Prove $f[A] subseteq f[B] Rightarrow A ⊆ B$, given f is injective.
$begingroup$
I have constructed a proof for the above question, but do not know where I must use the hypothesis of injectivity:
(Note: Let $X,Y$ be sets with $A,B ⊆ X$ and $f:X to Y $.)
Let $y∈f[A]$; then, there exists $a∈A$ s.t. $f(a) = y$. Since $f[A]⊆ f[B]$, then, we also have $y∈f[B]$ and so, $f(a)∈ f[B]$. We conclude $a∈B$. Thus, $a∈A Rightarrow a∈B$, so $A⊆B$.
discrete-mathematics proof-verification elementary-set-theory
edited Jan 4 at 16:40
Foobaz John
22.7k41452
asked Jan 4 at 16:29
Julia KimJulia Kim
174
$endgroup$
add a comment |
$begingroup$
I have constructed a proof for the above question, but do not know where I must use the hypothesis of injectivity:
(Note: Let $X,Y$ be sets with $A,B ⊆ X$ and $f:X to Y $.)
Let $y∈f[A]$; then, there exists $a∈A$ s.t. $f(a) = y$. Since $f[A]⊆ f[B]$, then, we also have $y∈f[B]$ and so, $f(a)∈ f[B]$. We conclude $a∈B$. Thus, $a∈A Rightarrow a∈B$, so $A⊆B$.
discrete-mathematics proof-verification elementary-set-theory
edited Jan 4 at 16:40
Foobaz John
22.7k41452
asked Jan 4 at 16:29
Julia KimJulia Kim
174
$endgroup$
$begingroup$
You should write the question properly. What's $f,A,B$?
$endgroup$
– Yanko
Jan 4 at 16:31
1
$begingroup$
How do you conclude that $ain B$? That's where you should use injectivity
$endgroup$
– Exodd
Jan 4 at 16:33
1
$begingroup$
The proof must be wrong as written since you never use the assumption of injectivity. That being said, you only need one sentence to fix this up.
$endgroup$
– Michael Burr
Jan 4 at 16:40
add a comment |
$begingroup$
I have constructed a proof for the above question, but do not know where I must use the hypothesis of injectivity:
(Note: Let $X,Y$ be sets with $A,B ⊆ X$ and $f:X to Y $.)
Let $y∈f[A]$; then, there exists $a∈A$ s.t. $f(a) = y$. Since $f[A]⊆ f[B]$, then, we also have $y∈f[B]$ and so, $f(a)∈ f[B]$. We conclude $a∈B$. Thus, $a∈A Rightarrow a∈B$, so $A⊆B$.
discrete-mathematics proof-verification elementary-set-theory
edited Jan 4 at 16:40
Foobaz John
22.7k41452
asked Jan 4 at 16:29
Julia KimJulia Kim
174
$endgroup$
I have constructed a proof for the above question, but do not know where I must use the hypothesis of injectivity:
(Note: Let $X,Y$ be sets with $A,B ⊆ X$ and $f:X to Y $.)
Let $y∈f[A]$; then, there exists $a∈A$ s.t. $f(a) = y$. Since $f[A]⊆ f[B]$, then, we also have $y∈f[B]$ and so, $f(a)∈ f[B]$. We conclude $a∈B$. Thus, $a∈A Rightarrow a∈B$, so $A⊆B$.
discrete-mathematics proof-verification elementary-set-theory
discrete-mathematics proof-verification elementary-set-theory
edited Jan 4 at 16:40
Foobaz John
22.7k41452
asked Jan 4 at 16:29
Julia KimJulia Kim
174
edited Jan 4 at 16:40
Foobaz John
22.7k41452
asked Jan 4 at 16:29
Julia KimJulia Kim
174
edited Jan 4 at 16:40
Foobaz John
22.7k41452
edited Jan 4 at 16:40
Foobaz John
22.7k41452
edited Jan 4 at 16:40
Foobaz John
22.7k41452
22.7k41452
asked Jan 4 at 16:29
Julia KimJulia Kim
174
asked Jan 4 at 16:29
Julia KimJulia Kim
174
asked Jan 4 at 16:29
Julia KimJulia Kim
174
174
$begingroup$
You should write the question properly. What's $f,A,B$?
$endgroup$
– Yanko
Jan 4 at 16:31
1
$begingroup$
How do you conclude that $ain B$? That's where you should use injectivity
$endgroup$
– Exodd
Jan 4 at 16:33
1
$begingroup$
The proof must be wrong as written since you never use the assumption of injectivity. That being said, you only need one sentence to fix this up.
$endgroup$
– Michael Burr
Jan 4 at 16:40
add a comment |
$begingroup$
You should write the question properly. What's $f,A,B$?
$endgroup$
– Yanko
Jan 4 at 16:31
1
$begingroup$
How do you conclude that $ain B$? That's where you should use injectivity
$endgroup$
– Exodd
Jan 4 at 16:33
1
$begingroup$
The proof must be wrong as written since you never use the assumption of injectivity. That being said, you only need one sentence to fix this up.
$endgroup$
– Michael Burr
Jan 4 at 16:40
$begingroup$
You should write the question properly. What's $f,A,B$?
$endgroup$
– Yanko
Jan 4 at 16:31
$begingroup$
You should write the question properly. What's $f,A,B$?
$endgroup$
– Yanko
Jan 4 at 16:31
1
1
$begingroup$
How do you conclude that $ain B$? That's where you should use injectivity
$endgroup$
– Exodd
Jan 4 at 16:33
$begingroup$
How do you conclude that $ain B$? That's where you should use injectivity
$endgroup$
– Exodd
Jan 4 at 16:33
1
1
$begingroup$
The proof must be wrong as written since you never use the assumption of injectivity. That being said, you only need one sentence to fix this up.
$endgroup$
– Michael Burr
Jan 4 at 16:40
$begingroup$
The proof must be wrong as written since you never use the assumption of injectivity. That being said, you only need one sentence to fix this up.
$endgroup$
– Michael Burr
Jan 4 at 16:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose that $xin A$. Then $y=f(x)in f(A)$ and because $f(A)subseteq f(B)$ we have that $yin f(B)$. So there exists $w$ in $B$ such that $f(w)=y=f(x)$. But $f$ is injective, whence $w=x$ and $xin B$ as desired.
answered Jan 4 at 16:39
Foobaz JohnFoobaz John
22.7k41452
$endgroup$
add a comment |
$begingroup$
Attempt:
Assume $B subset A$ ($B$ proper subset of $A$).
($A$ $B not = emptyset$)
There exists an $a in A$ $B.$
$f(a) in f(A$ $B) subset f(A)$, i.e.
$f(a) in f(A)$;
since $f$ is injective: $f(a) not in f(B)$,
contradicts the assumption $f(A)subset f(B).$
answered Jan 4 at 18:51
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
add a comment |
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2 Answers
2
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2 Answers
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active
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$begingroup$
Suppose that $xin A$. Then $y=f(x)in f(A)$ and because $f(A)subseteq f(B)$ we have that $yin f(B)$. So there exists $w$ in $B$ such that $f(w)=y=f(x)$. But $f$ is injective, whence $w=x$ and $xin B$ as desired.
answered Jan 4 at 16:39
Foobaz JohnFoobaz John
22.7k41452
$endgroup$
add a comment |
$begingroup$
Suppose that $xin A$. Then $y=f(x)in f(A)$ and because $f(A)subseteq f(B)$ we have that $yin f(B)$. So there exists $w$ in $B$ such that $f(w)=y=f(x)$. But $f$ is injective, whence $w=x$ and $xin B$ as desired.
answered Jan 4 at 16:39
Foobaz JohnFoobaz John
22.7k41452
$endgroup$
add a comment |
$begingroup$
Suppose that $xin A$. Then $y=f(x)in f(A)$ and because $f(A)subseteq f(B)$ we have that $yin f(B)$. So there exists $w$ in $B$ such that $f(w)=y=f(x)$. But $f$ is injective, whence $w=x$ and $xin B$ as desired.
answered Jan 4 at 16:39
Foobaz JohnFoobaz John
22.7k41452
$endgroup$
Suppose that $xin A$. Then $y=f(x)in f(A)$ and because $f(A)subseteq f(B)$ we have that $yin f(B)$. So there exists $w$ in $B$ such that $f(w)=y=f(x)$. But $f$ is injective, whence $w=x$ and $xin B$ as desired.
answered Jan 4 at 16:39
Foobaz JohnFoobaz John
22.7k41452
answered Jan 4 at 16:39
Foobaz JohnFoobaz John
22.7k41452
answered Jan 4 at 16:39
Foobaz JohnFoobaz John
22.7k41452
answered Jan 4 at 16:39
Foobaz JohnFoobaz John
22.7k41452
22.7k41452
add a comment |
add a comment |
$begingroup$
Attempt:
Assume $B subset A$ ($B$ proper subset of $A$).
($A$ $B not = emptyset$)
There exists an $a in A$ $B.$
$f(a) in f(A$ $B) subset f(A)$, i.e.
$f(a) in f(A)$;
since $f$ is injective: $f(a) not in f(B)$,
contradicts the assumption $f(A)subset f(B).$
answered Jan 4 at 18:51
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
add a comment |
$begingroup$
Attempt:
Assume $B subset A$ ($B$ proper subset of $A$).
($A$ $B not = emptyset$)
There exists an $a in A$ $B.$
$f(a) in f(A$ $B) subset f(A)$, i.e.
$f(a) in f(A)$;
since $f$ is injective: $f(a) not in f(B)$,
contradicts the assumption $f(A)subset f(B).$
answered Jan 4 at 18:51
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
add a comment |
$begingroup$
Attempt:
Assume $B subset A$ ($B$ proper subset of $A$).
($A$ $B not = emptyset$)
There exists an $a in A$ $B.$
$f(a) in f(A$ $B) subset f(A)$, i.e.
$f(a) in f(A)$;
since $f$ is injective: $f(a) not in f(B)$,
contradicts the assumption $f(A)subset f(B).$
answered Jan 4 at 18:51
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
Attempt:
Assume $B subset A$ ($B$ proper subset of $A$).
($A$ $B not = emptyset$)
There exists an $a in A$ $B.$
$f(a) in f(A$ $B) subset f(A)$, i.e.
$f(a) in f(A)$;
since $f$ is injective: $f(a) not in f(B)$,
contradicts the assumption $f(A)subset f(B).$
answered Jan 4 at 18:51
Peter SzilasPeter Szilas
11.5k2822
answered Jan 4 at 18:51
Peter SzilasPeter Szilas
11.5k2822
answered Jan 4 at 18:51
Peter SzilasPeter Szilas
11.5k2822
answered Jan 4 at 18:51
Peter SzilasPeter Szilas
11.5k2822
11.5k2822
add a comment |
add a comment |
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$begingroup$
You should write the question properly. What's $f,A,B$?
$endgroup$
– Yanko
Jan 4 at 16:31
1
$begingroup$
How do you conclude that $ain B$? That's where you should use injectivity
$endgroup$
– Exodd
Jan 4 at 16:33
1
$begingroup$
The proof must be wrong as written since you never use the assumption of injectivity. That being said, you only need one sentence to fix this up.
$endgroup$
– Michael Burr
Jan 4 at 16:40