Function injective iff
$begingroup$
Let $f:Xto Y$. Prove that
$$text{f is injective}iff forall A subseteq X, quad f(X-A)subseteq Y-f(A)$$
My try:
For $leftarrow$.
Let $f(x_1)= f(x_2)$ and take for $A={x_1}$ then if $x_1neq x_2$ then $x_2in X-A$ so $f(x_2)subseteq Y-f(x_1)=Y-f(x_2)$ so $x_1=x_2$
For $rightarrow$
If $yin f(X-A)$ then $f^{-1}(y) in X-A $
$yin f(X) $ and $f^{-1}(y)notin A$
Since $f(X)subseteq Y$ then
$$yin Y-f(A)$$
I'm not sure for $leftarrow$, but for $rightarrow$ I think I'm wrong because didn't use injective
functions elementary-set-theory
edited Dec 7 '18 at 7:04
Martin Sleziak
44.8k9118272
asked Nov 26 '16 at 21:34
user183294user183294
353
$endgroup$
add a comment |
$begingroup$
Let $f:Xto Y$. Prove that
$$text{f is injective}iff forall A subseteq X, quad f(X-A)subseteq Y-f(A)$$
My try:
For $leftarrow$.
Let $f(x_1)= f(x_2)$ and take for $A={x_1}$ then if $x_1neq x_2$ then $x_2in X-A$ so $f(x_2)subseteq Y-f(x_1)=Y-f(x_2)$ so $x_1=x_2$
For $rightarrow$
If $yin f(X-A)$ then $f^{-1}(y) in X-A $
$yin f(X) $ and $f^{-1}(y)notin A$
Since $f(X)subseteq Y$ then
$$yin Y-f(A)$$
I'm not sure for $leftarrow$, but for $rightarrow$ I think I'm wrong because didn't use injective
functions elementary-set-theory
edited Dec 7 '18 at 7:04
Martin Sleziak
44.8k9118272
asked Nov 26 '16 at 21:34
user183294user183294
353
$endgroup$
$begingroup$
Related: math.stackexchange.com/questions/511662/…
$endgroup$
– Asaf Karagila♦
Nov 26 '16 at 21:38
$begingroup$
It is better to use $setminus$ for the relative complements, not to mix up with the Minkowski addition.
$endgroup$
– A.Γ.
Nov 26 '16 at 21:43
$begingroup$
The claim: If $yin f(X-A)$ then $f^{-1}(y) in X-A$ is not true in general, since $f^{-1}(y)$ may contain many points. However, for an injective function $f^{-1}(y)$ is a singleton.
$endgroup$
– A.Γ.
Nov 26 '16 at 21:57
$begingroup$
See also Does $f(X setminus A)subseteq Ysetminus f(A), forall Asubseteq X$ imply $f$ is injective ? and If $f$ is 1-1, prove that $f(Asetminus B) = f(A)setminus f(B)$
$endgroup$
– Martin Sleziak
Nov 27 '16 at 14:33
add a comment |
$begingroup$
Let $f:Xto Y$. Prove that
$$text{f is injective}iff forall A subseteq X, quad f(X-A)subseteq Y-f(A)$$
My try:
For $leftarrow$.
Let $f(x_1)= f(x_2)$ and take for $A={x_1}$ then if $x_1neq x_2$ then $x_2in X-A$ so $f(x_2)subseteq Y-f(x_1)=Y-f(x_2)$ so $x_1=x_2$
For $rightarrow$
If $yin f(X-A)$ then $f^{-1}(y) in X-A $
$yin f(X) $ and $f^{-1}(y)notin A$
Since $f(X)subseteq Y$ then
$$yin Y-f(A)$$
I'm not sure for $leftarrow$, but for $rightarrow$ I think I'm wrong because didn't use injective
functions elementary-set-theory
edited Dec 7 '18 at 7:04
Martin Sleziak
44.8k9118272
asked Nov 26 '16 at 21:34
user183294user183294
353
$endgroup$
Let $f:Xto Y$. Prove that
$$text{f is injective}iff forall A subseteq X, quad f(X-A)subseteq Y-f(A)$$
My try:
For $leftarrow$.
Let $f(x_1)= f(x_2)$ and take for $A={x_1}$ then if $x_1neq x_2$ then $x_2in X-A$ so $f(x_2)subseteq Y-f(x_1)=Y-f(x_2)$ so $x_1=x_2$
For $rightarrow$
If $yin f(X-A)$ then $f^{-1}(y) in X-A $
$yin f(X) $ and $f^{-1}(y)notin A$
Since $f(X)subseteq Y$ then
$$yin Y-f(A)$$
I'm not sure for $leftarrow$, but for $rightarrow$ I think I'm wrong because didn't use injective
functions elementary-set-theory
functions elementary-set-theory
edited Dec 7 '18 at 7:04
Martin Sleziak
44.8k9118272
asked Nov 26 '16 at 21:34
user183294user183294
353
edited Dec 7 '18 at 7:04
Martin Sleziak
44.8k9118272
asked Nov 26 '16 at 21:34
user183294user183294
353
edited Dec 7 '18 at 7:04
Martin Sleziak
44.8k9118272
edited Dec 7 '18 at 7:04
Martin Sleziak
44.8k9118272
edited Dec 7 '18 at 7:04
Martin Sleziak
44.8k9118272
44.8k9118272
asked Nov 26 '16 at 21:34
user183294user183294
353
asked Nov 26 '16 at 21:34
user183294user183294
353
asked Nov 26 '16 at 21:34
user183294user183294
353
353
$begingroup$
Related: math.stackexchange.com/questions/511662/…
$endgroup$
– Asaf Karagila♦
Nov 26 '16 at 21:38
$begingroup$
It is better to use $setminus$ for the relative complements, not to mix up with the Minkowski addition.
$endgroup$
– A.Γ.
Nov 26 '16 at 21:43
$begingroup$
The claim: If $yin f(X-A)$ then $f^{-1}(y) in X-A$ is not true in general, since $f^{-1}(y)$ may contain many points. However, for an injective function $f^{-1}(y)$ is a singleton.
$endgroup$
– A.Γ.
Nov 26 '16 at 21:57
$begingroup$
See also Does $f(X setminus A)subseteq Ysetminus f(A), forall Asubseteq X$ imply $f$ is injective ? and If $f$ is 1-1, prove that $f(Asetminus B) = f(A)setminus f(B)$
$endgroup$
– Martin Sleziak
Nov 27 '16 at 14:33
add a comment |
$begingroup$
Related: math.stackexchange.com/questions/511662/…
$endgroup$
– Asaf Karagila♦
Nov 26 '16 at 21:38
$begingroup$
It is better to use $setminus$ for the relative complements, not to mix up with the Minkowski addition.
$endgroup$
– A.Γ.
Nov 26 '16 at 21:43
$begingroup$
The claim: If $yin f(X-A)$ then $f^{-1}(y) in X-A$ is not true in general, since $f^{-1}(y)$ may contain many points. However, for an injective function $f^{-1}(y)$ is a singleton.
$endgroup$
– A.Γ.
Nov 26 '16 at 21:57
$begingroup$
See also Does $f(X setminus A)subseteq Ysetminus f(A), forall Asubseteq X$ imply $f$ is injective ? and If $f$ is 1-1, prove that $f(Asetminus B) = f(A)setminus f(B)$
$endgroup$
– Martin Sleziak
Nov 27 '16 at 14:33
$begingroup$
Related: math.stackexchange.com/questions/511662/…
$endgroup$
– Asaf Karagila♦
Nov 26 '16 at 21:38
$begingroup$
Related: math.stackexchange.com/questions/511662/…
$endgroup$
– Asaf Karagila♦
Nov 26 '16 at 21:38
$begingroup$
It is better to use $setminus$ for the relative complements, not to mix up with the Minkowski addition.
$endgroup$
– A.Γ.
Nov 26 '16 at 21:43
$begingroup$
It is better to use $setminus$ for the relative complements, not to mix up with the Minkowski addition.
$endgroup$
– A.Γ.
Nov 26 '16 at 21:43
$begingroup$
The claim: If $yin f(X-A)$ then $f^{-1}(y) in X-A$ is not true in general, since $f^{-1}(y)$ may contain many points. However, for an injective function $f^{-1}(y)$ is a singleton.
$endgroup$
– A.Γ.
Nov 26 '16 at 21:57
$begingroup$
The claim: If $yin f(X-A)$ then $f^{-1}(y) in X-A$ is not true in general, since $f^{-1}(y)$ may contain many points. However, for an injective function $f^{-1}(y)$ is a singleton.
$endgroup$
– A.Γ.
Nov 26 '16 at 21:57
$begingroup$
See also Does $f(X setminus A)subseteq Ysetminus f(A), forall Asubseteq X$ imply $f$ is injective ? and If $f$ is 1-1, prove that $f(Asetminus B) = f(A)setminus f(B)$
$endgroup$
– Martin Sleziak
Nov 27 '16 at 14:33
$begingroup$
See also Does $f(X setminus A)subseteq Ysetminus f(A), forall Asubseteq X$ imply $f$ is injective ? and If $f$ is 1-1, prove that $f(Asetminus B) = f(A)setminus f(B)$
$endgroup$
– Martin Sleziak
Nov 27 '16 at 14:33
add a comment |
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$begingroup$
Related: math.stackexchange.com/questions/511662/…
$endgroup$
– Asaf Karagila♦
Nov 26 '16 at 21:38
$begingroup$
It is better to use $setminus$ for the relative complements, not to mix up with the Minkowski addition.
$endgroup$
– A.Γ.
Nov 26 '16 at 21:43
$begingroup$
The claim: If $yin f(X-A)$ then $f^{-1}(y) in X-A$ is not true in general, since $f^{-1}(y)$ may contain many points. However, for an injective function $f^{-1}(y)$ is a singleton.
$endgroup$
– A.Γ.
Nov 26 '16 at 21:57
$begingroup$
See also Does $f(X setminus A)subseteq Ysetminus f(A), forall Asubseteq X$ imply $f$ is injective ? and If $f$ is 1-1, prove that $f(Asetminus B) = f(A)setminus f(B)$
$endgroup$
– Martin Sleziak
Nov 27 '16 at 14:33