7192525
3,7192525
add a comment |
add a comment |
$begingroup$
As you have proven $A subset f^{-1}(f(A))$ in order for $A ne f^{-1}(f(A))$ there must be and $xin f^{-1}(f(A))$ so that $x not in A$.
But $x in f^{-1}(f(A))$ means $f(x) in f(A)$ which means there is an $y in A$ so that $f(y) = f(x)$. But we know $yne x$ because $x not in A$. That's not unusual. That just means $f$ isn't injective.
If we take any old non-injective function $f$, say $f(x) = x^2$ and any pair of $x,y$ so that $f(x) = f(y); x ne y$, say $x = -1$ and $y=1$.... then we just take any set $A$ so that $yin A$ and $x not in A$.... say $A= {1}$.
The $A= {1}$, $f(A) = {f(w)|win A} = {f(1)} = 1$. And $f^{-1}(f(A)) = {y| f(y) in f(A)} ={y|f(y) in {1}} = {1, -1}$.
And that's a valid example..
Or $A = {-3,3, 1, 7}$. Or $B = mathbb R^+$ or $C = mathbb Rsetminus {-1}$.
Then $f(A) = {9,1,49}; f^{-1}(f(A)) = {pm 3, pm 1, pm 49}$.
$f(B) = mathbb R^+; f^{-1}(f(B)) = mathbb R$, $f(C) = mathbb R^+; f^{-1}(f(C)) = mathbb R^+$.
Etc.
answered Jan 19 at 16:28
fleabloodfleablood
71.7k22686
$endgroup$
add a comment |
$begingroup$
As you have proven $A subset f^{-1}(f(A))$ in order for $A ne f^{-1}(f(A))$ there must be and $xin f^{-1}(f(A))$ so that $x not in A$.
But $x in f^{-1}(f(A))$ means $f(x) in f(A)$ which means there is an $y in A$ so that $f(y) = f(x)$. But we know $yne x$ because $x not in A$. That's not unusual. That just means $f$ isn't injective.
If we take any old non-injective function $f$, say $f(x) = x^2$ and any pair of $x,y$ so that $f(x) = f(y); x ne y$, say $x = -1$ and $y=1$.... then we just take any set $A$ so that $yin A$ and $x not in A$.... say $A= {1}$.
The $A= {1}$, $f(A) = {f(w)|win A} = {f(1)} = 1$. And $f^{-1}(f(A)) = {y| f(y) in f(A)} ={y|f(y) in {1}} = {1, -1}$.
And that's a valid example..
Or $A = {-3,3, 1, 7}$. Or $B = mathbb R^+$ or $C = mathbb Rsetminus {-1}$.
Then $f(A) = {9,1,49}; f^{-1}(f(A)) = {pm 3, pm 1, pm 49}$.
$f(B) = mathbb R^+; f^{-1}(f(B)) = mathbb R$, $f(C) = mathbb R^+; f^{-1}(f(C)) = mathbb R^+$.
Etc.
answered Jan 19 at 16:28
fleabloodfleablood
71.7k22686
$endgroup$
add a comment |
$begingroup$
As you have proven $A subset f^{-1}(f(A))$ in order for $A ne f^{-1}(f(A))$ there must be and $xin f^{-1}(f(A))$ so that $x not in A$.
But $x in f^{-1}(f(A))$ means $f(x) in f(A)$ which means there is an $y in A$ so that $f(y) = f(x)$. But we know $yne x$ because $x not in A$. That's not unusual. That just means $f$ isn't injective.
If we take any old non-injective function $f$, say $f(x) = x^2$ and any pair of $x,y$ so that $f(x) = f(y); x ne y$, say $x = -1$ and $y=1$.... then we just take any set $A$ so that $yin A$ and $x not in A$.... say $A= {1}$.
The $A= {1}$, $f(A) = {f(w)|win A} = {f(1)} = 1$. And $f^{-1}(f(A)) = {y| f(y) in f(A)} ={y|f(y) in {1}} = {1, -1}$.
And that's a valid example..
Or $A = {-3,3, 1, 7}$. Or $B = mathbb R^+$ or $C = mathbb Rsetminus {-1}$.
Then $f(A) = {9,1,49}; f^{-1}(f(A)) = {pm 3, pm 1, pm 49}$.
$f(B) = mathbb R^+; f^{-1}(f(B)) = mathbb R$, $f(C) = mathbb R^+; f^{-1}(f(C)) = mathbb R^+$.
Etc.
answered Jan 19 at 16:28
fleabloodfleablood
71.7k22686
$endgroup$
As you have proven $A subset f^{-1}(f(A))$ in order for $A ne f^{-1}(f(A))$ there must be and $xin f^{-1}(f(A))$ so that $x not in A$.
But $x in f^{-1}(f(A))$ means $f(x) in f(A)$ which means there is an $y in A$ so that $f(y) = f(x)$. But we know $yne x$ because $x not in A$. That's not unusual. That just means $f$ isn't injective.
If we take any old non-injective function $f$, say $f(x) = x^2$ and any pair of $x,y$ so that $f(x) = f(y); x ne y$, say $x = -1$ and $y=1$.... then we just take any set $A$ so that $yin A$ and $x not in A$.... say $A= {1}$.
The $A= {1}$, $f(A) = {f(w)|win A} = {f(1)} = 1$. And $f^{-1}(f(A)) = {y| f(y) in f(A)} ={y|f(y) in {1}} = {1, -1}$.
And that's a valid example..
Or $A = {-3,3, 1, 7}$. Or $B = mathbb R^+$ or $C = mathbb Rsetminus {-1}$.
Then $f(A) = {9,1,49}; f^{-1}(f(A)) = {pm 3, pm 1, pm 49}$.
$f(B) = mathbb R^+; f^{-1}(f(B)) = mathbb R$, $f(C) = mathbb R^+; f^{-1}(f(C)) = mathbb R^+$.
Etc.
answered Jan 19 at 16:28
fleabloodfleablood
71.7k22686
answered Jan 19 at 16:28
fleabloodfleablood
71.7k22686
answered Jan 19 at 16:28
fleabloodfleablood
71.7k22686
answered Jan 19 at 16:28
fleabloodfleablood
71.7k22686
71.7k22686
add a comment |
add a comment |
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2
$begingroup$
The trick is to choose an $f$ that is not injective (one to one)
$endgroup$
– Omnomnomnom
Jan 19 at 15:58
$begingroup$
Thank you, I kind of thought about it, but was too unsure.
$endgroup$
– Mathiaspilot123
Jan 19 at 15:59